void check(int start, int* count, int size, int * set)
{
if(start == size) {
(*count) += 1;
return;
}
for(int i = start; i < size ; i++)
{
if((set[start] == 0) && (((i+1) % (start +1) == 0) || (start + 1) % (i+1) == 0 ))
{
set[start] = 1;
check(start +1, count, size, set);
set[start] = 0;
}
}
}
int countArrangement(int n){
int* set = (int *)malloc(sizeof(int) * n);
memset(set, 0, sizeof(int) * n);
int count = 0;
check(0, &count, n, set);
return count;
}
this is the code translated from java to c but the problem is , when n is greater than 6 then the result is wrong
for 7 it increase by one but after that the result is always smaller than the expected value, I am not able to understand what am i missing at.
Your answer
1
2
3
8
10
36
42
128
216
600
660
3456
3744
9408
18900
Expected answer
1
2
3
8
10
36
41
132
250
700
750
4010
4237
10680
24679
that java code
public class Solution {
int count = 0;
public int countArrangement(int N) {
boolean[] visited = new boolean[N + 1];
calculate(N, 1, visited);
return count;
}
public void calculate(int N, int pos, boolean[] visited) {
if (pos > N)
count++;
for (int i = 1; i <= N; i++) {
if (!visited[i] && (pos % i == 0 || i % pos == 0)) {
visited[i] = true;
calculate(N, pos + 1, visited);
visited[i] = false;
}
}
}
}
Can you just point out the missing part?
this is the code translated from java to c
Well, not really.
You have not made a one-to-one translation. The C code is (perhaps by mistake) using a completely different algorithm.
Start by making a one-to-one translation. Once you have that working, you can start playing with algorithm changes. But don't do both in the same step.
A one-to-one translation would be more like:
void calculate(int N, int pos, int * visited, int* count)
{
if (pos > N)
(*count)++;
for (int i = 1; i <= N; i++) {
if (!visited[i] && (pos % i == 0 || i % pos == 0)) {
visited[i] = 1;
calculate(N, pos + 1, visited, count);
visited[i] = 0;
}
}
}
int countArrangement(int n)
{
int* set = calloc(n+1, sizeof *set);
int count = 0;
calculate(n, 1, set, &count);
free(set);
return count;
}
Notice how the C code for calculate is almost identical to the java version. No change of algorithm - only a few changes required due to language differences.
I will show you my problem. This is using leetcode and I'm trying to create an atoi method.
public int myAtoi(String s) {
System.out.println(s.matches("^[^ -0123456789].*")); //this is the regex I am debugging
if(s.matches("^[^ -0123456789].*")){
return 0;
}
int solution = 0;
s = s.replaceAll("[^-0123456789.]","");
solution = 0;
boolean negative = false;
if(s.charAt(0) == '-'){
s = s.replaceAll("-","");
negative = true;
}
if(s.matches("^[0-9]?[.][0-9]+")){
s = s.substring(0, s.indexOf('.'));
System.out.println(s);
}
for(int i = s.length(); i > 0; i--){
solution = solution + (s.charAt(s.length() - i) - 48) * (int)Math.pow(10,i - 1);
}
if(negative) solution = solution * -1;
if(negative && solution > 0) return (int) Math.pow(-2,31);
if(!negative && solution < 0) return (int) Math.pow(2,31) - 1;
return solution;
}
here is the output section screenshot provided incase I have missed something there but a text description also exists.
enter image description here
When the input is "+-12" the output is supposed to be (int) 0. This is due to the requirement being that "if the string does not start with a number, a space, or a negative sign" we return 0.
The line of code whch is supposed to handle this starts at 4 and looks like
if(s.matches("^[^ -0123456789].*")){
return 0;
}
What is wrong with my regex?
We don't really have to use regular expressions for solving this problem, because of the time complexity.
for instance, if(s.matches("^[0-9]?[.][0-9]+")){ does not run linearly, runs quadratically due to the lazy quantifier (?).
