Longest subarray with all elements same

Given an array arr[] of size N, the task is to find the largest subarray which consists of all equal elements.

Examples:

Input: arr[] = {1, 1, 2, 2, 2, 3, 3};
Output: 3
Explanation:
Longest subarray with equal elements is {2, 2, 2}

Input: arr[] = {1, 1, 2, 2, 2, 3, 3, 3, 3};
Output: 4
Explanation:
Longest subarray with equal elements is {3, 3, 3, 3}

Approach: The idea is to traverse the array and check that the current element is equal to the previous element or not. If yes then increment the length of the longest subarray by 1. Otherwise, the current longest subarray is equal to 1. Also, update the longest subarray with equal elements at each step of the iteration.



Below is the implementation of the above approach:

C++

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// C++ program to find largest
// subarray with all equal elements.
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find largest sub
// array with all equal elements.
int subarray(int arr[], int n)
{
  
    int ans = 1, temp = 1;
  
    // Traverse the array
    for (int i = 1; i < n; i++) {
  
        // If elemet is same as
        // previous increment temp value
        if (arr[i] == arr[i - 1]) {
            ++temp;
        }
        else {
            ans = max(ans, temp);
            temp = 1;
        }
    }
    ans = max(ans, temp);
  
    // Return the required answer
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 2, 2, 1, 1,
                  2, 2, 2, 3, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Function call
    cout << subarray(arr, n);
  
    return 0;
}

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Java

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// Java program to find largest
// subarray with all equal elements.
import java.util.*;
  
class GFG{
  
// Function to find largest sub
// array with all equal elements.
static int subarray(int arr[], int n)
{
    int ans = 1, temp = 1;
  
    // Traverse the array
    for(int i = 1; i < n; i++)
    {
          
       // If elemet is same as
       // previous increment temp value
       if (arr[i] == arr[i - 1]) 
       {
           ++temp;
       }
       else
       {
           ans = Math.max(ans, temp);
           temp = 1;
       }
    }
    ans = Math.max(ans, temp);
  
    // Return the required answer
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 2, 1, 1, 2,
                  2, 2, 3, 3 };
    int n = arr.length;
  
    // Function call
    System.out.print(subarray(arr, n));
}
}
  
// This code is contributed by AbhiThakur

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Python3

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# Python3 program to find largest
# subarray with all equal elements.
  
# Function to find largest sub
# array with all equal elements.
def subarray(arr, n):
      
    ans, temp = 1, 1
  
    # Traverse the array
    for i in range(1, n):
  
        # If element is same as previous
        # increment temp value
        if arr[i] == arr[i - 1]:
            temp = temp + 1
        else:
            ans = max(ans, temp)
            temp = 1
                  
    ans = max(ans, temp)
  
    # Return the required answer
    return ans
  
# Driver code
arr = [ 2, 2, 1, 1, 2,
        2, 2, 3, 3 ]
n = len(arr)
  
# Function call
print(subarray(arr, n))
  
# This code is contributed by jrishabh99

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C#

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// C# program to find largest
// subarray with all equal elements.
using System;
class GFG{
  
// Function to find largest sub
// array with all equal elements.
static int subarray(int[] arr, int n)
{
    int ans = 1, temp = 1;
  
    // Traverse the array
    for(int i = 1; i < n; i++) 
    {
          
       // If elemet is same as
       // previous increment temp value
       if (arr[i] == arr[i - 1])
       {
           ++temp;
       }
       else 
       {
           ans = Math.Max(ans, temp);
           temp = 1;
       }
    }
    ans = Math.Max(ans, temp);
      
    // Return the required answer
    return ans;
}
  
// Driver code
public static void Main()
{
    int[] arr = { 2, 2, 1, 1, 2,
                  2, 2, 3, 3 };
    int n = arr.Length;
  
    // Function call
    Console.Write(subarray(arr, n));
}
}
  
// This code is contributed by Nidhi_biet

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Output:

3

Time Complexity: O(N)

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