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how to determine if a number is a smart number in java?

I have this question I am trying to solve. I have tried coding for the past 4 hours.
An integer is defined to be a Smart number if it is an element in the infinite sequence
1, 2, 4, 7, 11, 16 …
Note that 2-1=1, 4-2=2, 7-4=3, 11-7=4, 16-11=5 so for k>1, the kth element of the sequence is equal to the k-1th element + k-1. For example, for k=6, 16 is the kth element and is equal to 11 (the k-1th element) + 5 ( k-1).
Write function named isSmart that returns 1 if its argument is a Smart number, otherwise it returns 0. So isSmart(11) returns 1, isSmart(22) returns 1 and isSmart(8) returns 0
I have tried the following code to
import java.util.Arrays;
public class IsSmart {
public static void main(String[] args) {
// TODO Auto-generated method stub
int x = isSmart(11);
System.out.println(x);
}
public static int isSmart(int n) {
int[] y = new int[n];
int j = 0;
for (int i = 1; i <= n; i++) {
y[j] = i;
j++;
}
System.out.println(Arrays.toString(y));
for (int i = 0; i <= y.length; i++) {
int diff = 0;
y[j] = y[i+1] - y[i] ;
y[i] = diff;
}
System.out.println(Arrays.toString(y));
for (int i = 0; i < y.length; i++) {
if(n == y[i])
return 1;
}
return 0;
}
}
When I test it with 11 it is giving me 0 but it shouldn't. Any idea how to correct my mistakes?

It can be done in a simpler way as follows
import java.util.Arrays;
public class IsSmart {
public static void main(String[] args) {
int x = isSmart(11);
System.out.println("Ans: "+x);
}
public static int isSmart(int n) {
//------------ CHECK THIS LOGIC ------------//
int[] y = new int[n];
int diff = 1;
for (int i = 1; i < n; i++) {
y[0] =1;
y[i] = diff + y[i-1];
diff++;
}
//------------ CHECK THIS LOGIC ------------//
System.out.println(Arrays.toString(y));
for (int i = 0; i < y.length; i++) {
if(n == y[i])
return 1;
}
return 0;
}
}

One of the problems is the way that your populating your array.
The array can be populated as such
for(int i = 0; i < n; i++) {
y[i] = (i == 0) ? 1 : y[i - 1] + i;
}
The overall application of the function isSmart can be simplified to:
public static int isSmart(int n) {
int[] array = new int[n];
for(int i = 0; i < n; i++) {
array[i] = (i == 0) ? 1 : array[i - 1] + i;
}
for (int i = 0; i < array.length; i++) {
if (array[i] == n) return 1;
}
return 0;
}

Note that you don't need to build an array:
public static int isSmart(int n) {
int smart = 1;
for (int i = 1; smart < n; i++) {
smart = smart + i;
}
return smart == n ? 1 : 0;
}

Here is a naive way to think of it to get you started - you need to fill out the while() loop. The important thing to notice is that:
The next value of the sequence will be the number of items in the sequence + the last item in the sequence.
import java.util.ArrayList;
public class Test {
public static void main(String[] args) {
System.out.println(isSmart(11));
}
public static int isSmart(int n) {
ArrayList<Integer> sequence = new ArrayList<Integer>();
// Start with 1 in the ArrayList
sequence.add(1);
// You need to keep track of the index, as well as
// the next value you're going to add to your list
int index = 1; // or number of elements in the sequence
int nextVal = 1;
while (nextVal < n) {
// Three things need to happen in here:
// 1) set nextVal equal to the sum of the current index + the value at the *previous* index
// 2) add nextVal to the ArrayList
// 3) incriment index by 1
}
// Now you can check to see if your ArrayList contains n (is Smart)
if (sequence.contains(n)) { return 1; }
return 0;
}
}

First think of a mathematical solution.
Smart numbers form a sequence:
a0 = 1
an+1 = n + an
This gives a function for smart numbers:
f(x) = ax² + bx + c
f(x + 1) = f(x) + x = ...
So the problem is to find for a given y a matching x.
You can do this by a binary search.
int isSmart(int n) {
int xlow = 1;
int xhigh = n; // Exclusive. For n == 0 return 1.
while (xlow < xhigh) {
int x = (xlow + xhigh)/2;
int y = f(x);
if (y == n) {
return 1;
}
if (y < n) {
xlow = x + 1;
} else {
xhigh = x;
}
}
return 0;
}
Yet smarter would be to use the solution for x and look whether it is an integer:
ax² + bx + c' = 0 where c' = c - n
x = ...

I was playing around with this and I noticed something. The smart numbers are
1 2 4 7 11 16 22 29 ...
If you subtract one you get
0 1 3 6 10 15 21 28 ...
0 1 2 3 4 5 6 7 ...
The above sequence happens to be the sum of the first n numbers starting with 0 which is n*(n+1)/2. So add 1 to that and you get a smart number.
Since n and n+1 are next door to each other you can derive them by reversing the process.
Take 29, subtract 1 = 28, * 2 = 56. The sqrt(56) rounded up is 8. So the 8th smart number (counting from 0) is 29.
Using that information you can detect a smart number without a loop by simply reversing the process.
public static int isSmart(int v) {
int vv = (v-1)*2;
int sq = (int)Math.sqrt(vv);
int chk = (sq*(sq+1))/2 + 1;
return (chk == v) ? 1 : 0;
}
Using a version which supports longs have verified this against the iterative process from 1 to 10,000,000,000.