We can just loop through once (order of N) and define some statements:
class Solution {
public static final int myAtoi(
String s
) {
s = s.trim();
char[] characters = s.toCharArray();
int sign = 1;
int index = 0;
if (
index < characters.length &&
(characters[index] == '-' || characters[index] == '+')
) {
if (characters[index] == '-') {
sign = -1;
}
++index;
}
int num = 0;
int bound = Integer.MAX_VALUE / 10;
while (
index < characters.length &&
characters[index] >= '0' &&
characters[index] <= '9'
) {
final int digit = characters[index] - '0';
if (num > bound || (num == bound && digit > 7)) {
return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
}
num *= 10;
num += digit;
++index;
}
return sign * num;
}
}
Here is a C++ version, if you might be interested:
// Most of headers are already included;
// Can be removed;
#include <iostream>
#include <cstdint>
#include <vector>
#include <string>
// The following block might trivially improve the exec time;
// Can be removed;
static const auto imporve_runtime = []() {
std::ios::sync_with_stdio(false);
std::cin.tie(NULL);
std::cout.tie(NULL);
return 0;
}();
#define MAX INT_MAX
#define MIN INT_MIN
using ValueType = std::int_fast32_t;
struct Solution {
static const int myAtoi(
const std::string str
) {
const ValueType len = std::size(str);
ValueType sign = 1;
ValueType index = 0;
while (index < len && str[index] == ' ') {
index++;
}
if (index == len) {
return 0;
}
if (str[index] == '-') {
sign = -1;
++index;
} else if (str[index] == '+') {
++index;
}
std::int_fast64_t num = 0;
while (index < len && num < MAX && std::isdigit(str[index])) {
ValueType digit = str[index] - '0';
num *= 10;
num += digit;
index++;
}
if (num > MAX) {
return sign == 1 ? MAX : MIN;
}
return sign * num;
}
};
// int main() {
// std::cout << Solution().myAtoi("words and 987") << "\n";
// std::cout << Solution().myAtoi("4193 with words") << "\n";
// std::cout << Solution().myAtoi(" -42") << "\n";
// }
Regarding your question
What is wrong with my regex?
If you'd like to see how a regular expression solution works, maybe this concise Python version would help (also runs on O(N ^ 2)):
import re
class Solution:
def myAtoi(self, s: str) -> int:
MAX, MIN = 2147483647, -2147483648
DIGIT_PATTERN = re.compile(r'^\s*[+-]?\d+')
s = re.findall(DIGIT_PATTERN, s)
try:
res = int(''.join(s))
except:
return 0
if res > MAX:
return MAX
if res < MIN:
return MIN
return res
We can workaround the expression of ^\s*[+-]?\d+ by dividing it into two subexpressions so that we would be able to get rid of the lazy quantifier and design an order of N solution, yet that would be unnecessary (and is also against the KISS principle).
Create void method that will put on screen square with patern like that :
xoxo
xoxo
xoxo
xoxo
First argument of the method will define amount of characters used to create a square side, second which character is first.
This is my solution but im wondering if i can do it with less code.
static void square(char a, int b) {
if (a == 'x') {
for (int i = 0; i < b; i++) {
int sum = 0;
do {
System.out.print("x");
sum++;
if (sum == b)
break;
System.out.print("o");
sum++;
}
while (sum != b);
System.out.println();
}
} else {
for (int i = 0; i < b; i++) {
int sum = 0;
do {
System.out.print("o");
sum++;
if (sum == b)
break;
System.out.print("x");
sum++;
}
while (sum != b);
System.out.println();
}
}
}
How to make pattern to look like
xoxox
oxoxo
xoxox
oxoxo
xoxox
And how to make this using only for loops or arrays.
O(n^2) loop and constant O(k) space. Start from 0 and just keep alternating the characters till you reach the end (b^2).
char oth = (a == 'x') ? 'o' : 'x';
for (int i = 0; i < (b * b); i++) {
System.out.print(i % 2 == 0 ? a : oth);
if ((i + 1) % b == 0) {
System.out.println();
}
}
O(n) loop and O(n) space. Construct the two pattern rows to print and alternate them.
char oth = (a == 'x') ? 'o' : 'x';
String x = (a == 'x') ? "xo" : "ox";
// Construct the two repeating patterns which will alternate
String first = String.join("", Collections.nCopies(b / 2, x));
String second = first;
if (b % 2 == 1) {
second = new StringBuilder(first).reverse().toString();
first += a;
second += oth;
}
for (int i = 0; i < b; i++) {
System.out.println(i % 2 == 0 ? first : second);
}
See output here: https://ideone.com/E3bVFK
Have fun
public static void square(int side, char order){
char x = (order=='x')?'x':'o';
char o = (order=='x')?'o':'x';
for ( int i = 0; i < side; i++ ) {
for ( int j = 0; j < side; j++ )
System.out.print((j%2==0)?x:o);
System.out.println();
}
}
Output would be exactly what you were asking for.
