Longest substring where all the characters appear at least K times | Set 3 - GeeksforGeeks

02 May, 2022
Given a string str and an integer K, the task is to find the length of the longest substring S such that every character in S appears at least K times.
Examples:
Input: str = “aabbba”, K = 3Output: 6Explanation: In substring “aabbba”, each character repeats at least k times and its length is 6.
Input: str = “ababacb”, K = 3Output: 0Explanation: There is no substring where each character repeats at least k times.
Naive Approach: The simplest approach to solve the given problem is discussed in Set 1. Time Complexity: O(N2)Auxiliary Space: O(26)
Divide and Conquer Approach: The divide and conquer approach for the given problem is discussed in the Set 2. Time Complexity: O(N*log N)Auxiliary Space: O(26)
Efficient Approach: The above two approaches can be optimized further by using Sliding Window technique. Follow the steps below to solve the problem:
Store the number of unique characters in the string str in a variable, say unique.
Initialize an array freq[] of size 26 with {0} and store the frequency of each character in this array.
Iterate over the range [1, unique] using the variable curr_unique. In each iteration, curr_unique is the maximum number of unique characters that must be present in the window.Reinitialize the array freq[] with {0} to store the frequency of each character in this window.Initialize start and end as 0, to store the starting and the ending point of the window respectively.Use two variables cnt, for storing the number of unique characters and countK, for storing the number of characters with at least K repeating characters in the current window.Now, iterate a loop while end < N, and perform the following:If the value of cnt is less than or equals to curr_unique, then expand the window from the right by adding a character to the end of the window. And increment its frequency by 1 in freq[].Otherwise, reduce the window from the left by removing a character from start and decrementing its frequency by 1 in freq[].At every step, update the values of cnt and countK.If the value of cnt is same as curr_unique and each character occurs at least K times, then update the overall maximum length and store it in ans.
Reinitialize the array freq[] with {0} to store the frequency of each character in this window.
Initialize start and end as 0, to store the starting and the ending point of the window respectively.
Use two variables cnt, for storing the number of unique characters and countK, for storing the number of characters with at least K repeating characters in the current window.
Now, iterate a loop while end < N, and perform the following:If the value of cnt is less than or equals to curr_unique, then expand the window from the right by adding a character to the end of the window. And increment its frequency by 1 in freq[].Otherwise, reduce the window from the left by removing a character from start and decrementing its frequency by 1 in freq[].At every step, update the values of cnt and countK.If the value of cnt is same as curr_unique and each character occurs at least K times, then update the overall maximum length and store it in ans.
If the value of cnt is less than or equals to curr_unique, then expand the window from the right by adding a character to the end of the window. And increment its frequency by 1 in freq[].
Otherwise, reduce the window from the left by removing a character from start and decrementing its frequency by 1 in freq[].
At every step, update the values of cnt and countK.
If the value of cnt is same as curr_unique and each character occurs at least K times, then update the overall maximum length and store it in ans.
After completing the above steps, print the value of ans as the result.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program for the above approach#include <bits/stdc++.h>using namespace std; // Function to find the length of// the longest substringint longestSubstring(string s, int k){    // Store the required answer    int ans = 0;     // Create a frequency map of the    // characters of the string    int freq[26] = { 0 };     // Store the length of the string    int n = s.size();     // Traverse the string, s    for (int i = 0; i < n; i++)         // Increment the frequency of        // the current character by 1        freq[s[i] - 'a']++;     // Stores count of unique characters    int unique = 0;     // Find the number of unique    // characters in string    for (int i = 0; i < 26; i++)        if (freq[i] != 0)            unique++;     // Iterate in range [1, unique]    for (int curr_unique = 1;         curr_unique <= unique;         curr_unique++) {         // Initialize frequency of all        // characters as 0        memset(freq, 0, sizeof(freq));         // Stores the start and the        // end of the window        int start = 0, end = 0;         // Stores the current number of        // unique characters and characters        // occuring atleast K times        int cnt = 0, count_k = 0;         while (end < n) {            if (cnt <= curr_unique) {                int ind = s[end] - 'a';                 // New unique character                if (freq[ind] == 0)                    cnt++;                 freq[ind]++;                 // New character which                // occurs atleast k times                if (freq[ind] == k)                    count_k++;                 // Expand window by                // incrementing end by 1                end++;            }            else {                int ind = s[start] - 'a';                 // Check if this character                // is present atleast k times                if (freq[ind] == k)                    count_k--;                 freq[ind]--;                 // Check if this character                // is unique                if (freq[ind] == 0)                    cnt--;                 // Shrink the window by                // incrementing start by 1                start++;            }             // If there are curr_unique            // characters and each character            // is atleast k times            if (cnt == curr_unique                && count_k == curr_unique)                 // Update the overall                // maximum length                ans = max(ans, end - start);        }    }     // return the answer    return ans;} // Driver Codeint main(){    string S = "aabbba";    int K = 3;    cout << longestSubstring(S, K) << endl;     return 0;}
