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Length of longest subarray having frequency of every element equal to K

Last Updated : 23 Jul, 2025
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Given an array arr[] consisting of N integers and an integer K, the task is to find the length of the longest subarray such that each element occurs K times.

Examples:

Input: arr[] = {3, 5, 2, 2, 4, 6, 4, 6, 5}, K = 2
Output: 8
Explanation: The subarray: {5, 2, 2, 4, 6, 4, 6, 5} of length 8 has frequency of every element as 2.

Input: arr[] = {5, 5, 5, 5}, K = 3
Output: 3
Explanation: The subarray: {5, 5, 5} of length 3 has frequency of every element as 3.

 

Approach: Follow the steps below to solve the problem:

  1. Generate all possible subarrays from the given array.
  2. For each subarray, initialize two unordered maps map1 and map2, to store the frequency of each element and store the count of elements with the respective frequencies.
  3. If for any subarray, the size of map2 is equal to 1 and the frequency of the current element is K, that means every element individually appears K times in the current subarray.
  4. Finally, return the maximum size of all such subarrays.

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
// required maximum subarray
int max_subarray_len(int arr[],
                     int N, int K)
{
    // Initialize answer to 0
    int ans = 0;

    // Generate all subarrays having i as the
    // starting index and j as the ending index
    for (int i = 0; i < N; i++) {

        // Stores frequency of subarray elements
        unordered_map<int, int> map1;

        // Stores subarray elements with
        // respective frequencies
        unordered_map<int, int> map2;

        for (int j = i; j < N; j++) {

            // Stores previous
            // frequency of arr[j]
            int prev_freq;

            // If arr[j] hasn't
            // occurred previously
            if (map1.find(arr[j])
                == map1.end()) {

                // Set frequency as 0
                prev_freq = 0;
            }

            else {

                // Update previous frequency
                prev_freq = map1[arr[j]];
            }

            // Increasing frequency
            // of arr[j] by 1
            map1[arr[j]]++;

            // If frequency is stored
            if (map2.find(prev_freq)
                != map2.end()) {

                // If previous frequency is 1
                if (map2[prev_freq] == 1) {

                    // Rove previous frequency
                    map2.erase(prev_freq);
                }
                else {

                    // Decrease previous frequency
                    map2[prev_freq]--;
                }
            }

            int new_freq = prev_freq + 1;

            // Increment new frequency
            map2[new_freq]++;

            // If updated frequency is equal to K
            if (map2.size() == 1
                && (new_freq) == K) {
                ans = max(
                    ans, j - i + 1);
            }
        }
    }

    // Return the maximum size
    // of the subarray
    return ans;
}

// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 3, 5, 2, 2, 4,
                  6, 4, 6, 5 };

    int K = 2;

    // Size of Array
    int N = sizeof(arr)
            / sizeof(arr[0]);

    // Function Call
    cout << max_subarray_len(
        arr, N, K);

    return 0;
}
Java 
// Java program for the above approach
import java.util.*;

class GFG
{

// Function to find the length of
// required maximum subarray
static int max_subarray_len(int arr[],
                     int N, int K)
{
    // Initialize answer to 0
    int ans = 0;

    // Generate all subarrays having i as the
    // starting index and j as the ending index
    for (int i = 0; i < N; i++) 
    {

        // Stores frequency of subarray elements
        HashMap<Integer,Integer> map1 = new HashMap<>();

        // Stores subarray elements with
        // respective frequencies
        HashMap<Integer,Integer> map2 = new HashMap<>();

        for (int j = i; j < N; j++) 
        {

            // Stores previous
            // frequency of arr[j]
            int prev_freq = 0;

            // If arr[j] hasn't
            // occurred previously
            if (!map1.containsKey(arr[j])) 
            {

                // Set frequency as 0
                prev_freq = 0;
            }

            else 
            {

                // Update previous frequency
                prev_freq = map1.get(arr[j]);
            }

            // Increasing frequency
            // of arr[j] by 1
            if(map1.containsKey(arr[j]))
            {
                map1.put(arr[j], map1.get(arr[j]) + 1);
            }
            else
            {
                map1.put(arr[j], 1);
            }