how to get fibonacci series within range without recursion

I am trying to calculate the fibonacci numbers within specific range(wide range of numbers in thousands)
I have wrote this but for I do not know to to modify it to make it within a range for example i need to get fibonacci numbers between 5027 and 8386589
class Fibonacci
{
public static void main(String args[])
{
int n1=0,n2=1,n3,i,count=10;
System.out.print(n1+" "+n2);//printing 0 and 1
for(i=2;i<count;++i)
{
n3=n1+n2;
System.out.print(" "+n3);
n1=n2;
n2=n3;
}
}
}

int fib(int low, int high){
// Initialize first three Fibonacci Numbers
int n1 = 0, n2 = 1, n3 = 1;
// Count fibonacci numbers in given range
int result = 0;
while (n1 <= high){
if (n1 >= low)
result++;
f1 = f2;
f2 = f3;
f3 = f1 + f2;
}
return result;
}

Try using a while loop instead of a for loop and include an if-statement
while(n3<8386589){
if(n3>5027){
System.out.print(n3+" ");
}
n3=n1+n2;
n1=n2;
n2=n3;
}

FWIW, here's my version (also using a while loop):
private static void Fibonacci(long lower, long upper)
{
long curr = 1, prev = 1;
while (curr <= upper)
{
long temp = curr;
curr = prev + curr;
prev = temp;
if (curr >= lower && curr <= upper)
{
System.out.println(curr);
}
}
}

The some idea just using BigInteger for bigger values :
private static BigInteger function_f(int n) {
// if n = 0 => f(n) = 0
if(n == 0)
return new BigInteger("0");
// Initialization of variables
// if n = 1 => f(n) = 1 (case included)
BigInteger result = new BigInteger("1");
BigInteger last_fn = new BigInteger("0");
BigInteger before_last_fn = new BigInteger("0");
// Do the loop for n > 1
for (int i = 2; i <= n; i++) {
// f(n - 2)
before_last_fn = last_fn;
// f(n - 1)
last_fn = result;
// f(n - 1) + f(n - 2)
result = last_fn.add(before_last_fn);
}
// Return the result
return result;
}

How to improve the algorithm for the Max Common Array Slice?

I was asked to take a HackerRank code test, and the exercise I was asked is the Max Common Array Slice. The problem goes as follows:
You are given a sequence of n integers a0, a1, . . . , an−1 and the
task is to find the maximum slice of the array which contains no more
than two different numbers.
Input 1 :
[1, 1, 1, 2, 2, 2, 1, 1, 2, 2, 6, 2, 1, 8]
Result 1 : Answer is 10 because the array slice of (0, 9) is the
largest slice of the array with no more than two different numbers.
There are 10 items in this slice which are "1, 1, 1, 2, 2, 2, 1, 1, 2, 2".
2 different numbers for this slice are 1 and 2.
Input 2:
[53, 800, 0, 0, 0, 356, 8988, 1, 1]
Result 2: Answer is 4 because the slice of (1, 4) is the largest slice
of the array with no more than two different numbers. The slice (2, 5)
would also be valid and would still give a result of 4.
There are 4 items in this slice which are "800,0,0,0".
2 different numbers for this slice are 800 and 0.
Maximum common array slice of the array which contains no more than
two different numbers implementation in Java takes a comma delimited
array of numbers from STDIN and the output is written back to console.
The implementation I provide (below) works, however 3 test cases timeout in HR. Clearly, HR hides the test cases, so I could not see exactly the conditions the timeout was triggered or even the length of the timeout.
I'm not surprised of the timeout, though, given the asymptotic complexity of my solution. But my question is: how could my solution be improved?
Thanks in advance to all those that will help!
import java.io.*;
import java.lang.*;
import java.util.*;
import java.util.stream.*;
public class Solution {
public static void main(String args[] ) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = br.readLine();
List<Integer> inputSequence = parseIntegerSequence(line);
int largestSliceSize = calculateLargestSlice(inputSequence);
System.out.println(largestSliceSize);
}
private static List<Integer> parseIntegerSequence(String input) {
if (input == null)
return new ArrayList();
return Arrays.asList(input.split("\\s*,\\s*"))
.stream()
.filter(item -> item.matches("^\\s*-?[0-9]+\\s*$"))
.map(item -> Integer.parseInt(item))
.collect(Collectors.toList());
}
private static int calculateLargestSlice(List<Integer> inputSequence) {
Map<Integer, Integer> temporaryMap = new HashMap<>();
int result = 0;
int startIndex = 0;
int uniqueItemCount = 0;
Integer[] array = inputSequence.toArray(new Integer[inputSequence.size()]);
while (startIndex < array.length) { // loop over the entire input sequence
temporaryMap.put(array[startIndex], startIndex);
uniqueItemCount++;
for (int j = startIndex + 1; j < array.length; j++) {
if (temporaryMap.get(array[j]) == null) {
if (uniqueItemCount != 2) {
temporaryMap.put(array[j], j);
uniqueItemCount++;
if (j == array.length - 1) {
result = Math.max(result, j - startIndex + 1);
startIndex = array.length;
break;
}
} else {
result = Math.max(result, j - startIndex);
int item = array[j-1];
int firstFoundIndex = 0;
for( int k=j-1; k>=0; k-- )
{
if( array[k] != item )
{
firstFoundIndex = k+1;
break;
}
}
startIndex = firstFoundIndex;
temporaryMap.clear();
uniqueItemCount = 0;
break;
}
} else if (temporaryMap.get(array[j]) != null) {
if (j == array.length - 1) {
result = Math.max(result, j - startIndex + 1);
startIndex = array.length;
break;
}
}
}
}
return result;
}
}