I am writing my own big integer class in java without imports and need a method for doubling a number of any size that is represented by a string. The code I have for this now works, but begins to take a long time once the numbers get bigger and bigger. I essentially create two arrays: the main array and the countdown array which both start as the same thing. Then, I run a while loop and increment the main array up and increment the countdown array down. When the countdown array reaches "0", I terminate the loop and the result is a new array with the new number doubled in size. Then of course I have if statements checking whether the arrays need to change the ten's place, etc.... here's what I have... Is there any way I can make it more efficient and quick?
public static String doubleDecimalString (String main) {
String countdown = main;
String finalBuild = "";
boolean runLoop = true;
//if zero is supplied, skip all the nonsense and just return 0
//else, loop through and find the true double
//was having trobule getting single digits to double correctly so i had to hard code this for now.
if (main.equals("0")) {
return main;
} else if (main.equals("5")) {
return "10";
} else if (main.equals("6")) {
return "12";
} else if (main.equals("7")) {
return "14";
} else if (main.equals("8")) {
return "16";
} else if (main.equals("9")) {
return "18";
} else {
//Array for ORIGINAL NUMBER
int[] mainPiece = new int[main.length()+2];
int arrayLength = mainPiece.length;
for ( int i = 0; i < main.length(); i++ ) {
mainPiece[i+2] = Integer.parseInt(main.substring( i, i+1));
}
mainPiece[0] = -1;
mainPiece[1] = -1;
//Array for COUNTDOWN NUMBER
int[] countdownPiece = new int[main.length()+2];
for ( int i = 0; i < main.length(); i++ ) {
countdownPiece[i+2] = Integer.parseInt(main.substring( i, i+1));
}
countdownPiece[0] = -1;
countdownPiece[1] = -1;
while ( runLoop ) {
//Increment and decrement the two arrays
mainPiece[arrayLength-1] += 1;
countdownPiece[arrayLength-1] -= 1;
//UPDATE MAIN ARRAY
if ( mainPiece[arrayLength-1] == 10 ) {
for (int x = arrayLength-1; x > 0; x--) {
if ( (mainPiece[x] == 10) && (mainPiece[x-1] != 9) ) {
mainPiece[x] = 0;
mainPiece[x -1] += 1;
} else if ( (mainPiece[x] == 10) && (mainPiece[x-1] == 9) ) {
mainPiece[x] = 0;
mainPiece[x -1] += 1;
x = arrayLength;
}
if ( (mainPiece[2] == 10) ) {
mainPiece[1] = 1;
mainPiece[2] = 0;
}
}
} // end main array
//UPDATE SIDE ARRAY
if ( countdownPiece[arrayLength-1] == -1 ) {
for (int x = arrayLength-1; x > 0; x--) {
if ( (countdownPiece[x] == -1) && (countdownPiece[x-1] > 0) && (x > 1) ) {
countdownPiece[x] = 9;
countdownPiece[x -1] -= 1;
} else if ( (countdownPiece[x] == -1) && (countdownPiece[x-1] == 0) && (x > 1) ) {
countdownPiece[x] = 9;
countdownPiece[x -1] -= 1;
x = arrayLength;
}
}
} //end side array
//tests whether the pieces need to be switched to -1 for scanning
for (int x = 0; x < arrayLength - 1; x++) {
if ( (countdownPiece[x] == -1 ) && (countdownPiece[x+1] == 0 ) ) {
countdownPiece[x+1] = -1;
}
}
//if the side array has reached "0" then the loop will stop and the main array will return the new doubled value
if ( (countdownPiece[arrayLength-1] == -1) && (countdownPiece[arrayLength-2] == -1) ) {
break;
}
} //end while loop
//transform array into string
finalBuild = "";
for (int T = 0; T < arrayLength; T++) {
finalBuild += (mainPiece[T] != -1) ? mainPiece[T] : "";
}
return finalBuild;
}
}
How about something like this (it basically does a multiply by two and accounts for carries):
private String doubleNumber(String number)
{
int doubleDig = 0;
int carry = 0;
StringBuilder sb = new StringBuilder();
for (int i = number.length() - 1; i >= 0; --i)
{
char c = number.charAt(i);
int origNum = Character.getNumericValue(c);
doubleDig = origNum * 2 + carry;
carry = doubleDig / 10;
doubleDig = doubleDig % 10;
sb.append(doubleDig);
}
if (carry > 0)
{
sb.append(carry);
}
return sb.reverse().toString();
}
Obviously this only handles integers.
