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Largest subarray with frequency of all elements same

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  • Difficulty Level : Medium
  • Last Updated : 27 Sep, 2022
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Given an array arr[] of N integers, the task is to find the size of the largest subarray with frequency of all elements the same.

Examples: 

Input: arr[] = {1, 2, 2, 5, 6, 5, 6} 
Output:
Explanation: 
The subarray = {2, 2, 5, 6, 5, 6} has frequency of every element is 2.

Input: arr[] = {1, 1, 1, 1, 1} 
Output:
Explanation: 
The subarray = {1, 1, 1, 1, 1} has frequency of every element is 5.

Approach: The idea is to generate all possible subarrays and check for each subarray whether any subarray has frequency of all elements or not. Below are the steps:

  1. Generate all possible subarrays.
  2. For each subarray, take two maps, one map to stores the frequency of every element and second map stores number of elements with given frequency.
  3. If for any subarray, size of second map becomes equal to 1, that means every element have same frequency in the subarray.
  4. Return the maximum size of all such subarrays.

Below is the implementation of above approach:

C++




// C++ program for the above approach
 
#include <iostream>
#include <unordered_map>
using namespace std;
 
// Function to find maximum subarray size
int max_subarray_size(int N, int arr[])
{
    int ans = 0;
 
    // Generating all subarray
    // i -> starting index
    // j -> end index
    for (int i = 0; i < N; i++)
    {
        // Map 1 to hash frequency
        // of all elements in subarray
        unordered_map<int, int> map1;
 
        // Map 2 to hash frequency
        // of all frequencies of
        // elements
        unordered_map<int, int> map2;
 
        for (int j = i; j < N; j++)
        {
            // ele_count is the previous
            // frequency of arr[j]
            int ele_count;
 
            // Finding previous frequency of
            // arr[j] in map 1
            if (map1.find(arr[j]) == map1.end())
            {
                ele_count = 0;
            }
            else
            {
                ele_count = map1[arr[j]];
            }
 
            // Increasing frequency of arr[j]
            // by 1
            map1[arr[j]]++;
 
            // Check if previous frequency
            // is present in map 2
            if (map2.find(ele_count) != map2.end())
            {
                // Delete previous frequency
                // if hash is equal to 1
                if (map2[ele_count] == 1)
                {
                    map2.erase(ele_count);
                }
                else
                {
                    // Decrement the hash of
                    // previous frequency
                    map2[ele_count]--;
                }
            }
 
            // Incrementing hash of new
            // frequency in map 2
            map2[ele_count + 1]++;
 
            // Check if map2 size is 1
            // and updating answer
            if (map2.size() == 1)
                ans = max(ans, j - i + 1);
        }
    }
 
    // Return the maximum size of subarray
    return ans;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 2, 2, 5, 6, 5, 6 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << max_subarray_size(N, arr);
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
class GFG
{
 
  // Function to find maximum subarray size
  static int max_subarray_size(int N, int[] arr)
  {
    int ans = 0;
 
    // Generating all subarray
    // i -> starting index
    // j -> end index
    for(int i = 0; i < N; i++)
    {
 
      // Map 1 to hash frequency
      // of all elements in subarray
      HashMap<Integer,
      Integer> map1 = new HashMap<>();
 
      // Map 2 to hash frequency
      // of all frequencies of
      // elements
      HashMap<Integer,
      Integer> map2 = new HashMap<>();
 
      for(int j = i; j < N; j++)
      {
 
        // ele_count is the previous
        // frequency of arr[j]
        int ele_count;
 
        // Finding previous frequency of
        // arr[j] in map 1
        if (!map1.containsKey(arr[j]))
        {
          ele_count = 0;
        }
        else
        {
          ele_count = 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);
        }
 
        // Check if previous frequency
        // is present in map 2
        if (map2.containsKey(ele_count))
        {
 
          // Delete previous frequency
          // if hash is equal to 1
          if (map2.get(ele_count) == 1)
          {
            map2.remove(ele_count);
          }
          else
          {
 
            // Decrement the hash of
            // previous frequency
            map2.put(ele_count,map2.get(ele_count) - 1);
          }
        }
 
        // Incrementing hash of new
        // frequency in map 2
        if (map2.containsKey(ele_count + 1))
        {
          map2.put(ele_count + 1, map2.get(ele_count + 1) + 1);
        }
        else
        {
          map2.put(ele_count + 1, 1);
        }
 
        // Check if map2 size is 1
        // and updating answer
        if (map2.size() == 1)
          ans = Math.max(ans, j - i + 1);
      }
    }
 
    // Return the maximum size of subarray
    return ans;
  }
 
  // Driver Code
  public static void main(String []args)
  {
 
    // Given array arr[]
    int[] arr = { 1, 2, 2, 5, 6, 5, 6 };
    int N = arr.length;
 
    // Function Call
    System.out.println(max_subarray_size(N, arr));
  }
}
 
// This code is contributed by rutvik_56.

