Find the number of Substrings with even number of K Last Updated : 18 Apr, 2023 Comments Improve Suggest changes Like Article Like Report Given a string str of length N and an integer K, the task is to count the number of substrings with even numbers of K. It means the count of 'K' in each substring will be even and must not be zero. Examples: Input: N = 4, str = "2131", K = 1Output: 2 Explanation: There are two substrings "2131" and '131" having an even number of K. Input: N = 4, str = "1121", K = 2Output: 0Explanation: No, substring exists having an even number of K. Approach: This can be solved with the following idea: We can precompute an array that counts the occurrence of K. After computing this we can traverse the for loops two times to find the ranges and checking the occurrence between the range is odd or even. For more clarity follow the process. Below are the steps of implementation: Declare a new array oneoccur[] of the same size as the input string.Run a for loop to traverse the input string.For each index add the value of the previous value of the precomputing array if the element is other than K and if the current element is K add the previous element incremented by 1.Now we have all occurrences of K till all indexes.Now we have to find all the ranges possible. For that traverse a nested for loop.In each range find the number of occurring K in this range.Check whether the number is even or not and check if it is not equal to 0.If it is even and not equal to zero, then increment the counter variable.After traversing all possible ranges Output the cnt. Below is the implementation of the above approach: C++ // C++ code of the above approach: #include <bits/stdc++.h> using namespace std; // To compute occurrence of K void oneoccur(string arr, int n, int prefixSum[], char K) { // Initialize if (arr[0] == K) { prefixSum[0] = 1; } else { prefixSum[0] = 0; } for (int i = 1; i < n; i++) { // Adding previous element // incremented by 1 if (arr[i] == K) prefixSum[i] = prefixSum[i - 1] + 1; // Adding previous element else prefixSum[i] = prefixSum[i - 1]; } } // Function to find count // number of substrings void findevenone(string s, int n, char K) { // Precomputing onecnt[] array int onencnt[n]; // Precomputing the occurrence // of K in onecnt[] oneoccur(s, n, onencnt, K); // Counter variable int cnt = 0; // For loop for traversing all the // rages for possible subarray for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { int onecnt = 0; // Counting the occurrence of // '1' occurrence of '1' from // i to j is pre[j]-pre[i-1] // if i==0 then this is pre[j] if (i == 0) { if (onencnt[j] % 2 == 0 && onencnt[j] != 0) { // Incrementing count cnt++; } } else { if ((onencnt[j] - onencnt[i - 1]) % 2 == 0 && (onencnt[j] - onencnt[i - 1]) != 0) { // Incrementing count cnt++; } } } } // Output it cout << cnt << endl; return; } // Driver code int main() { string str = "2131"; char K = '1'; int size = str.size(); // Function call findevenone(str, size, K); return 0; } Java import java.util.*; public class Main { // To compute occurrence of K static void oneoccur(String arr, int n, int prefixSum[], char K) { // Initialize if (arr.charAt(0) == K) { prefixSum[0] = 1; } else { prefixSum[0] = 0; } for (int i = 1; i < n; i++) { // Adding previous element incremented by 1 if (arr.charAt(i) == K) { prefixSum[i] = prefixSum[i - 1] + 1; } else { prefixSum[i] = prefixSum[i - 1]; } } } // Function to find count of number of substrings static void findevenone(String s, int n, char K) { // Precomputing onecnt[] array int onencnt[] = new int[n]; // Precomputing the occurrence of K in onecnt[] oneoccur(s, n, onencnt, K); // Counter variable int cnt = 0; // For loop for traversing all the ranges for possible subarray for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { int onecnt = 0; // Counting the occurrence of '1' from i to j // The occurrence of '1' from i to j is pre[j]-pre[i-1] // if i==0 then this is pre[j] if (i == 0) { if (onencnt[j] % 2 == 0 && onencnt[j] != 0) { // Incrementing count cnt++; } } else { if ((onencnt[j] - onencnt[i - 1]) % 2 == 0 && (onencnt[j] - onencnt[i - 1]) != 0) { // Incrementing count cnt++; } } } } // Output it System.out.println(cnt); } // Driver code public