            // If frequency is stored
            if (map2.containsKey(prev_freq)) 
            {

                // If previous frequency is 1
                if (map2.get(prev_freq) == 1)
                {

                    // Rove previous frequency
                    map2.remove(prev_freq);
                }
                else 
                {

                    // Decrease previous frequency
                    map2.put(prev_freq, map2.get(prev_freq)-1);
                    
                    
                }
            }

            int new_freq = prev_freq + 1;

            // Increment new frequency
            if(map2.containsKey(new_freq))
            {
                map2.put(new_freq, map2.get(new_freq) + 1);
            }
            else{
                map2.put(new_freq, 1);
            }

            // If updated frequency is equal to K
            if (map2.size() == 1
                && (new_freq) == K) {
                ans = Math.max(
                    ans, j - i + 1);
            }
        }
    }

    // Return the maximum size
    // of the subarray
    return ans;
}

// Driver Code
public static void main(String[] args)
{
    // Given array arr[]
    int arr[] = { 3, 5, 2, 2, 4,
                  6, 4, 6, 5 };

    int K = 2;

    // Size of Array
    int N = arr.length;

    // Function Call
    System.out.print(max_subarray_len(
        arr, N, K));
}
}

// This code is contributed by Rajput-Ji
Python3 
# Python3 program for the above 
# approach
from collections import defaultdict

# Function to find the length of
# required maximum subarray
def max_subarray_len(arr, N, K):

    # Initialize answer to 0
    ans = 0

    # Generate all subarrays having
    # i as the starting index and j 
    # as the ending index
    for i in range(N):

        # Stores frequency of subarray 
        # elements
        map1 = defaultdict(int)

        # Stores subarray elements with
        # respective frequencies
        map2 = defaultdict(int)

        for j in range(i, N):

            # If arr[j] hasn't
            # occurred previously
            if (arr[j] not in map1):

                # Set frequency as 0
                prev_freq = 0

            else:

                # Update previous frequency
                prev_freq = map1[arr[j]]

            # Increasing frequency
            # of arr[j] by 1
            map1[arr[j]] += 1

            # If frequency is stored
            if prev_freq in map2:

                # If previous frequency is 1
                if (map2[prev_freq] == 1):

                    # Rove previous frequency
                    del map2[prev_freq]

                else:

                    # Decrease previous frequency
                    map2[prev_freq] -= 1

            new_freq = prev_freq + 1

            # Increment new frequency
            map2[new_freq] += 1

            # If updated frequency is equal 
            # to K
            if (len(map2) == 1 and 
               (new_freq) == K):
                ans = max(ans, j - i + 1)
                
    # Return the maximum size
    # of the subarray
    return ans

# Driver Code
if __name__ == "__main__":

    # Given array arr[]
    arr = [3, 5, 2, 2, 4,
           6, 4, 6, 5]

    K = 2

    # Size of Array
    N = len(arr)

    # Function Call
    print(max_subarray_len(
          arr, N, K))

# This code is contributed by Chitranayal
C# 
// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG
{

// Function to find the length of
// required maximum subarray
static int max_subarray_len(int []arr,
                     int N, int K)
{
    // Initialize answer to 0
    int ans = 0;

    // Generate all subarrays having i as the
    // starting index and j as the ending index
    for (int i = 0; i < N; i++) 
    {

        // Stores frequency of subarray elements
        Dictionary<int,int> map1 = new Dictionary<int,int>();

        // Stores subarray elements with
        // respective frequencies
        Dictionary<int,int> map2 = new Dictionary<int,int>();

        for (int j = i; j < N; j++) 
        {

            // Stores previous
            // frequency of arr[j]
            int prev_freq = 0;

            // If arr[j] hasn't
            // occurred previously
            if (!map1.ContainsKey(arr[j])) 
            {

                // Set frequency as 0
                prev_freq = 0;
            }

            else 
            {

                // Update previous frequency
                prev_freq = map1[arr[j]];
            }