Python3




# Python3 program for the above approach
 
# Function to find maximum subarray size
def max_subarray_size(N, arr):
 
    ans = 0
 
    # Generating all subarray
    # i -> starting index
    # j -> end index
    for i in range(N):
         
        # Map 1 to hash frequency
        # of all elements in subarray
        map1 = {}
 
        # Map 2 to hash frequency
        # of all frequencies of
        # elements
        map2 = {}
 
        for j in range(i, N):
         
            # ele_count is the previous
            # frequency of arr[j]
 
            # Finding previous frequency of
            # arr[j] in map 1
            if (arr[j] not in map1):
                ele_count = 0
            else:
                ele_count = map1[arr[j]]
 
            # Increasing frequency of arr[j]
            # by 1
            if arr[j] in map1:
                map1[arr[j]] += 1
            else:
                map1[arr[j]] = 1
 
            # Check if previous frequency
            # is present in map 2
            if (ele_count in map2):
         
                # Delete previous frequency
                # if hash is equal to 1
                if (map2[ele_count] == 1):
                    del map2[ele_count]
                else:
                 
                    # Decrement the hash of
                    # previous frequency
                    map2[ele_count] -= 1
 
            # Incrementing hash of new
            # frequency in map 2
            if ele_count + 1 in map2:
                map2[ele_count + 1] += 1
            else:
                map2[ele_count + 1] = 1
 
            # Check if map2 size is 1
            # and updating answer
            if (len(map2) == 1):
                ans = max(ans, j - i + 1)
 
    # Return the maximum size of subarray
    return ans
 
# Driver Code
 
# Given array arr[]
arr = [ 1, 2, 2, 5, 6, 5, 6 ]
 
N = len(arr)
 
# Function Call
print(max_subarray_size(N, arr))
 
# This code is contributed by divyeshrabadiya07

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to find maximum subarray size
static int max_subarray_size(int N, int[] arr)
{
    int ans = 0;
     
    // Generating all subarray
    // i -> starting index
    // j -> end index
    for(int i = 0; i < N; i++)
    {
         
        // Map 1 to hash frequency
        // of all elements in subarray
        Dictionary<int,
                   int> map1 = new Dictionary<int,
                                              int>();
  
        // Map 2 to hash frequency
        // of all frequencies of
        // elements
        Dictionary<int,
                   int> map2 = new Dictionary<int,
                                              int>();
  
        for(int j = i; j < N; j++)
        {
             
            // ele_count is the previous
            // frequency of arr[j]
            int ele_count;
  
            // Finding previous frequency of
            // arr[j] in map 1
            if (!map1.ContainsKey(arr[j]))
            {
                ele_count = 0;
            }
            else
            {
                ele_count = map1[arr[j]];
            }
  
            // Increasing frequency of arr[j]
            // by 1
            if (map1.ContainsKey(arr[j]))
            {
                map1[arr[j]]++;
            }
            else
            {
                map1.Add(arr[j], 1);
            }
  
            // Check if previous frequency
            // is present in map 2
            if (map2.ContainsKey(ele_count))
            {
                 
                // Delete previous frequency
                // if hash is equal to 1
                if (map2[ele_count] == 1)
                {
                    map2.Remove(ele_count);
                }
                else
                {
                     
                    // Decrement the hash of
                    // previous frequency
                    map2[ele_count]--;
                }
            }
  
            // Incrementing hash of new
            // frequency in map 2
            if (map2.ContainsKey(ele_count + 1))
            {
                map2[ele_count + 1]++;
            }
            else
            {
                map2.Add(ele_count + 1, 1);
            }
  
            // Check if map2 size is 1
            // and updating answer
            if (map2.Count == 1)
                ans = Math.Max(ans, j - i + 1);
        }
    }
  
