Optimal solution for the maximum single sell profit algorithm - java

The input array is:
A[0] = 23171
A[1] = 21015
A[2] = 21123
A[3] = 21366
A[4] = 21013
A[5] = 21367
Mission is to find maximum profit. E.g A[3] - A[2] = 243
and my code is:
class Solution {
int profit = 0;
public int solution(int[] A) {
for (int i = 0;i < A.length; i++){
for (int j = i + 1; j < A.length; j++){
if(A[j] - A[i] > profit)
profit = A[j] - A[i];
}
}
return profit;
}
}
The result is suppose to be 365 but it blows up on larger inputs.
This code has a time complexity of O(N2) but is possible to do with O(N). I can't really see how to avoid nesting here... Any pointers in the right direction appreciated.

You only need to get max value and min value from your array and substract them both, so in a O(N) iteration, get the min and the max values.
class Solution {
public int solution(int[] A) {
int min = Integer.MAX_VALUE;
int max = Integer.MIN_VALUE;
for (int i = 0;i < A.length; i++){
if(A[i] > max) max = A[i];
if(A[i] < min) min = A[i];
}
return max - min;
}
}

I think that most of you got it wrong. The problem in the post is the maximum single sell profit problem which is a typical interview question.
The most optimal solution:
public int dynamicProgrammingSingleSellProfit(int[] arr) {
if(arr.length == 0) {
return 0;
}
int profit = 0;
int cheapest = arr[0];
for (int i = 0; i < arr.length; i++) {
cheapest = Math.min(cheapest, arr[i]);
profit = Math.max(profit, arr[i] - cheapest);
}
return profit;
}
It has O(n) time and O(1) space complexity.
If you examine the original question the op is looking for profit and since we can't travel in time (yet) you can't just compare the minimum and the maximum in the array.

Related

Largest number in k swaps

Given a number K and string str of digits denoting a positive integer, build the largest number possible by performing swap operations on the digits of str at most K times.
Example 1:
Input:
K = 4
str = "1234567"
Output:
7654321
Explanation:
Three swaps can make the
input 1234567 to 7654321, swapping 1
with 7, 2 with 6 and finally 3 with 5
I am trying to solve it using two loops. For every index i, I am finding the largest integer between (i+1)th index to (N-1)th index, where N is the size of string. If the largest number is greater than arr[i], then swap it. Below is the code I have written.
public static String findMaximumNum(String str, int k) {
int N = str.length();
int[] arr = new int[N];
for (int i = 0; i < N; i++) {
arr[i] = Integer.valueOf(str.charAt(i) + "");
}
int swaps = 0;
for (int i = 0; i < N - 1; i++) {
if(swaps == k)
break;
int maxIndex = findMaxInRange(arr, i + 1, N - 1);
if (arr[i] < arr[maxIndex]) {
swap(arr, i, maxIndex);
swaps++;
}
}
String out = "";
for (int i = 0; i < N; i++) {
out = out + arr[i] + "";
}
return out;
}
private static int findMaxInRange(int[] arr, int i, int j) {
int max = Integer.MIN_VALUE;
int maxIndex = i;
for (int k = i; k <= j; k++) {
if (arr[k] >= max) {
max = arr[k];
maxIndex = k;
}
}
return maxIndex;
}
private static void swap(int[] arr, int i, int j) {
System.out.println("swapping "+arr[i]+" and "+arr[j]+" from "+Arrays.toString(arr));
int ch = arr[i];
arr[i] = arr[j];
arr[j] = ch;
}
public static void main(String[] args) {
System.out.println(findMaximumNum("61892795431", 4));
}
It is failing for few test cases. One of the test cases where it is failing is
Input:
4
61892795431
Its Correct output is:
99876215431
And MyCode's output is:
99876125431
I am not able to figure out how the output is '99876215431' and what is wrong in my approach. Please help me understand. Thanks a lot in advance :)
The basic steps how to solve this problem:
0. cast string to array of integers
make a loop K times
in this loop go from i+1 (LOOP VAR) to end of a collection and search for higher value
when we find higher value then collection[i], we will remember its value and index on witch it is. Important thing to note is that we want to swap biggest number but i also has to be last possible number.
at the end of iteration we swap the elements (i with best index)
we are done so all its left is convert our int list back to string.
code: (its python because java is pain)
def sort(swaps, string):
l = list(map(int, list(string)))
print(l)
for i in range(swaps):
best = l[i] + 1
bestIndex = i
for j in range(i+1, len(l)):
if best <= l[j]:
best = l[j]
bestIndex = j
print(i, bestIndex)
l[i], l[bestIndex] = l[bestIndex], l[i]
return "".join(map(str, l))
print(sort(4, "61892795431"))
Your code is correct. The problem comes from the parameter 4 (max number of swaps). If you use 10, the sorting is completed successfully.
Maybe the deeper problems comes from the fact that you are comparing the swaps of your algorithm with the swaps that you would do efficiently to sort the numbers. Your algorithm may work but probably it is not the most efficient, so the number of swaps needed is above the optimum.