I would use StringBuilder or List to build your doubled value.
Use a carry variable to store the carry amount and initialize to 0.
Start at the least significant digit, double the digits and add the carry.
Then set the carry to digit / 10, then the digit to digit % 10
Append digit to your builder or list.
After you loop through all your digits, check if carry is > 0 and append if needed.
Reverse the StringBuilder or list and join and you have your answer.
public class Doubler {
public static void main(String[] args) {
System.out.println(doubleDec("9123123123087987342348798234298723948723987234982374928374239847239487.23233099"));
}
public static String doubleDec(String dec) {
StringBuilder builder = new StringBuilder();
int carry = 0;
for (int i = dec.length() - 1; i > -1 ; i--) {
char charDigit = dec.charAt(i);
if (charDigit == '.') {
builder.append(charDigit);
} else {
int digit = Character.getNumericValue(charDigit);
if (digit == -1) {
throw new IllegalStateException("Invalid character in decimal string.");
}
digit = digit * 2 + carry;
carry = digit / 10;
digit = digit % 10;
builder.append(digit);
}
}
if (carry != 0) {
builder.append(carry);
}
return builder.reverse().toString();
}
}
// 18246246246175974684697596468597447897447974469964749856748479694478974.46466198
I'm trying to solve a question and my question here is why doesn't my solution work?. Here's the question and below's the answer.
Question taken from leetcode: http://oj.leetcode.com/problems/decode-ways/
A message containing letters from A-Z is being encoded to numbers using the following mapping:
'A' -> 1
'B' -> 2
...
'Z' -> 26
Given an encoded message containing digits, determine the total number of ways to decode it.
For example,Given encoded message "12", it could be decoded as "AB" (1 2) or "L" (12). The number of ways decoding "12" is 2.
My solution:
The point with my solution is going backwards and multiplying the number of options if a split is found. By split I mean that digits can be interpreted in two ways. For example: 11 can interpreted in two ways 'aa' or 'k'.
public class Solution {
public int numDecodings(String s) {
if (s.isEmpty() || s.charAt(0) == '0') return 0;
int decodings = 1;
boolean used = false; // Signifies that the prev was already use as a decimal
for (int index = s.length()-1 ; index > 0 ; index--) {
char curr = s.charAt(index);
char prev = s.charAt(index-1);
if (curr == '0') {
if (prev != '1' && prev != '2') {
return 0;
}
index--; // Skip prev because it is part of curr
used = false;
} else {
if (prev == '1' || (prev == '2' && curr <= '6')) {
decodings = decodings * 2;
if (used) {
decodings = decodings - 1;
}
used = true;
} else {
used = false;
}
}
}
return decodings;
}
}
The failure is on the following input:
Input:"4757562545844617494555774581341211511296816786586787755257741178599337186486723247528324612117156948"
Output: 3274568
Expected: 589824
This is a really interesting problem. First, I will show how I would solve this problem. We will see that it is not that complicated when using recursion, and that the problem can be solved using dynamic programming. We will produce a general solution that does not hardcode an upper limit of 26 for each code point.
A note on terminology: I will use the term code point (CP) not in the Unicode sense, but to refer to one of the code numbers 1 though 26. Each code point is represented as a variable number of characters. I will also use the terms encoded text (ET) and clear text (CT) in their obvious meanings. When talking about a sequence or array, the first element is called the head. The remaining elements are the tail.
Theoretical Prelude
The EC "" has one decoding: the CT "".
The EC "3" can be destructured into '3' + "", and has one decoding.
The EC "23" can be destructured as '2' + "3" or '23' + "". Each of the tails has one decoding, so the whole EC has two decodings.