    // Return the maximum size of subarray
    return ans;
}
 
// Driver Code
static void Main()
{
     
    // Given array arr[]
    int[] arr = { 1, 2, 2, 5, 6, 5, 6 };
  
    int N = arr.Length;
  
    // Function Call
    Console.WriteLine(max_subarray_size(N, arr));
}
}
 
// This code is contributed by divyesh072019

Javascript




<script>
// JavaScript program for the above approach
 
// Function to find maximum subarray size
function max_subarray_size(N, arr)
{
    let ans = 0;
 
    // Generating all subarray
    // i -> starting index
    // j -> end index
    for (let i = 0; i < N; i++)
    {
        // Map 1 to hash frequency
        // of all elements in subarray
        let map1 = new Map();
 
        // Map 2 to hash frequency
        // of all frequencies of
        // elements
        let map2 = new Map();
 
        for (let j = i; j < N; j++)
        {
            // ele_count is the previous
            // frequency of arr[j]
            let ele_count;
 
            // Finding previous frequency of
            // arr[j] in map 1
            if (!map1.has(arr[j]))
            {
                ele_count = 0;
            }
            else
            {
                ele_count = map1.get(arr[j]);
            }
 
            // Increasing frequency of arr[j] by 1
            map1.set(arr[j],ele_count+1)
 
            // Check if previous frequency
            // is present in map 2
             
            if (map2.has(ele_count))
            {
                // Delete previous frequency
                // if hash is equal to 1
                if (map2.get(ele_count) == 1)
                {
                    map2.delete(ele_count);
                }
                else
                {
                    // Decrement the hash of
                    // previous frequency
                    map2.set(ele_count,map2.get(ele_count)-1)
                }
            }
 
            // Incrementing hash of new
            // frequency in map 2
            if(map2.get(ele_count+1) !== undefined){
                map2.set(ele_count+1,map2.get(ele_count+1)+1);
            }
            else map2.set(ele_count+1,1);
 
            // Check if map2 size is 1
            // and updating answer
            if (map2.size == 1){
                ans = Math.max(ans, j - i + 1);
            }
             
        }
    }
 
    // Return the maximum size of subarray
    return ans;
}
 
// Driver Code
 
// Given array arr
const arr = [ 1, 2, 2, 5, 6, 5, 6 ];
 
const N = arr.length;
 
// Function Call
document.write(max_subarray_size(N, arr));
 
// This code is contributed by shinjanpatra.
</script>

Output

6

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

Naive Approach: Simple solution is to generate all subarrays and check whether each element has same frequency or not.

C++14




#include <bits/stdc++.h>
using namespace std;
 
//global variable to store the answer
int ans=1;
 
//function to check equal for all equal frequencies
int checkFreq(int *arr,int r,int l)
{
    int i,count=1;
    //vector to store subarray
    vector<int>temp;
    //vector to store all frequencies of this subarray
    vector<int>freq;
     
    freq.clear();
    temp.clear();
     
    //insert all subarray elements
    for(i=r;i<=l;i++)
    temp.push_back(arr[i]);
     
    sort(temp.begin(),temp.end());
     
    //counting equal consecutive elements to store frequencies
    for(i=0;i<temp.size();i++)
    {
        if(temp[i]==temp[i+1])
        count++;
        else{
            freq.push_back(count);
            count=1;
        }
    }
    //checking for all equal elements
    for(i=1;i<freq.size();i++)
    {
        if(freq[0]!=freq[i])
        return -1;
    }
    return temp.size();
}
 
//code to generate all subarrays in n^2
void generateSubArrays(int *arr,int start,int end,int len)
{
    if(end==len)
    return;
    else if(start>end)
    generateSubArrays(arr,0,end+1,len);
    else{
        ans=max(ans,checkFreq(arr,start,end));
        generateSubArrays(arr,start+1,end,len);
    }
}
 
 
//drivers code
int main()
{
    int arr[]={ 1, 10, 5, 4, 4, 4, 10 };
    int n=sizeof(arr)/sizeof(arr[0]);
    generateSubArrays(arr,0,0,n);
    cout<<ans;
    return 0;
}

Java




/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
 
class GFG
{
   
// Java code for the same approach
 
// global variable to store the answer
static int ans = 1;
 
// function to check equal for all equal frequencies
static int checkFreq(int arr[], int r, int l)
{
    int i, count = 1;
   