Search minimum absolute value difference in java

How can I optimize my algorithm to find the minimum absolute value difference in a given array. Here is my approach which checks each and every element and returning the value.
static int minAbsVal(int[] myArray){
ArrayList<Integer> diffs= new ArrayList(); // all absolute value differences
for (int i = 0; i < myArray.length; i++) {
for (int j = 0; j < myArray.length; j++) {
if (myArray[j]!=myArray[i]){
int absVal =Math.abs( myArray[i]- myArray[j]);
diffs.add(absVal);
}
}
}
int min = diffs.get(0);
for (int i = 1; i < diffs.size(); i++) {
if (min>diffs.get(i)){
min = diffs.get(i);
}
}
return min;
}
Check this solution. If array contains duplicates than it returns 0 if you wanna avoid than you can add conditions also.
static int minAbsVal(int[] myArray){
Arrays.sort(myArray);
ArrayList<Integer> diffs= new ArrayList<Integer>(); // all absolute value differences
for (int i = 0; i < myArray.length-1; i++) {
diffs.add(Math.abs( myArray[i+1] - myArray[i]));
}
Collections.sort(diffs);
return diffs.get(0);
}
Or you can also use :-
static int minAbsVal(int[] myArray){
Arrays.sort(myArray);
int minimum=Math.abs( myArray[1] - myArray[0]);
for (int i = 1; i < myArray.length-1; i++) {
int diff = Math.abs( myArray[i+1] - myArray[i]);
if (diff == 0)
return 0;
else if (minimum > diff)
minimum = diff;
}
return minimum;
}
Check the following solution,
static int minimumAbsoluteDifference(int n, int[] arr) {
int a = Math.abs(arr[1]-arr[0]);
Arrays.sort(arr);
for(int x=0;x<n-1;x++){
if(Math.abs(arr[x]-arr[x+1])<a){
a = Math.abs(arr[x]-arr[x+1]);
if(a==0){
return a;
}
}
}
return a;
}
Dependson the size of your array a good approach would be to divide it in different smaller arrays and look for the minimum of each in pararell and then compare those results .
Anyways you dont need so many iterations, you can just set the baseValue to the first element of the array and when iterating the array compare the baseValue vs the current, if the current is smaller then asign baseValue to current (I would rename baseValue to minValue in that case)

