Given a string str and an integer K, the task is to find the length of the longest sub-string S’ such that every character in S’ appears at least K times.
Input: s = “xyxyyz”, k = 2
Output: 5
“xyxyy” is the longest sub-string where
every character appears at least twice.
Input: s = “neveropen”, k = 2
Output: 2
Approach: Consider all the possible sub-strings and for every sub-string, calculate the frequency of each of its character and check whether all the characters appear at least K times. For all such valid sub-strings, find the largest length possible.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MAX 26// Function that return true if frequency// of all the present characters is at least kbool atLeastK(int freq[], int k){ for (int i = 0; i < MAX; i++) { // If the character is present and // its frequency is less than k if (freq[i] != 0 && freq[i] < k) return false; } return true;}// Function to set every frequency to zerovoid setZero(int freq[]){ for (int i = 0; i < MAX; i++) freq[i] = 0;}// Function to return the length of the longest// sub-string such that it contains every// character at least k timesint findlength(string str, int n, int k){ // To store the required maximum length int maxLen = 0; int freq[MAX]; // Starting index of the sub-string for (int i = 0; i < n; i++) { setZero(freq); // Ending index of the sub-string for (int j = i; j < n; j++) { freq[str[j] - 'a']++; // If the frequency of every character // of the current sub-string is at least k if (atLeastK(freq, k)) { // Update the maximum possible length maxLen = max(maxLen, j - i + 1); } } } return maxLen;}// Driver codeint main(){ string str = "xyxyyz"; int n = str.length(); int k = 2; cout << findlength(str, n, k); return 0;} |
Java
// Java Implementation of the above approachclass GFG {static final int MAX = 26;// Function that return true if frequency// of all the present characters is at least kstatic boolean atLeastK(int freq[], int k){ for (int i = 0; i < MAX; i++) { // If the character is present and // its frequency is less than k if (freq[i] != 0 && freq[i] < k) return false; } return true;}// Function to set every frequency to zerostatic void setZero(int freq[]){ for (int i = 0; i < MAX; i++) freq[i] = 0;}// Function to return the length of the longest// sub-string such that it contains every// character at least k timesstatic int findlength(String str, int n, int k){ // To store the required maximum length int maxLen = 0; int freq[] = new int[MAX]; // Starting index of the sub-string for (int i = 0; i < n; i++) { setZero(freq); // Ending index of the sub-string for (int j = i; j < n; j++) { freq[str.charAt(j) - 'a']++; // If the frequency of every character // of the current sub-string is at least k if (atLeastK(freq, k)) { // Update the maximum possible length maxLen = Math.max(maxLen, j - i + 1); } } } return maxLen;}// Driver codepublic static void main(String args[]) { String str = "xyxyyz"; int n = str.length(); int k = 2; System.out.println(findlength(str, n, k));}}// This code has been contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach MAX = 26# Function that return true if frequency # of all the present characters is at least k def atLeastK(freq, k) : for i in range(MAX) : # If the character is present and # its frequency is less than k if (freq[i] != 0 and freq[i] < k) : return False; return True; # Function to set every frequency to zero def setZero(freq) : for i in range(MAX) : freq[i] = 0; # Function to return the length of the longest # sub-string such that it contains every # character at least k times def findlength(string, n, k) : # To store the required maximum length maxLen = 0; freq = [0]*MAX; # Starting index of the sub-string for i in range(n) : setZero(freq); # Ending index of the sub-string for j in range(i,n) : freq[ord(string[j]) - ord('a')] += 1; # If the frequency of every character # of the current sub-string is at least k if (atLeastK(freq, k)) : # Update the maximum possible length maxLen = max(maxLen, j - i + 1); return maxLen; # Driver code if __name__ == "__main__" : string = "xyxyyz"; n = len(string); k = 2; print(findlength(string, n, k)); # This code is contributed by AnkitRai01 |
C#
// C# Implementation of the above approachusing System; class GFG {static int MAX = 26;// Function that return true if frequency// of all the present characters is at least kstatic Boolean atLeastK(int []freq, int k){ for (int i = 0; i < MAX; i++) { // If the character is present and // its frequency is less than k if (freq[i] != 0 && freq[i] < k) return false; } return true;}// Function to set every frequency to zerostatic void setZero(int []freq){ for (int i = 0; i < MAX; i++) freq[i] = 0;}// Function to return the length of the longest// sub-string such that it contains every// character at least k timesstatic int findlength(String str, int n, int k){ // To store the required maximum length int maxLen = 0; int []freq = new int[MAX]; // Starting index of the sub-string for (int i = 0; i < n; i++) { setZero(freq); // Ending index of the sub-string for (int j = i; j < n; j++) { freq[str[j] - 'a']++; // If the frequency of every character // of the current sub-string is at least k if (atLeastK(freq, k)) { // Update the maximum possible length maxLen = Math.Max(maxLen, j - i + 1); } } } return maxLen;}// Driver codepublic static void Main(String []args) { String str = "xyxyyz"; int n = str.Length; int k = 2; Console.WriteLine(findlength(str, n, k));}}// This code is contributed by Princi Singh |
Javascript
<script>// Javascript implementation of the approachvar MAX = 26;// Function that return true if frequency// of all the present characters is at least kfunction atLeastK(freq, k){ for (var i = 0; i < MAX; i++) { // If the character is present and // its frequency is less than k if (freq[i] != 0 && freq[i] < k) return false; } return true;}// Function to set every frequency to zerofunction setZero(freq){ for (var i = 0; i < MAX; i++) freq[i] = 0;}// Function to return the length of the longest// sub-string such that it contains every// character at least k timesfunction findlength(str, n, k){ // To store the required maximum length var maxLen = 0; var freq = Array(MAX).fill(0); // Starting index of the sub-string for (var i = 0; i < n; i++) { setZero(freq); // Ending index of the sub-string for (var j = i; j < n; j++) { freq[str[j].charCodeAt(0) - 'a'.charCodeAt(0)]++; // If the frequency of every character // of the current sub-string is at least k if (atLeastK(freq, k)) { // Update the maximum possible length maxLen = Math.max(maxLen, j - i + 1); } } } return maxLen;}// Driver codevar str = "xyxyyz";var n = str.length;var k = 2;document.write( findlength(str, n, k));</script> |
Output:
5
Time Complexity: O(n2)
Auxiliary Space: O(MAX), where MAX is the maximum number of unique characters in the string.
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