static void main(String[] args) { String str = "2131"; char K = '1'; int size = str.length(); // Function call findevenone(str, size, K); } } Python3 # Python code of the above approach # To compute occurrence of K def oneoccur(arr, n, prefixSum, K): # Initialize if arr[0] == K: prefixSum[0] = 1 else: prefixSum[0] = 0 for i in range(1, n): # Adding previous element incremented by 1 if arr[i] == K: prefixSum[i] = prefixSum[i - 1] + 1 # Adding previous element else: prefixSum[i] = prefixSum[i - 1] # Function to find count number of substrings def findevenone(s, n, K): # Precomputing onecnt[] array onecnt = [0] * n # Precomputing the occurrence of K in onecnt[] oneoccur(s, n, onecnt, K) # Counter variable cnt = 0 # For loop for traversing all the ranges for possible subarray for i in range(n): for j in range(i, n): # Counting the occurrence of '1' from i to j is pre[j]-pre[i-1] # if i==0 then this is pre[j] if i == 0: if onecnt[j] % 2 == 0 and onecnt[j] != 0: # Incrementing count cnt += 1 else: if (onecnt[j] - onecnt[i - 1]) % 2 == 0 and (onecnt[j] - onecnt[i - 1]) != 0: # Incrementing count cnt += 1 # Output it print(cnt) # Driver code if __name__ == '__main__': str = "2131" K = '1' size = len(str) # Function call findevenone(str, size, K) # This code is contributed by Susobhan Akhuli C# using System; class Program { // To compute occurrence of K static void OneOccur(string arr, int n, int[] prefixSum, char K) { // Initialize if (arr[0] == K) { prefixSum[0] = 1; } else { prefixSum[0] = 0; } for (int i = 1; i < n; i++) { // Adding previous element incremented by 1 if (arr[i] == K) prefixSum[i] = prefixSum[i - 1] + 1; // Adding previous element else prefixSum[i] = prefixSum[i - 1]; } } // Function to find count of number of substrings static void FindEvenOne(string s, int n, char K) { // Precomputing onecnt[] array int[] oneCnt = new int[n]; // Precomputing the occurrence of K in onecnt[] OneOccur(s, n, oneCnt, K); // Counter variable int cnt = 0; // For loop for traversing all the ranges for // possible subarray for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { int oneCntInRange = 0; // Counting the occurrence of '1' // Occurrence of '1' from i to j is // oneCnt[j] - oneCnt[i - 1] If i == 0 then // this is oneCnt[j] if (i == 0) { if (oneCnt[j] % 2 == 0 && oneCnt[j] != 0) { // Incrementing count cnt++; } } else { if ((oneCnt[j] - oneCnt[i - 1]) % 2 == 0 && (oneCnt[j] - oneCnt[i - 1]) != 0) { // Incrementing count cnt++; } } } } // Output it Console.WriteLine(cnt); } // Driver code static void Main() { string str = "2131"; char K = '1'; int size = str.Length; // Function call FindEvenOne(str, size, K); } } JavaScript // JS code of the above approach: // To compute occurrence of K function oneoccur( arr, n, prefixSum, K) { // Initialize if (arr[0] == K) { prefixSum[0] = 1; } else { prefixSum[0] = 0; } for (let i = 1; i < n; i++) { // Adding previous element // incremented by 1 if (arr[i] == K) prefixSum[i] = prefixSum[i - 1] + 1; // Adding previous element else prefixSum[i] = prefixSum[i - 1]; } } // Function to find count // number of substrings function findevenone( s, n, K) { // Precomputing onecnt[] array let onencnt=new Array(n); // Precomputing the occurrence // of K in onecnt[] oneoccur(s, n, onencnt, K); // Counter variable let cnt = 0; // For loop for traversing all the // rages for possible subarray for (let i = 0; i < n; i++) { for (let j = i; j < n; j++) { let onecnt = 0; // Counting the occurrence of // '1' occurrence of '1' from // i to j is pre[j]-pre[i-1] // if i==0 then this is pre[j] if (i == 0) { if (onencnt[j] % 2 == 0 && onencnt[j] != 0) { // Incrementing count cnt++; } } else { if ((onencnt[j] - onencnt[i - 1]) % 2 == 0 && (onencnt[j] - onencnt[i - 1]) != 0) { // Incrementing count cnt++; } } } } // Output it console.log(cnt); return; } // Driver code let str = "2131"; let K = '1'; let size = str.length; // Function call findevenone(str, size, K); Output2 Time complexity: O(N2)Auxiliary Space: O(N) Comment More infoAdvertise with us Next Article Analysis of Algorithms K kg_codex Follow Improve Article Tags : Strings DSA Practice Tags : Strings Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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