Max subarray hints needed

Find the contiguous subarray within an array (containing at least one
number) which has the largest sum.
For example, given the array [-2,1,-3,4,-1,2,1,-5,4], the contiguous
subarray [4,-1,2,1] has the largest sum = 6.
I am unable to solve this problem, but I would just like some hints.
It it said this can be solved using Dynamic Programming, but I am struggling to see a connection.
Would the DP connection be taking the sum of the whole array?
A reference for your problem can be found here.
You must use the Kadane's algorithm to solve such a case, which goes something like this:
Initialize:
max_so_far = 0
max_ending_here = 0
Loop for each element of the array
(a) max_ending_here = max_ending_here + a[i]
(b) if(max_ending_here < 0)
max_ending_here = 0
(c) if(max_so_far < max_ending_here)
max_so_far = max_ending_here
return max_so_far
A sample code for your reference:
static int maxSubArraySum(int a[])
{
int size = a.length;
int max_so_far = Integer.MIN_VALUE, max_ending_here = 0;
for (int i = 0; i < size; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
PLease find Java Code.
public static void findMaxSubArray(int arr[]){
int max_end_her=arr[0];
int max_so_far=arr[0];
int x=arr[0];
for(int i=1;i<arr.length;i++){
max_end_her=Math.max(x, max_end_her+arr[i]);
max_so_far=Math.max(max_so_far, max_end_her);
}
System.out.println(max_so_far);
}

Find min in Arraylist of object, can't compare the elements in an arraylist

I'm trying to invoke the getScore() method in my StudentScore class to determine the min and max over elements in an ArrayList, within the printStat() method presented below. I'm getting an ArrayIndexOutOfBoundException. What does that mean, and how can I fix the problem?
public class ScoreCalculator{
private int[] scoreCounter;
ArrayList<StudentScore> scores ;
public ScoreCalculator(int maxScore) {
scoreCounter = new int[maxScore];
scores = new ArrayList<StudentScore>(maxScore);
}
public void printStat() {
System.out.println("Score Report for " + scores.size() + " Students ");
min = 0;
max = 0;
int j=0;
for ( j = 0; j < scores.size(); j++) {
if (scores.get(j).getScore() < scores.get(j - 1).getScore()) {
min = scores.get(j).getScore();
}
if (scores.get(j).getScore() > scores.get(j - 1).getScore()) {
max = scores.get(j).getScore();
}
}
System.out.println(min);
System.out.println(max);
}
If your loop starts form j=0, and you access the element at j-1 in the list, you will see where the issue is.
When j=0, you try to access -1. There is no index -1. Hence the error. Starting from j=1 solves this.
Does this help in any way?
please,
change from
for(j=0;j<scores.size();j++){
to
for(j=1;j<scores.size();j++){
You cannot find the min/max of a sequence by only comparing neighboring values.
You have to compare values to the min/max-so-far.
if (scores.isEmpty())
throw new IllegalStateException("No scores found");
int min = scores.get(0).getScore();
int max = min;
for (int j = 1; j < scores.size(); j++) {
int score = scores.get(j).getScore();
if (score < min)
min = score;
if (score > max)
max = score;
}

Time Complexity of the Algorithms

I have created two algorithms that calculate the prefix averages with a given array. I wanted to derive the time complexities of both algorithms, but I have been struggling a bit. I watched this YouTube video:
https://www.youtube.com/watch?v=udwxWq9wZgg&safe=active. I did not understand how to count the operations in a for loop and a nested for loop.
At 2:27, I managed to count the operations in the for loop in PrefixAverages2. It was 3n+1. However, I cannot understand from 5:50 onwards.
Thanks in advance.
public double[] PrefixAverages1(double input[])
{
double A[] = new double[input.length];
double s;
for(int i=0; i <= input.length - 1 ;i++)
{
s = input[0];
for(int j=1; j <= i ;j++)
{
s = s + input[j];
}
A[i] = s / (i+1);
}
return A;
}
public double[] PrefixAverages2(double input[])
{
double A[] = new double[input.length];
double s = 0;
for( int i=0; i <= input.length - 1 ; i++)
{
s = s + input[i];
A[i] = s / (i+1);
}
return A;
}
for(int i=0; i <= input.length - 1 ;i++)
for(int j=1; j <= i ;j++)
This is quadratic, for a given i, inner loop goes about i-times, so you have to sum over i, so basically you have something like sum_{i=1}^{i=l} i, which is the sum of the first l integers, so l(l+1)/2, then quadratic.
For the second algorithm you just have one loop so its complexity is linear.

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