The EC "123" can be destructured as '1' + "23" or '12' + "3". The tails have two and one decodings respectively. The whole EC has three decodings. The destructuring '123' + "" is not valid, because 123 > 26, our upper limit.
… and so on for ECs of any length.
So given a string like "123", we can obtain the number of decodings by finding all valid CPs at the beginning, and summing up the number of decodings of each tail.
The most difficult part of this is to find valid heads. We can get the maximal length of the head by looking at a string representation of the upper limit. In our case, the head can be up to two characters long. But not all heads of appropriate lengths are valid, because they have to be ≤ 26 as well.
Naive Recursive Implementation
Now we have done all the necessary work for a simple (but working) recursive implementation:
static final int upperLimit = 26;
static final int maxHeadSize = ("" + upperLimit).length();
static int numDecodings(String encodedText) {
// check base case for the recursion
if (encodedText.length() == 0) {
return 1;
}
// sum all tails
int sum = 0;
for (int headSize = 1; headSize <= maxHeadSize && headSize <= encodedText.length(); headSize++) {
String head = encodedText.substring(0, headSize);
String tail = encodedText.substring(headSize);
if (Integer.parseInt(head) > upperLimit) {
break;
}
sum += numDecodings(tail);
}
return sum;
}
Cached Recursive Implementation
Obviously this isn't very efficient, because (for longer ETs), the same tail will be analyzed multiple times. Also, we create a lot of temporary strings, but we'll let that be for now. One thing we can easily do is to memoize the number of decodings of a specific tail. For that, we use an array of the same length as the input string:
static final int upperLimit = 26;
static final int maxHeadSize = ("" + upperLimit).length();
static int numDecodings(String encodedText) {
return numDecodings(encodedText, new Integer[1 + encodedText.length()]);
}
static int numDecodings(String encodedText, Integer[] cache) {
// check base case for the recursion
if (encodedText.length() == 0) {
return 1;
}
// check if this tail is already known in the cache
if (cache[encodedText.length()] != null) {
return cache[encodedText.length()];
}
// cache miss -- sum all tails
int sum = 0;
for (int headSize = 1; headSize <= maxHeadSize && headSize <= encodedText.length(); headSize++) {
String head = encodedText.substring(0, headSize);
String tail = encodedText.substring(headSize);
if (Integer.parseInt(head) > upperLimit) {
break;
}
sum += numDecodings(tail, cache); // pass the cache through
}
// update the cache
cache[encodedText.length()] = sum;
return sum;
}
Note that we use an Integer[], not an int[]. This way, we can check for non-existent entries using a test for null. This solution is not only correct, it is also comfortably fast – naive recursion runs in O(number of decodings) time, while the memoized version runs in O(string length) time.
Towards a DP Solution
When you run above code in your head, you will notice that the first invocation with the whole string will have a cache miss, then calculate the number of decodings for the first tail, which also misses the cache every time. We can avoid this by evaluating the tails first, starting from the end of the input. Because all tails will have been evaluated before the whole string is, we can remove the checks for cache misses. Now we also don't have any reason for recursion, because all previous results are already in the cache.
static final int upperLimit = 26;
static final int maxHeadSize = ("" + upperLimit).length();
static int numDecodings(String encodedText) {
int[] cache = new int[encodedText.length() + 1];
// base case: the empty string at encodedText.length() is 1:
cache[encodedText.length()] = 1;
for (int position = encodedText.length() - 1; position >= 0; position--) {
// sum directly into the cache
for (int headSize = 1; headSize <= maxHeadSize && headSize + position <= encodedText.length(); headSize++) {
String head = encodedText.substring(position, position + headSize);
if (Integer.parseInt(head) > upperLimit) {
break;
}
cache[position] += cache[position + headSize];
}
}
return cache[0];
}
This algorithm could be optimized further by noticing that we only ever query the last maxHeadSize elements in the cache. So instead of an array, we could use a fixed-sized queue. At that point, we would have a dynamic programming solution that runs in *O(input length) time and O(maxHeadSize) space.