// Java program for the above approachimport java.util.*;class GFG{ // Function to find the length of// the longest subStringstatic void longestSubString(char[] s, int k){       // Store the required answer    int ans = 0;     // Create a frequency map of the    // characters of the String    int freq[] = new int[26];     // Store the length of the String    int n = s.length;     // Traverse the String, s    for (int i = 0; i < n; i++)         // Increment the frequency of        // the current character by 1        freq[s[i] - 'a']++;     // Stores count of unique characters    int unique = 0;     // Find the number of unique    // characters in String    for (int i = 0; i < 26; i++)        if (freq[i] != 0)            unique++;     // Iterate in range [1, unique]    for (int curr_unique = 1;         curr_unique <= unique;         curr_unique++)    {         // Initialize frequency of all        // characters as 0        Arrays.fill(freq, 0);         // Stores the start and the        // end of the window        int start = 0, end = 0;         // Stores the current number of        // unique characters and characters        // occuring atleast K times        int cnt = 0, count_k = 0;        while (end < n)        {            if (cnt <= curr_unique)            {                int ind = s[end] - 'a';                 // New unique character                if (freq[ind] == 0)                    cnt++;                freq[ind]++;                 // New character which                // occurs atleast k times                if (freq[ind] == k)                    count_k++;                 // Expand window by                // incrementing end by 1                end++;            }            else            {                int ind = s[start] - 'a';                 // Check if this character                // is present atleast k times                if (freq[ind] == k)                    count_k--;                freq[ind]--;                 // Check if this character                // is unique                if (freq[ind] == 0)                    cnt--;                 // Shrink the window by                // incrementing start by 1                start++;            }             // If there are curr_unique            // characters and each character            // is atleast k times            if (cnt == curr_unique                && count_k == curr_unique)                 // Update the overall                // maximum length                ans = Math.max(ans, end - start);        }    }     // Print the answer    System.out.print(ans);} // Driver Codepublic static void main(String[] args){    String S = "aabbba";    int K = 3;    longestSubString(S.toCharArray(), K);}} // This code is contributed by 29AjayKumar
# Python3 program for the above approach # Function to find the length of# the longest substringdef longestSubstring(s, k) :     # Store the required answer    ans = 0     # Create a frequency map of the    # characters of the string    freq = [0]*26     # Store the length of the string    n = len(s)     # Traverse the string, s    for i in range(n) :         # Increment the frequency of        # the current character by 1        freq[ord(s[i]) - ord('a')] += 1     # Stores count of unique characters    unique = 0     # Find the number of unique    # characters in string    for i in range(26) :        if (freq[i] != 0) :            unique += 1     # Iterate in range [1, unique]    for curr_unique in range(1, unique + 1) :         # Initialize frequency of all        # characters as 0        Freq = [0]*26         # Stores the start and the        # end of the window        start, end = 0, 0         # Stores the current number of        # unique characters and characters        # occuring atleast K times        cnt, count_k = 0, 0         while (end < n) :            if (cnt <= curr_unique) :                ind = ord(s[end]) - ord('a')                 # New unique character                if (Freq[ind] == 0) :                    cnt += 1                 Freq[ind] += 1                 # New character which                # occurs atleast k times                if (Freq[ind] == k) :                    count_k += 1                 # Expand window by                # incrementing end by 1                end += 1                         else :                ind = ord(s[start]) - ord('a')                 # Check if this character                # is present atleast k times                if (Freq[ind] == k) :                    count_k -= 1                 Freq[ind] -= 1                 # Check if this character                # is unique                if (Freq[ind] == 0) :                    cnt -= 1                 # Shrink the window by                # incrementing start by 1                start += 1             # If there are curr_unique            # characters and each character            # is atleast k times            if ((cnt == curr_unique) and (count_k == curr_unique)) :                 # Update the overall                # maximum length                ans = max(ans, end - start)     # Print the answer    print(ans) S = "aabbba"K = 3longestSubstring(S, K) # This code is contributed by divyesh072019.