    // vector to store subarray
    ArrayList<Integer> temp = new ArrayList<>();
   
    // vector to store all frequencies of this subarray
    ArrayList<Integer> freq = new ArrayList<>();
     
     
    // insert all subarray elements
    for(i = r; i <= l; i++)
        temp.add(arr[i]);
     
    Collections.sort(temp);
     
    // counting equal consecutive elements to store frequencies
    for(i = 0; i < temp.size()-1; i++)
    {
        if(temp.get(i) == temp.get(i+1))
        count++;
        else{
            freq.add(count);
            count=1;
        }
    }
    // checking for all equal elements
    for(i = 1; i < freq.size(); i++)
    {
        if(freq.get(0) != freq.get(i))
        return -1;
    }
    return temp.size();
}
 
// code to generate all subarrays in n^2
static void generateSubArrays(int arr[], int start, int end, int len)
{
    if(end == len)
        return;
    else if(start > end)
        generateSubArrays(arr, 0, end + 1, len);
    else{
        ans = Math.max(ans, checkFreq(arr, start, end));
        generateSubArrays(arr, start + 1, end, len);
    }
}
 
// Driver Code
public static void main(String args[])
{
    int arr[] = { 1, 10, 5, 4, 4, 4, 10 };
    int n = arr.length;
    generateSubArrays(arr,0,0,n);
    System.out.println(ans);
}
}
 
// This code is contributed by shinjanpatra

Python3




# Python code for the approach
 
# global variable to store the answer
from pickle import GLOBAL
 
ans = 1
 
# function to check equal for all equal frequencies
def checkFreq(arr, r, l):
 
    i, count = 1, 1
     
    # vector to store subarray
    temp = []
     
    # vector to store all frequencies of this subarray
    freq = []
     
    freq.clear()
    temp.clear()
     
    # insert all subarray elements
    for i in range(r, l + 1):
        temp.append(arr[i])
     
    temp.sort()
     
    # counting equal consecutive elements to store frequencies
    for i in range(len(temp) - 1):
 
        if(temp[i] == temp[i + 1]):
            count += 1
        else:
            freq.append(count)
            count = 1
 
    # checking for all equal elements
    for i in range(1, len(freq)):
     
        if(freq[0] != freq[i]):
            return -1
 
    return len(temp)
 
# code to generate all subarrays in n^2
def generateSubArrays(arr, start, end, Len):
 
    global ans
 
    if(end == Len):
        return
    elif(start > end):
        generateSubArrays(arr, 0, end + 1, Len)
    else:
        ans = max(ans,checkFreq(arr, start, end))
        generateSubArrays(arr, start + 1, end, Len)
     
# drivers code
arr = [ 1, 10, 5, 4, 4, 4, 10 ]
n = len(arr)
generateSubArrays(arr, 0, 0, n)
print(ans)
 
# This code is contributed by shinjanpatra

C#




using System;
using System.Collections.Generic;
 
public static class GFG {
 
  // global variable to store the answer
  public static int ans = 1;
 
  // function to check equal for all equal frequencies
  public static int checkFreq(int[] arr, int r, int l)
  {
    int i, count = 1;
 
    // List to store subarray
    List<int> temp = new List<int>();
 
    // List to store all frequencies of this subarray
    List<int> freq = new List<int>();
 
 
    // insert all subarray elements
    for (i = r; i <= l; i++)
      temp.Add(arr[i]);
 
    temp.Sort();
 
    // counting equal consecutive elements to store
    // frequencies
    for (i = 0; i < temp.Count - 1; i++) {
      if (temp[i] == temp[i + 1])
        count++;
      else {
        freq.Add(count);
        count = 1;
      }
    }
    // checking for all equal elements
    for (i = 1; i < freq.Count; i++) {
      if (freq[0] != freq[i])
        return -1;
    }
    return temp.Count;
  }
 
  // code to generate all subarrays in n^2
  public static void generateSubArrays(int[] arr,
                                       int start, int end,
                                       int len)
  {
    if (end == len)
      return;
    else if (start > end)
      generateSubArrays(arr, 0, end + 1, len);
    else {
      ans = Math.Max(ans, checkFreq(arr, start, end));
      generateSubArrays(arr, start + 1, end, len);
    }
  }
 
  static public void Main()
  {
    int[] arr = { 1, 10, 5, 4, 4, 4, 10 };
    int n = arr.Length;
    generateSubArrays(arr, 0, 0, n);
    Console.WriteLine(ans);
  }
}
 