Specialization for upperLimit = 26
The above algorithms were kept as general as possible, but we can go and manually specialize it for a specific upperLimit. This can be useful because it allows us to do various optimizations. However, this introduces “magic numbers” that make the code harder to maintain. Such manual specializations should therefore be avoided in non-critical software (and the above algorithm is already as fast as it gets).
static int numDecodings(String encodedText) {
// initialize the cache
int[] cache = {1, 0, 0};
for (int position = encodedText.length() - 1; position >= 0; position--) {
// rotate the cache
cache[2] = cache[1];
cache[1] = cache[0];
cache[0] = 0;
// headSize == 1
if (position + 0 < encodedText.length()) {
char c = encodedText.charAt(position + 0);
// 1 .. 9
if ('1' <= c && c <= '9') {
cache[0] += cache[1];
}
}
// headSize == 2
if (position + 1 < encodedText.length()) {
char c1 = encodedText.charAt(position + 0);
char c2 = encodedText.charAt(position + 1);
// 10 .. 19
if ('1' == c1) {
cache[0] += cache[2];
}
// 20 .. 26
else if ('2' == c1 && '0' <= c2 && c2 <= '6') {
cache[0] += cache[2];
}
}
}
return cache[0];
}
Comparision with your code
The code is superficially similar. However, your parsing around characters is more convoluted. You have introduced a used variable that, if set, will decrement the decode count in order to account for double-character CPs. This is wrong, but I am not sure why. The main problem is that you are doubling the count at almost every step. As we have seen, the previous counts are added, and may very well be different.
This indicates that you wrote the code without proper preparation. You can write many kinds of software without having to think too much, but you can't do without careful analysis when designing an algorithm. For me, it is often helpful to design an algorithm on paper, and draw diagrams of each step (along the lines of the “Theoretical Prelude” of this answer). This is especially useful when you are thinking too much about the language you are going to implement in, and too little about possibly wrong assumptions.
I suggest that you read up on “proofs by induction” to understand how to write a correct recursive algorithm. Once you have a recursive solution, you can always translate it into an iterative version.
So here is some what simpler way out for your problem. This is pretty close to calculating Fibonacci, with the difference that there are condition checks on each smaller size subproblem.
The space complexity is O(1) and time is O(n)
The code is in C++.
int numDecodings(string s)
{
if( s.length() == 0 ) return 0;
int j = 0;
int p1 = (s[j] != '0' ? 1 : 0); // one step prev form j=1
int p2 = 1; // two step prev from j=1, empty
int p = p1;
for( int j = 1; j < s.length(); j++ )
{
p = 0;
if( s[j] != '0' )
p += p1;
if( isValidTwo(s, j-1, j) )
p += p2;
if( p==0 ) // no further decoding necessary,
break; // as the prefix 0--j is has no possible decoding.
p2 = p1; // update prev for next j+1;
p1 = p;
}
return p;
}
bool isValidTwo(string &s, int i, int j)
{
int val= 10*(s[i]-'0')+s[j]-'0';
if ( val <= 9 )
return false;
if ( val > 26 )
return false;
return true;
}
Here is my code to solve the problem. I use DP , I think it's clear to understand.
Written in Java
public class Solution {
public int numDecodings(String s) {
if(s == null || s.length() == 0){
return 0;
}
int n = s.length();
int[] dp = new int[n+1];
dp[0] = 1;
dp[1] = s.charAt(0) != '0' ? 1 : 0;
for(int i = 2; i <= n; i++){
int first = Integer.valueOf(s.substring(i-1,i));
int second = Integer.valueOf(s.substring(i-2,i));
if(first >= 1 && first <= 9){
dp[i] += dp[i-1];
}
if(second >= 10 && second <= 26){
dp[i] += dp[i-2];
}
}
return dp[n];
}
}
Since I struggled with this problem myself, here is my solution and reasoning. Probably I will mostly repeat what amon wrote, but maybe someone will find it helpful. Also it's c# not java.
Let's say that we have input "12131" and want to obtain all possible decoded strings.
Straightforward recursive solution would do iterate from left to right, obtain valid 1 and 2 digits heads, and invoke function recursively for tail.
We can visualize it using a tree:
There are 5 leaves and this is number of all possible decoded strings. There are also 3 empty leaves, because number 31 cannot be decoded into letter, so these leaves are invalid.
