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

  • Difficulty Level : Hard
  • Last Updated : 27 Aug, 2021

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:

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Input: str = “aabbba”, K = 3
Output: 6
Explanation: In substring “aabbba”, each character repeats at least k times and its length is 6.



Input: str = “ababacb”, K = 3
Output: 0
Explanation: 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.
  • After completing the above steps, print the value of ans as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the length of
// the longest substring
int 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);
        }
    }
 
    // Print the answer
    cout << ans;
}
 
// Driver Code
int main()
{
    string S = "aabbba";
    int K = 3;
    longestSubstring(S, K);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to find the length of
// the longest subString
static 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 Code
public static void main(String[] args)
{
    String S = "aabbba";
    int K = 3;
    longestSubString(S.toCharArray(), K);
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 program for the above approach
 
# Function to find the length of
# the longest substring
def 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 = 3
longestSubstring(S, K)
 
# This code is contributed by divyesh072019.

C#




// C# program to implement
// the above approach
using System;
 
class GFG
{
   
// Function to find the length of
// the longest substring
static 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 Code
public static void Main()
{
   string S = "aabbba";
    int K = 3;
    longestSubstring(S, K);
}
}
 
// This code is contributed by splevel62.

Javascript




<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>

 
 

Output: 
6

 

Time Complexity: O(N)
Auxiliary Space: O(1) 




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