// C# program to implement// the above approachusing System; class GFG{   // Function to find the length of// the longest substringstatic void longestSubstring(string s, int k){       // Store the required answer    int ans = 0;     // Create a frequency map of the    // characters of the string    int[] freq = new int[26];     // Store the length of the string    int n = s.Length;     // Traverse the string, s    for (int i = 0; i < n; i++)         // Increment the frequency of        // the current character by 1        freq[s[i] - 'a']++;     // Stores count of unique characters    int unique = 0;     // Find the number of unique    // characters in string    for (int i = 0; i < 26; i++)        if (freq[i] != 0)            unique++;     // Iterate in range [1, unique]    for (int curr_unique = 1;         curr_unique <= unique;         curr_unique++)    {         // Initialize frequency of all        // characters as 0        for (int i = 0; i < freq.Length; i++)        {            freq[i] = 0;        }                 // Stores the start and the        // end of the window        int start = 0, end = 0;         // Stores the current number of        // unique characters and characters        // occuring atleast K times        int cnt = 0, count_k = 0;        while (end < n)        {            if (cnt <= curr_unique)            {                int ind = s[end] - 'a';                 // New unique character                if (freq[ind] == 0)                    cnt++;                freq[ind]++;                 // New character which                // occurs atleast k times                if (freq[ind] == k)                    count_k++;                 // Expand window by                // incrementing end by 1                end++;            }            else            {                int ind = s[start] - 'a';                 // Check if this character                // is present atleast k times                if (freq[ind] == k)                    count_k--;                freq[ind]--;                 // Check if this character                // is unique                if (freq[ind] == 0)                    cnt--;                 // Shrink the window by                // incrementing start by 1                start++;            }             // If there are curr_unique            // characters and each character            // is atleast k times            if (cnt == curr_unique                && count_k == curr_unique)                 // Update the overall                // maximum length                ans = Math.Max(ans, end - start);        }    }     // Print the answer    Console.Write(ans);} // Driver Codepublic static void Main(){   string S = "aabbba";    int K = 3;    longestSubstring(S, K);}} // This code is contributed by splevel62.
<script>     // Javascript program to implement    // the above approach         // Function to find the length of    // the longest substring    function longestSubstring(s, k)    {         // Store the required answer        let ans = 0;         // Create a frequency map of the        // characters of the string        let freq = new Array(26);        freq.fill(0);         // Store the length of the string        let n = s.length;         // Traverse the string, s        for (let i = 0; i < n; i++)             // Increment the frequency of            // the current character by 1            freq[s[i].charCodeAt() -               'a'.charCodeAt()]++;         // Stores count of unique characters        let unique = 0;         // Find the number of unique        // characters in string        for (let i = 0; i < 26; i++)            if (freq[i] != 0)                unique++;         // Iterate in range [1, unique]        for (let curr_unique = 1;             curr_unique <= unique;             curr_unique++)        {             // Initialize frequency of all            // characters as 0            for (let i = 0; i < freq.length; i++)            {                freq[i] = 0;            }             // Stores the start and the            // end of the window            let start = 0, end = 0;             // Stores the current number of            // unique characters and characters            // occuring atleast K times            let cnt = 0, count_k = 0;            while (end < n)            {                if (cnt <= curr_unique)                {                    let ind = s[end].charCodeAt() -                              'a'.charCodeAt();                     // New unique character                    if (freq[ind] == 0)                        cnt++;                    freq[ind]++;                     // New character which                    // occurs atleast k times                    if (freq[ind] == k)                        count_k++;                     // Expand window by                    // incrementing end by 1                    end++;                }                else                {                    let ind = s[start].charCodeAt() -                    'a'.charCodeAt();                     // Check if this character                    // is present atleast k times                    if (freq[ind] == k)                        count_k--;                    freq[ind]--;                     // Check if this character                    // is unique                    if (freq[ind] == 0)                        cnt--;                     // Shrink the window by                    // incrementing start by 1                    start++;                }                 // If there are curr_unique                // characters and each character                // is atleast k times                if (cnt == curr_unique                    && count_k == curr_unique)                     // Update the overall                    // maximum length                    ans = Math.max(ans, end - start);            }        }         // Print the answer        document.write(ans);    }         let S = "aabbba";    let K = 3;    longestSubstring(S, K); </script>
6

Time Complexity: O(N)Auxiliary Space: O(1) 
splevel62
29AjayKumar
divyesh072019
suresh07
sagar0719kumar
adityadas285
frequency-counting
sliding-window
substring
Hash
Searching
Strings
sliding-window
Searching
Hash
Strings
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