// This code is contributed by akashish__

Javascript




<script>
 
// JavaScript code for the same approach
 
//global variable to store the answer
let ans=1;
 
//function to check equal for all equal frequencies
function checkFreq(arr,r,l)
{
    let i,count=1;
    //vector to store subarray
    let temp = [];
    //vector to store all frequencies of this subarray
    let freq = [];
     
     
    //insert all subarray elements
    for(i=r;i<=l;i++)
        temp.push(arr[i]);
     
    temp.sort();
     
    //counting equal consecutive elements to store frequencies
    for(i=0;i<temp.length;i++)
    {
        if(temp[i]==temp[i+1])
        count++;
        else{
            freq.push(count);
            count=1;
        }
    }
    //checking for all equal elements
    for(i=1;i<freq.length;i++)
    {
        if(freq[0]!=freq[i])
        return -1;
    }
    return temp.length;
}
 
//code to generate all subarrays in n^2
function generateSubArrays(arr,start,end,len)
{
    if(end==len)
        return;
    else if(start>end)
        generateSubArrays(arr,0,end+1,len);
    else{
        ans=Math.max(ans,checkFreq(arr,start,end));
        generateSubArrays(arr,start+1,end,len);
    }
}
 
 
//drivers code
 
 
let arr = [ 1, 10, 5, 4, 4, 4, 10 ];
let n = arr.length;
generateSubArrays(arr,0,0,n);
console.log(ans);
 
// code is contributed by shinjanpatra
 
</script>

Output:

4

Time Complexity: O(n^3 log n)

Auxiliary Space: O(n)

Efficient solution: Another efficient way is to store frequencies of all elements in a map consecutively and check its frequency each time with starting of the map.

C++




#include <bits/stdc++.h>
using namespace std;
 
int longestSubArray(int a[],int n){
    int i;
    //minimum subarray will always be 1
int ans = 1;
 
for(int i = 0; i < n; i++) {
    //map to contain all occurrences
map < int, int > mp;
 
for(int j = i; j < n; j++) {
    //storing frequencies at key a[j]
mp[a[j]] = mp[a[j]] + 1;
//checker for each subarrays
bool ok = true;
 
//count for each frequency
int count = mp.begin() -> second;
 
//traversing current map
for(map < int, int > :: iterator it = mp.begin(); it!= mp.end(); ++it) {
 
if(count != it -> second) {
ok = false;
break;
}
}
if (ok) {
    //storing maximum value
ans = max(ans, j - i + 1);
}
}
}
return ans;
}
 
//drivers code
int main()
{
    int arr[]={1,2,8,8,4,4};
    int n=sizeof(arr)/sizeof(arr[0]);
    cout<<longestSubArray(arr,n);
}

Java




/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
 
class GFG {
  static int longestSubArray(int a[],int n){
    int i;
 
    // minimum subarray will always be 1
    int ans = 1;
 
    for(i = 0; i < n; i++)
    {
 
      // map to contain all occurrences
      TreeMap <Integer, Integer> mp = new TreeMap<>();
 
      for(int j = i; j < n; j++)
      {
 
        // storing frequencies at key a[j]
        if(mp.containsKey(a[j])){
          mp.put(a[j],mp.get(a[j])+1);
        }
        else mp.put(a[j], 1);
 
        // checker for each subarrays
        Boolean ok = true;
 
        // count for each frequency
        int count = mp.firstEntry().getValue();
 
        // traversing current map
        for(Map.Entry mapEl : mp.entrySet()) {
 
          if(count != (int)mapEl.getValue()) {
            ok = false;
            break;
          }
        }
        if (ok)
        {
 
          // storing maximum value
          ans = Math.max(ans, j - i + 1);
        }
      }
    }
    return ans;
  }
 
  /* Driver program to test above function*/
  public static void main(String args[])
  {
    // Input
    int arr[] = {1,2,8,8,4,4};
    int n = arr.length;
    System.out.println(longestSubArray(arr, n));
  }
}
 
// This code is contributed by shinjanpatra

Output:

4

Time Complexity: O(n^2 )

Auxiliary Space: O(n)

Related Topic: Subarrays, Subsequences, and Subsets in Array


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