Algorithm might look like this:
public IList<string> Decode(string s)
{
var result = new List<string>();
if (s.Length <= 2)
{
if (s.Length == 1)
{
if (s[0] != '0')
result.Add(this.ToASCII(s));
}
else if (s.Length == 2)
{
if (s[0] != '0' && s[1] != '0')
result.Add(this.ToASCII(s.Substring(0, 1)) + this.ToASCII(s.Substring(1, 1)));
if (s[0] != '0' && int.Parse(s) > 0 && int.Parse(s) <= 26)
result.Add(this.ToASCII(s));
}
}
else
{
for (int i = 1; i <= 2; ++i)
{
string head = s.Substring(0, i);
if (head[0] != '0' && int.Parse(head) > 0 && int.Parse(head) <= 26)
{
var tails = this.Decode(s.Substring(i));
foreach (var tail in tails)
result.Add(this.ToASCII(head) + tail);
}
}
}
return result;
}
public string ToASCII(string str)
{
int number = int.Parse(str);
int asciiChar = number + 65 - 1; // A in ASCII = 65
return ((char)asciiChar).ToString();
}
We have to take care of numbers starting with 0 ("0", "03", etc.), and greater than 26.
Because in this problem we need only count decoding ways, and not actual strings, we can simplify this code:
public int DecodeCount(string s)
{
int count = 0;
if (s.Length <= 2)
{
if (s.Length == 1)
{
if (s[0] != '0')
count++;
}
else if (s.Length == 2)
{
if (s[0] != '0' && s[1] != '0')
count++;
if (s[0] != '0' && int.Parse(s) > 0 && int.Parse(s) <= 26)
count++;
}
}
else
{
for (int i = 1; i <= 2; ++i)
{
string head = s.Substring(0, i);
if (head[0] != '0' && int.Parse(head) > 0 && int.Parse(head) <= 26)
count += this.DecodeCount(s.Substring(i));
}
}
return count;
}
The problem with this algorithm is that we compute results for the same input string multiple times. For example there are 3 nodes ending with 31: ABA31, AU31, LA31. Also there are 2 nodes ending with 131: AB131, L131.
We know that if node ends with 31 it has only one child, since 31 can be decoded only in one way to CA. Likewise, we know that if string ends with 131 it has 2 children, because 131 can be decoded into ACA or LA. Thus, instead of computing it all over again we can cache it in map, where key is string (eg: "131"), and value is number of decoded ways:
public int DecodeCountCached(string s, Dictionary<string, int> cache)
{
if (cache.ContainsKey(s))
return cache[s];
int count = 0;
if (s.Length <= 2)
{
if (s.Length == 1)
{
if (s[0] != '0')
count++;
}
else if (s.Length == 2)
{
if (s[0] != '0' && s[1] != '0')
count++;
if (s[0] != '0' && int.Parse(s) > 0 && int.Parse(s) <= 26)
count++;
}
}
else
{
for (int i = 1; i <= 2; ++i)
{
string head = s.Substring(0, i);
if (head[0] != '0' && int.Parse(head) > 0 && int.Parse(head) <= 26)
count += this.DecodeCountCached(s.Substring(i), cache);
}
}
cache[s] = count;
return count;
}
We can refine this even further. Instead of using strings as a keys, we can use length, because what is cached is always tail of input string. So instead of caching strings: "1", "31", "131", "2131", "12131" we can cache lengths of tails: 1, 2, 3, 4, 5:
public int DecodeCountDPTopDown(string s, Dictionary<int, int> cache)
{
if (cache.ContainsKey(s.Length))
return cache[s.Length];
int count = 0;
if (s.Length <= 2)
{
if (s.Length == 1)
{
if (s[0] != '0')
count++;
}
else if (s.Length == 2)
{
if (s[0] != '0' && s[1] != '0')
count++;
if (s[0] != '0' && int.Parse(s) > 0 && int.Parse(s) <= 26)
count++;
}
}
else
{
for (int i = 1; i <= 2; ++i)
{
string head = s.Substring(0, i);
if (s[0] != '0' && int.Parse(head) > 0 && int.Parse(head) <= 26)
count += this.DecodeCountDPTopDown(s.Substring(i), cache);
}
}
cache[s.Length] = count;
return count;
}
This is recursive top-down dynamic programming approach. We start from the begining, and then recursively compute solutions for tails, and memoize those results for further use.