This is my answer in Java and it passed all the HackerRank test cases. Please feel free to comment if you find something wrong.
public static int maxCommonArraySlice(List<Integer> inputSequence) {
if(inputSequence.size() < 2) return inputSequence.size(); // I'm doubting this line should be <= 2
List<Integer> twoInts = new LinkedList<>();
twoInts.add(inputSequence.get(0));
int start = 0;
int end = inputSequence.size();
int count = 0;
int max_length = 0;
while(start < end) {
if(twoInts.contains(inputSequence.get(start))) {
count++;
start++;
}
else if(twoInts.size() == 1) {
twoInts.add(inputSequence.get(start));
}
else { // twoInts.size() is 2
count = 0;
start--;
twoInts.set(0, inputSequence.get(start));
twoInts.set(1, inputSequence.get(start + 1));
}
if(count > max_length) {
max_length = count;
}
}
return max_length;
}
public static void main(String[] args) {
List<Integer> input = new LinkedList<Integer>(Arrays.asList(53,800,0,0,0,356,8988,1,1));
System.out.println(maxCommonArraySlice(input));
}

I think this would work:
public int solution(int[] arr) {
int lastSeen = -1;
int secondLastSeen = -1;
int lbs = 0;
int tempCount = 0;
int lastSeenNumberRepeatedCount = 0;
for (int current : arr) {
if (current == lastSeen || current == secondLastSeen) {
tempCount ++;
} else {
// if the current number is not in our read list it means new series has started, tempCounter value in this case will be
// how many times lastSeen number repeated before this new number encountered + 1 for current number.
tempCount = lastSeenNumberRepeatedCount + 1;
}
if (current == lastSeen) {
lastSeenNumberRepeatedCount++;
} else {
lastSeenNumberRepeatedCount = 1;
secondLastSeen = lastSeen;
lastSeen = current;
}
lbs = Math.max(tempCount, lbs);
}
return lbs;
}
Reference

This is a python solution, as per requested by OP
def solution(arr):
if (len(arr) <= 2): print arr
lres = 0
rres = 0
l = 0
r = 1
last = arr[1]
prev = arr[0]
while(r <= len(arr)-1):
if prev != last:
if arr[r] == prev:
prev = last
last = arr[r]
elif arr[r] != last:
l = r-1
while(arr[l-1] == last):
l -= 1
last = arr[r]
prev = arr[r-1]
else:
if arr[r] != prev:
last = arr[r]
if r - l > rres-lres:
rres = r
lres = l
r += 1
print arr[lres:rres+1]
For current segment let's say that last is the last value added and prev is the second distinct value in the segment. (initially they might be equal).
Let's keep to pointers l and r to left and right ends of the current segment with at most two distinct elements. And let's say we consider element arr[r].
If current segment [l,r-1] contains only one distinct element, we can safely add arr[r], with possibly updating last and prev.
Now if arr[r] equals to last, then we don't need to do anything. If arr[r] equals to prev, we need to swap prev and last. If it equals to neither of those two, we need to update l left pointer, by tracing back from r-1 until we find an element which is not equal to last, then update last and prev.

Dynamic programming approach to TSP in Java

I'm a beginner, and I'm trying to write a working travelling salesman problem using dynamic programming approach.
This is the code for my compute function:
public static int compute(int[] unvisitedSet, int dest) {
if (unvisitedSet.length == 1)
return distMtx[dest][unvisitedSet[0]];
int[] newSet = new int[unvisitedSet.length-1];
int distMin = Integer.MAX_VALUE;
for (int i = 0; i < unvisitedSet.length; i++) {
for (int j = 0; j < newSet.length; j++) {
if (j < i) newSet[j] = unvisitedSet[j];
else newSet[j] = unvisitedSet[j+1];
}
int distCur;
if (distMtx[dest][unvisitedSet[i]] != -1) {
distCur = compute(newSet, unvisitedSet[i]) + distMtx[unvisitedSet[i]][dest];
if (distMin > distCur)
distMin = distCur;
}
else {
System.out.println("No path between " + dest + " and " + unvisitedSet[i]);
}
}
return distMin;
}
The code is not giving me the correct answers, and I'm trying to figure out where the error is occurring. I think my error occurs when I add:
distCur = compute(newSet, unvisitedSet[i]) + distMtx[unvisitedSet[i]][dest];
So I've been messing around with that part, moving the addition to the very end right before I return distMin and so on... But I couldn't figure it out.
Although I'm sure it can be inferred from the code, I will state the following facts to clarify.
distMtx stores all the intercity distances, and distances are symmetric, meaning if distance from city A to city B is 3, then the distance from city B to city A is also 3. Also, if two cities don't have any direct paths, the distance value is -1.
Any help would be very much appreciated!
Thanks!
Edit:
The main function reads the intercity distances from a text file. Because I'm assuming the number of cities will always be less than 100, global int variable distMtx is [100][100].
Once the matrix is filled with the necessary information, an array of all the cities are created. The names of the cities are basically numbers. So if I have 4 cities, set[4] = {0, 1, 2, 3}.
In the main function, after distMtx and set is created, first call to compute() is called:
int optRoute = compute(set, 0);
System.out.println(optRoute);
Sample input:
-1 3 2 7
3 -1 10 1
2 10 -1 4
7 1 4 -1
Expected output:
10