            // Increasing frequency
            // of arr[j] by 1
            if(map1.ContainsKey(arr[j]))
            {
                map1[arr[j]] = map1[arr[j]] + 1;
            }
            else
            {
                map1.Add(arr[j], 1);
            }

            // If frequency is stored
            if (map2.ContainsKey(prev_freq)) 
            {

                // If previous frequency is 1
                if (map2[prev_freq] == 1)
                {

                    // Rove previous frequency
                    map2.Remove(prev_freq);
                }
                else 
                {

                    // Decrease previous frequency
                    map2[prev_freq]= map2[prev_freq]-1;
                    
                    
                }
            }

            int new_freq = prev_freq + 1;

            // Increment new frequency
            if(map2.ContainsKey(new_freq))
            {
                map2[new_freq] = map2[new_freq] + 1;
            }
            else{
                map2.Add(new_freq, 1);
            }

            // If updated frequency is equal to K
            if (map2.Count == 1
                && (new_freq) == K) {
                ans = Math.Max(
                    ans, j - i + 1);
            }
        }
    }

    // Return the maximum size
    // of the subarray
    return ans;
}

// Driver Code
public static void Main(String[] args)
{
    // Given array []arr
    int []arr = { 3, 5, 2, 2, 4,
                  6, 4, 6, 5 };

    int K = 2;

    // Size of Array
    int N = arr.Length;

    // Function Call
    Console.Write(max_subarray_len(
        arr, N, K));
}
}

// This code is contributed by 29AjayKumar
JavaScript 
<script>
// Javascript program for the above approach


// Function to find the length of
// required maximum subarray
function max_subarray_len(arr,N,K)
{
    // Initialize answer to 0
    let ans = 0;
 
    // Generate all subarrays having i as the
    // starting index and j as the ending index
    for (let i = 0; i < N; i++)
    {
 
        // Stores frequency of subarray elements
        let map1 = new Map();
 
        // Stores subarray elements with
        // respective frequencies
        let map2 = new Map();
 
        for (let j = i; j < N; j++)
        {
 
            // Stores previous
            // frequency of arr[j]
            let prev_freq = 0;
 
            // If arr[j] hasn't
            // occurred previously
            if (!map1.has(arr[j]))
            {
 
                // Set frequency as 0
                prev_freq = 0;
            }
 
            else
            {
 
                // Update previous frequency
                prev_freq = map1.get(arr[j]);
            }
 
            // Increasing frequency
            // of arr[j] by 1
            if(map1.has(arr[j]))
            {
                map1.set(arr[j], map1.get(arr[j]) + 1);
            }
            else
            {
                map1.set(arr[j], 1);
            }
 
            // If frequency is stored
            if (map2.has(prev_freq))
            {
 
                // If previous frequency is 1
                if (map2.get(prev_freq) == 1)
                {
 
                    // Rove previous frequency
                    map2.delete(prev_freq);
                }
                else
                {
 
                    // Decrease previous frequency
                    map2.set(prev_freq, map2.get(prev_freq)-1);
                     
                     
                }
            }
 
            let new_freq = prev_freq + 1;
 
            // Increment new frequency
            if(map2.has(new_freq))
            {
                map2.set(new_freq, map2.get(new_freq) + 1);
            }
            else
            {
                map2.set(new_freq, 1);
            }
 
            // If updated frequency is equal to K
            if (map2.size == 1
                && (new_freq) == K) 
            {
                ans = Math.max(
                    ans, j - i + 1);
            }
        }
    }
 
    // Return the maximum size
    // of the subarray
    return ans;
}

// Driver Code
let arr=[ 3, 5, 2, 2, 4,
                  6, 4, 6, 5 ];
                  
let K = 2;
// Size of Array
let N = arr.length;

// Function Call
document.write(max_subarray_len(
arr, N, K));

    

// This code is contributed by rag2127
</script>

Output: 
8

 


Time Complexity: O(N2)
Auxiliary Space: O(N)

Related Topic: Subarrays, Subsequences, and Subsets in Array


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Analysis of Algorithms

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