We can translate it to bottom-up iterative DP solution. We start from the end and cache results for tiles like in previous solution. Instead of map we can use array because keys are integers:
public int DecodeCountBottomUp(string s)
{
int[] chache = new int[s.Length + 1];
chache[0] = 0; // for empty string;
for (int i = 1; i <= s.Length; ++i)
{
string tail = s.Substring(s.Length - i, i);
if (tail.Length == 1)
{
if (tail[0] != '0')
chache[i]++;
}
else if (tail.Length == 2)
{
if (tail[0] != '0' && tail[1] != '0')
chache[i]++;
if (tail[0] != '0' && int.Parse(tail) > 0 && int.Parse(tail) <= 26)
chache[i]++;
}
else
{
if (tail[0] != '0')
chache[i] += chache[i - 1];
if (tail[0] != '0' && int.Parse(tail.Substring(0, 2)) > 0 && int.Parse(tail.Substring(0, 2)) <= 26)
chache[i] += chache[i - 2];
}
}
return chache.Last();
}
Some people simplify it even further, initializing cache[0] with value 1, so they can get rid of conditions for tail.Length==1 and tail.Length==2. For me it is unintuitive trick though, since clearly for empty string there is 0 decode ways not 1, so in such case additional condition must be added to handle empty input:
public int DecodeCountBottomUp2(string s)
{
if (s.Length == 0)
return 0;
int[] chache = new int[s.Length + 1];
chache[0] = 1;
chache[1] = s.Last() != '0' ? 1 : 0;
for (int i = 2; i <= s.Length; ++i)
{
string tail = s.Substring(s.Length - i, i);
if (tail[0] != '0')
chache[i] += chache[i - 1];
if (tail[0] != '0' && int.Parse(tail.Substring(0, 2)) > 0 && int.Parse(tail.Substring(0, 2)) <= 26)
chache[i] += chache[i - 2];
}
return chache.Last();
}
My solution is based on the idea that the arrangement of items(char/digit) within a particular substring is completely independent of the same within a different substring.
So we need to multiply each of those independent ways to get the total number of ways.
// nc is the number of consecutive 1's or 2's in a substring.
// Returns the number of ways these can be arranged within
// themselves to a valid expr.
int ways(int nc){
int n = pow(2, (nc/2)); //this part can be memorized using map for optimization
int m = n;
if (nc%2) {
m *= 2;
}
return n + m - 1;
}
bool validTens(string A, int i){
return (A[i] == '1' || (A[i] == '2' && A[i+1] <= '6'));
}
int numDecodings(string A) {
int ans = 1;
int nc;
if ((A.length() == 0)||(A[0] == '0')) return 0;
for(int i = 1; i < A.length();i++){
if(A[i] == '0' && validTens(A, i-1) == false) return 0; //invalid string
while(i < A.length() && validTens(A, i-1)) {
if(A[i] == '0'){
//think of '110' or '1210', the last two digits must be together
if(nc > 0) nc--;
}
else nc++;
i++;
}
ans *= ways(nc);
nc = 0;
}
return ans;
}
Java solution with space and time complexity O(n)
public int numDecodings(String s) {
int n = s.length();
if (n > 0 && s.charAt(0) == '0')
return 0;
int[] d = new int[n + 1];
d[0] = 1;
d[1] = s.charAt(0) != '0' ? 1 : 0;
for (int i = 2; i <= n; i++) {
if (s.charAt(i - 1) > '0')
d[i] = d[i] + d[i - 1];
if (s.charAt(i - 2) == '2' && s.charAt(i - 1) < '7')
d[i] = d[i - 2] + d[i];
if (s.charAt(i - 2) == '1' && s.charAt(i - 1) <= '9')
d[i] = d[i - 2] + d[i];
}
return d[n];
}
Here is an O(N) C++ DP implementation.
int numDecodings(string s) {
if(s[0] == '0') return 0; // Invalid Input
int n = s.length();
// dp[i] denotes the number of ways to decode the string of length 0 to i
vector<int> dp(n+1, 0);
// base case : string of 0 or 1 characters will have only 1 way to decode
dp[0] = dp[1] = 1;
for(int i = 2; i <= n; i++) {
// considering the previous number
if(s[i-1] > '0') dp[i] += dp[i-1];
// considering the previous two numbers
if(s[i-2] == '1' || (s[i-2] == '2' && s[i-1] < '7')) dp[i] += dp[i-2];
}
return dp[n];
}