Here's a working iterative solution to the TSP with dynamic programming. What would make your life easier is to store the current state as a bitmask instead of in an array. This has the advantage that the state representation is compact and can be cached easily.
I made a video detailing the solution to this problem on Youtube, please enjoy! Code was taken from my github repo
/**
* An implementation of the traveling salesman problem in Java using dynamic
* programming to improve the time complexity from O(n!) to O(n^2 * 2^n).
*
* Time Complexity: O(n^2 * 2^n)
* Space Complexity: O(n * 2^n)
*
**/
import java.util.List;
import java.util.ArrayList;
import java.util.Collections;
public class TspDynamicProgrammingIterative {
private final int N, start;
private final double[][] distance;
private List<Integer> tour = new ArrayList<>();
private double minTourCost = Double.POSITIVE_INFINITY;
private boolean ranSolver = false;
public TspDynamicProgrammingIterative(double[][] distance) {
this(0, distance);
}
public TspDynamicProgrammingIterative(int start, double[][] distance) {
N = distance.length;
if (N <= 2) throw new IllegalStateException("N <= 2 not yet supported.");
if (N != distance[0].length) throw new IllegalStateException("Matrix must be square (n x n)");
if (start < 0 || start >= N) throw new IllegalArgumentException("Invalid start node.");
this.start = start;
this.distance = distance;
}
// Returns the optimal tour for the traveling salesman problem.
public List<Integer> getTour() {
if (!ranSolver) solve();
return tour;
}
// Returns the minimal tour cost.
public double getTourCost() {
if (!ranSolver) solve();
return minTourCost;
}
// Solves the traveling salesman problem and caches solution.
public void solve() {
if (ranSolver) return;
final int END_STATE = (1 << N) - 1;
Double[][] memo = new Double[N][1 << N];
// Add all outgoing edges from the starting node to memo table.
for (int end = 0; end < N; end++) {
if (end == start) continue;
memo[end][(1 << start) | (1 << end)] = distance[start][end];
}
for (int r = 3; r <= N; r++) {
for (int subset : combinations(r, N)) {
if (notIn(start, subset)) continue;
for (int next = 0; next < N; next++) {
if (next == start || notIn(next, subset)) continue;
int subsetWithoutNext = subset ^ (1 << next);
double minDist = Double.POSITIVE_INFINITY;
for (int end = 0; end < N; end++) {
if (end == start || end == next || notIn(end, subset)) continue;
double newDistance = memo[end][subsetWithoutNext] + distance[end][next];
if (newDistance < minDist) {
minDist = newDistance;
}
}
memo[next][subset] = minDist;
}
}
}
// Connect tour back to starting node and minimize cost.
for (int i = 0; i < N; i++) {
if (i == start) continue;
double tourCost = memo[i][END_STATE] + distance[i][start];
if (tourCost < minTourCost) {
minTourCost = tourCost;
}
}
int lastIndex = start;
int state = END_STATE;
tour.add(start);
// Reconstruct TSP path from memo table.
for (int i = 1; i < N; i++) {
int index = -1;
for (int j = 0; j < N; j++) {
if (j == start || notIn(j, state)) continue;
if (index == -1) index = j;
double prevDist = memo[index][state] + distance[index][lastIndex];
double newDist = memo[j][state] + distance[j][lastIndex];
if (newDist < prevDist) {
index = j;
}
}
tour.add(index);
state = state ^ (1 << index);
lastIndex = index;
}
tour.add(start);
Collections.reverse(tour);
ranSolver = true;
}
private static boolean notIn(int elem, int subset) {
return ((1 << elem) & subset) == 0;
}
// This method generates all bit sets of size n where r bits
// are set to one. The result is returned as a list of integer masks.
public static List<Integer> combinations(int r, int n) {
List<Integer> subsets = new ArrayList<>();
combinations(0, 0, r, n, subsets);
return subsets;
}
// To find all the combinations of size r we need to recurse until we have
// selected r elements (aka r = 0), otherwise if r != 0 then we still need to select
// an element which is found after the position of our last selected element
private static void combinations(int set, int at, int r, int n, List<Integer> subsets) {
// Return early if there are more elements left to select than what is available.
int elementsLeftToPick = n - at;
if (elementsLeftToPick < r) return;
// We selected 'r' elements so we found a valid subset!
if (r == 0) {
subsets.add(set);
} else {
for (int i = at; i < n; i++) {
// Try including this element
set |= 1 << i;
combinations(set, i + 1, r - 1, n, subsets);
// Backtrack and try the instance where we did not include this element
set &= ~(1 << i);
}
}
}
public static void main(String[] args) {
// Create adjacency matrix
int n = 6;
double[][] distanceMatrix = new double[n][n];
for (double[] row : distanceMatrix) java.util.Arrays.fill(row, 10000);
distanceMatrix[5][0] = 10;
distanceMatrix[1][5] = 12;
distanceMatrix[4][1] = 2;
distanceMatrix[2][4] = 4;
distanceMatrix[3][2] = 6;
distanceMatrix[0][3] = 8;
int startNode = 0;
TspDynamicProgrammingIterative solver = new TspDynamicProgrammingIterative(startNode, distanceMatrix);
// Prints: [0, 3, 2, 4, 1, 5, 0]
System.out.println("Tour: " + solver.getTour());
// Print: 42.0
System.out.println("Tour cost: " + solver.getTourCost());
}
}

I know this is pretty old question but it might help somebody in the future.
Here is very well written paper on TSP with dynamic programming approach
https://github.com/evandrix/SPOJ/blob/master/DP_Main112/Solving-Traveling-Salesman-Problem-by-Dynamic-Programming-Approach-in-Java.pdf

I think you have to make some changes in your program.
Here there is an implementation
http://www.sanfoundry.com/java-program-implement-traveling-salesman-problem-using-nearest-neighbour-algorithm/

A better algorithm to find the next palindrome of a number string

Firstly here is the problem:
A positive integer is called a palindrome if its representation in the decimal system is the same when read from left to right and from right to left. For a given positive integer K of not more than 1000000 digits, write the value of the smallest palindrome larger than K to output. Numbers are always displayed without leading zeros.
Input: The first line contains integer t, the number of test cases. Integers K are given in the next t lines.
Output: For each K, output the smallest palindrome larger than K.
Example
Input:
2
808
2133
Output:
818
2222
Secondly here is my code:
// I know it is bad practice to not cater for erroneous input,
// however for the purpose of the execise it is omitted
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.lang.Exception;
import java.math.BigInteger;
public class Main
{
public static void main(String [] args){
try{
Main instance = new Main(); // create an instance to access non-static
// variables
// Use java.util.Scanner to scan the get the input and initialise the
// variable
Scanner sc=null;
BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
String input = "";
int numberOfTests = 0;
String k; // declare any other variables here
if((input = r.readLine()) != null){
sc = new Scanner(input);
numberOfTests = sc.nextInt();
}
for (int i = 0; i < numberOfTests; i++){
if((input = r.readLine()) != null){
sc = new Scanner(input);
k=sc.next(); // initialise the remainder of the variables sc.next()
instance.palindrome(k);
} //if
}// for
}// try
catch (Exception e)
{
e.printStackTrace();
}
}// main
public void palindrome(String number){
StringBuffer theNumber = new StringBuffer(number);
int length = theNumber.length();
int left, right, leftPos, rightPos;
// if incresing a value to more than 9 the value to left (offset) need incrementing
int offset, offsetPos;
boolean offsetUpdated;
// To update the string with new values
String insert;
boolean hasAltered = false;
for(int i = 0; i < length/2; i++){
leftPos = i;
rightPos = (length-1) - i;
offsetPos = rightPos -1; offsetUpdated = false;
// set values at opposite indices and offset
left = Integer.parseInt(String.valueOf(theNumber.charAt(leftPos)));
right = Integer.parseInt(String.valueOf(theNumber.charAt(rightPos)));
offset = Integer.parseInt(String.valueOf(theNumber.charAt(offsetPos)));
if(left != right){
// if r > l then offest needs updating
if(right > left){
// update and replace
right = left;
insert = Integer.toString(right);
theNumber.replace(rightPos, rightPos + 1, insert);
offset++; if (offset == 10) offset = 0;
insert = Integer.toString(offset);
theNumber.replace(offsetPos, offsetPos + 1, insert);
offsetUpdated = true;
// then we need to update the value to left again
while (offset == 0 && offsetUpdated){
offsetPos--;
offset =
Integer.parseInt(String.valueOf(theNumber.charAt(offsetPos)));
offset++; if (offset == 10) offset = 0;
// replace
insert = Integer.toString(offset);
theNumber.replace(offsetPos, offsetPos + 1, insert);
}
// finally incase right and offset are the two middle values
left = Integer.parseInt(String.valueOf(theNumber.charAt(leftPos)));
if (right != left){
right = left;
insert = Integer.toString(right);
theNumber.replace(rightPos, rightPos + 1, insert);
}
}// if r > l
else
// update and replace
right = left;
insert = Integer.toString(right);
theNumber.replace(rightPos, rightPos + 1, insert);
}// if l != r
}// for i
System.out.println(theNumber.toString());
}// palindrome
}
Finally my explaination and question.
My code compares either end and then moves in
if left and right are not equal
if right is greater than left
(increasing right past 9 should increase the digit
to its left i.e 09 ---- > 10) and continue to do
so if require as for 89999, increasing the right
most 9 makes the value 90000
before updating my string we check that the right
and left are equal, because in the middle e.g 78849887
we set the 9 --> 4 and increase 4 --> 5, so we must cater for this.
The problem is from spoj.pl an online judge system. My code works for all the test can provide but when I submit it, I get a time limit exceeded error and my answer is not accepted.
Does anyone have any suggestions as to how I can improve my algorithm. While writing this question i thought that instead of my while (offset == 0 && offsetUpdated) loop i could use a boolean to to make sure i increment the offset on my next [i] iteration. Confirmation of my chang or any suggestion would be appreciated, also let me know if i need to make my question clearer.

This seems like a lot of code. Have you tried a very naive approach yet? Checking whether something is a palindrome is actually very simple.
private boolean isPalindrome(int possiblePalindrome) {
String stringRepresentation = String.valueOf(possiblePalindrome);
if ( stringRepresentation.equals(stringRepresentation.reverse()) ) {
return true;
}
}
Now that might not be the most performant code, but it gives you a really simple starting point:
private int nextLargestPalindrome(int fromNumber) {
for ( int i = fromNumber + 1; ; i++ ) {
if ( isPalindrome( i ) ) {
return i;
}
}
}
Now if that isn't fast enough you can use it as a reference implementation and work on decreasing the algorithmic complexity.
There should actually be a constant-time (well it is linear on the number of digits of the input) way to find the next largest palindrome. I will give an algorithm that assumes the number is an even number of digits long (but can be extended to an odd number of digits).
Find the decimal representation of the input number ("2133").
Split it into the left half and right half ("21", "33");
Compare the last digit in the left half and the first digit in the right half.
a. If the right is greater than the left, increment the left and stop. ("22")
b. If the right is less than the left, stop.
c. If the right is equal to the left, repeat step 3 with the second-last digit in the left and the second digit in the right (and so on).
Take the left half and append the left half reversed. That's your next largest palindrome. ("2222")
Applied to a more complicated number:
1. 1234567887654322
2. 12345678 87654322
3. 12345678 87654322
^ ^ equal
3. 12345678 87654322
^ ^ equal
3. 12345678 87654322
^ ^ equal
3. 12345678 87654322
^ ^ equal
3. 12345678 87654322
^ ^ equal
3. 12345678 87654322
^ ^ equal
3. 12345678 87654322
^ ^ equal
3. 12345678 87654322
^ ^ greater than, so increment the left
3. 12345679
4. 1234567997654321 answer
This seems a bit similar to the algorithm you described, but it starts at the inner digits and moves to the outer.

There is no reason to fiddle with individual digits when the only needed operation is one simple addition. The following code is based on Raks' answer.
The code stresses simplicity over execution speed, intentionally.
import static org.junit.Assert.assertEquals;
import java.math.BigInteger;
import org.junit.Test;
public class NextPalindromeTest {
public static String nextPalindrome(String num) {
int len = num.length();
String left = num.substring(0, len / 2);
String middle = num.substring(len / 2, len - len / 2);
String right = num.substring(len - len / 2);
if (right.compareTo(reverse(left)) < 0)
return left + middle + reverse(left);
String next = new BigInteger(left + middle).add(BigInteger.ONE).toString();
return next.substring(0, left.length() + middle.length())
+ reverse(next).substring(middle.length());
}
private static String reverse(String s) {
return new StringBuilder(s).reverse().toString();
}
#Test
public void testNextPalindrome() {
assertEquals("5", nextPalindrome("4"));
assertEquals("11", nextPalindrome("9"));
assertEquals("22", nextPalindrome("15"));
assertEquals("101", nextPalindrome("99"));
assertEquals("151", nextPalindrome("149"));
assertEquals("123454321", nextPalindrome("123450000"));
assertEquals("123464321", nextPalindrome("123454322"));
}
}

Well I have constant order solution(atleast of order k, where k is number of digits in the number)
Lets take some examples
suppose n=17208
divide the number into two parts from middle
and reversibly write the most significant part onto the less significant one.
ie, 17271
if the so generated number is greater than your n it is your palindrome, if not just increase the center number(pivot) ie, you get 17371
other examples
n=17286
palidrome-attempt=17271(since it is less than n increment the pivot, 2 in this case)
so palidrome=17371
n=5684
palidrome1=5665
palidrome=5775
n=458322
palindrome=458854
now suppose n = 1219901
palidrome1=1219121
incrementing the pivot makes my number smaller here
so increment the number adjacent pivot too
1220221
and this logic could be extended

public class NextPalindrome
{
int rev, temp;
int printNextPalindrome(int n)
{
int num = n;
for (int i = num+1; i >= num; i++)
{
temp = i;
rev = 0;
while (temp != 0)
{
int remainder = temp % 10;
rev = rev * 10 + remainder;
temp = temp / 10;
}
if (rev == i)
{
break;
}
}
return rev;
}
public static void main(String args[])
{
NextPalindrome np = new NextPalindrome();
int nxtpalin = np.printNextPalindrome(11);
System.out.println(nxtpalin);
}
}

Here is my code in java. Whole idea is from here.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter number of tests: ");
int t = sc.nextInt();
for (int i = 0; i < t; i++) {
System.out.println("Enter number: ");
String numberToProcess = sc.next(); // ne proveravam dal su brojevi
nextSmallestPalindrom(numberToProcess);
}
}
private static void nextSmallestPalindrom(String numberToProcess) {
int i, j;
int length = numberToProcess.length();
int[] numberAsIntArray = new int[length];
for (int k = 0; k < length; k++)
numberAsIntArray[k] = Integer.parseInt(String
.valueOf(numberToProcess.charAt(k)));
numberToProcess = null;
boolean all9 = true;
for (int k = 0; k < length; k++) {
if (numberAsIntArray[k] != 9) {
all9 = false;
break;
}
}
// case 1, sve 9ke
if (all9) {
whenAll9(length);
return;
}
int mid = length / 2;
if (length % 2 == 0) {
i = mid - 1;
j = mid;
} else {
i = mid - 1;
j = mid + 1;
}
while (i >= 0 && numberAsIntArray[i] == numberAsIntArray[j]) {
i--;
j++;
}
// case 2 already polindrom
if (i == -1) {
if (length % 2 == 0) {
i = mid - 1;
j = mid;
} else {
i = mid;
j = i;
}
addOneToMiddleWithCarry(numberAsIntArray, i, j, true);
} else {
// case 3 not polindrom
if (numberAsIntArray[i] > numberAsIntArray[j]) { // 3.1)
while (i >= 0) {
numberAsIntArray[j] = numberAsIntArray[i];
i--;
j++;
}
for (int k = 0; k < numberAsIntArray.length; k++)
System.out.print(numberAsIntArray[k]);
System.out.println();
} else { // 3.2 like case 2
if (length % 2 == 0) {
i = mid - 1;
j = mid;
} else {
i = mid;
j = i;
}
addOneToMiddleWithCarry(numberAsIntArray, i, j, false);
}
}
}
private static void whenAll9(int length) {
for (int i = 0; i <= length; i++) {
if (i == 0 || i == length)
System.out.print('1');
else
System.out.print('0');
}
}
private static void addOneToMiddleWithCarry(int[] numberAsIntArray, int i,
int j, boolean palindrom) {
numberAsIntArray[i]++;
numberAsIntArray[j] = numberAsIntArray[i];
while (numberAsIntArray[i] == 10) {
numberAsIntArray[i] = 0;
numberAsIntArray[j] = numberAsIntArray[i];
i--;
j++;
numberAsIntArray[i]++;
numberAsIntArray[j] = numberAsIntArray[i];
}
if (!palindrom)
while (i >= 0) {
numberAsIntArray[j] = numberAsIntArray[i];
i--;
j++;
}
for (int k = 0; k < numberAsIntArray.length; k++)
System.out.print(numberAsIntArray[k]);
System.out.println();
}
}

Try this
public static String genNextPalin(String base){
//check if it is 1 digit
if(base.length()==1){
if(Integer.parseInt(base)==9)
return "11";
else
return (Integer.parseInt(base)+1)+"";
}
boolean check = true;
//check if it is all 9s
for(char a: base.toCharArray()){
if(a!='9')
check = false;
}
if(check){
String num = "1";
for(int i=0; i<base.length()-1; i++)
num+="0";
num+="1";
return num;
}
boolean isBasePalin = isPalindrome(base);
int mid = base.length()/2;
if(isBasePalin){
//if base is palin and it is odd increase mid and return
if(base.length()%2==1){
BigInteger leftHalf = new BigInteger(base.substring(0,mid+1));
String newLeftHalf = leftHalf.add(BigInteger.ONE).toString();
String newPalin = genPalin2(newLeftHalf.substring(0,mid),newLeftHalf.charAt(mid));
return newPalin;
}
else{
BigInteger leftHalf = new BigInteger(base.substring(0,mid));
String newLeftHalf = leftHalf.add(BigInteger.ONE).toString();
String newPalin = genPalin(newLeftHalf.substring(0,mid));
return newPalin;
}
}
else{
if(base.length()%2==1){
BigInteger leftHalf = new BigInteger(base.substring(0,mid));
BigInteger rightHalf = new BigInteger(reverse(base.substring(mid+1,base.length())));
//check if leftHalf is greater than right half
if(leftHalf.compareTo(rightHalf)==1){
String newPalin = genPalin2(base.substring(0,mid),base.charAt(mid));
return newPalin;
}
else{
BigInteger leftHalfMid = new BigInteger(base.substring(0,mid+1));
String newLeftHalfMid = leftHalfMid.add(BigInteger.ONE).toString();
String newPalin = genPalin2(newLeftHalfMid.substring(0,mid),newLeftHalfMid.charAt(mid));
return newPalin;
}
}
else{
BigInteger leftHalf = new BigInteger(base.substring(0,mid));
BigInteger rightHalf = new BigInteger(reverse(base.substring(mid,base.length())));
//check if leftHalf is greater than right half
if(leftHalf.compareTo(rightHalf)==1){
return genPalin(base.substring(0,mid));
}
else{
BigInteger leftHalfMid = new BigInteger(base.substring(0,mid));
String newLeftHalfMid = leftHalfMid.add(BigInteger.ONE).toString();
return genPalin(newLeftHalfMid);
}
}
}
}
public static String genPalin(String base){
return base + new StringBuffer(base).reverse().toString();
}
public static String genPalin2(String base, char middle){
return base + middle +new StringBuffer(base).reverse().toString();
}
public static String reverse(String in){
return new StringBuffer(in).reverse().toString();
}
static boolean isPalindrome(String str) {
int n = str.length();
for( int i = 0; i < n/2; i++ )
if (str.charAt(i) != str.charAt(n-i-1))
return false;
return true;
}

HI Here is another simple algorithm using python,
def is_palindrome(n):
if len(n) <= 1:
return False
else:
m = len(n)/2
for i in range(m):
j = i + 1
if n[i] != n[-j]:
return False
return True
def next_palindrome(n):
if not n:
return False
else:
if is_palindrome(n) is True:
return n
else:
return next_palindrome(str(int(n)+1))
print next_palindrome('1000010')

I have written comments to clarify what each step is doing in this python code.
One thing that need to be taken into consideration is that input can be very large that we can not simply perform integer operations on it. So taking input as string and then manipulating it would be much easier.
tests = int(input())
results = []
for i in range(0, tests):
pal = input().strip()
palen = len(pal)
mid = int(palen/2)
if palen % 2 != 0:
if mid == 0: # if the number is of single digit e.g. next palindrome for 5 is 6
ipal = int(pal)
if ipal < 9:
results.append(int(pal) + 1)
else:
results.append(11) # for 9 next palindrome will be 11
else:
pal = list(pal)
pl = l = mid - 1
pr = r = mid + 1
flag = 'n' # represents left and right half of input string are same
while pl >= 0:
if pal[pl] > pal[pr]:
flag = 'r' # 123483489 in this case pal[pl] = 4 and pal[pr] = 3 so we just need to copy left half in right half
break # 123484321 will be the answer
elif pal[pl] < pal[pr]:
flag = 'm' # 123487489 in this case pal[pl] = 4 and pal[pr] = 9 so copying left half in right half will make number smaller
break # in this case we need to take left half increment by 1 and the copy in right half 123494321 will be the anwere
else:
pl = pl -1
pr = pr + 1
if flag == 'm' or flag == 'n': # increment left half by one and copy in right half
if pal[mid] != '9': # if mid element is < 9 the we can simply increment the mid number only and copy left in right half
pal[mid] = str(int(pal[mid]) + 1)
while r < palen:
pal[r] = pal[l]
r = r + 1
l = l - 1
results.append(''.join(pal))
else: # if mid element is 9 this will effect entire left half because of carry
pal[mid] = '0' # we need to take care of large inputs so we can not just directly add 1 in left half
pl = l
while pal[l] == '9':
pal[l] = '0'
l = l - 1
if l >= 0:
pal[l] = str(int(pal[l]) + 1)
while r < palen:
pal[r] = pal[pl]
r = r + 1
pl = pl - 1
if l < 0:
pal[0] = '1'
pal[palen - 1] = '01'
results.append(''.join(pal))
else:
while r < palen: # when flag is 'r'
pal[r] = pal[l]
r = r + 1
l = l - 1
results.append(''.join(pal))
else: # even length almost similar concept here with flags having similar significance as in case of odd length input
pal = list(pal)
pr = r = mid
pl = l = mid - 1
flag = 'n'
while pl >= 0:
if pal[pl] > pal[pr]:
flag = 'r'
break
elif pal[pl] < pal[pr]:
flag = 'm'
break
else:
pl = pl -1
pr = pr + 1
if flag == 'r':
while r < palen:
pal[r] = pal[l]
r = r + 1
l = l - 1
results.append(''.join(pal))
else:
if pal[l] != '9':
pal[l] = str(int(pal[l]) + 1)
while r < palen:
pal[r] = pal[l]
r = r + 1
l = l - 1
results.append(''.join(pal))
else:
pal[mid] = '0'
pl = l
while pal[l] == '9':
pal[l] = '0'
l = l - 1
if l >= 0:
pal[l] = str(int(pal[l]) + 1)
while r < palen:
pal[r] = pal[pl]
r = r + 1
pl = pl - 1
if l < 0:
pal[0] = '1'
pal[palen - 1] = '01'
results.append(''.join(pal))
for xx in results:
print(xx)

We can find next palindrome easily like below.
private void findNextPalindrom(int i) {
i++;
while (!checkPalindrom(i)) {
i++;
}
Log.e(TAG, "findNextPalindrom:next palindrom is===" + i);
}
private boolean checkPalindrom(int num) {
int temp = num;
int rev = 0;
while (num > 0) {
int rem = num % 10;
rev = rev * 10 + rem;
num = num / 10;
}
return temp == rev;
}

Simple codes and test output:
class NextPalin
{
public static void main( String[] args )
{
try {
int[] a = {2, 23, 88, 234, 432, 464, 7887, 7657, 34567, 99874, 7779222, 2569981, 3346990, 229999, 2299999 };
for( int i=0; i<a.length; i++)
{
int add = findNextPalin(a[i]);
System.out.println( a[i] + " + " + add + " = " + (a[i]+add) );
}
}
catch( Exception e ){}
}
static int findNextPalin( int a ) throws Exception
{
if( a < 0 ) throw new Exception();
if( a < 10 ) return a;
int count = 0, reverse = 0, temp = a;
while( temp > 0 ){
reverse = reverse*10 + temp%10;
count++;
temp /= 10;
}
//compare 'half' value
int halfcount = count/2;
int base = (int)Math.pow(10, halfcount );
int reverseHalfValue = reverse % base;
int currentHalfValue = a % base;
if( reverseHalfValue == currentHalfValue ) return 0;
if( reverseHalfValue > currentHalfValue ) return (reverseHalfValue - currentHalfValue);
if( (((a-currentHalfValue)/base)%10) == 9 ){
//cases like 12945 or 1995
int newValue = a-currentHalfValue + base*10;
int diff = findNextPalin(newValue);
return base*10 - currentHalfValue + diff;
}
else{
return (base - currentHalfValue + reverseHalfValue );
}
}
}
$ java NextPalin
2 + 2 = 4
23 + 9 = 32
88 + 0 = 88
234 + 8 = 242
432 + 2 = 434
464 + 0 = 464
7887 + 0 = 7887
7657 + 10 = 7667
34567 + 76 = 34643
99874 + 25 = 99899
7779222 + 555 = 7779777
2569981 + 9771 = 2579752
3346990 + 443 = 3347433
229999 + 9933 = 239932
2299999 + 9033 = 2309032