Minimum index i such that all the elements from index i to given index are equal

Given an array arr[] of integers and an integer pos, the task is to find the minimum index i such that all the elements from index i to index pos are equal.

Examples:

Input: arr[] = {2, 1, 1, 1, 5, 2}, pos = 3
Output: 1
Elements in index range [1, 3] are all equal to 1.



Input: arr[] = {2, 1, 1, 1, 5, 2}, pos = 5
Output: 5

Approach: Starting from index pos – 1, traverse the array in reverse and for the first index i such that arr[i] != arr[pos] print i + 1 which is the required index.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the minimum required index
int minIndex(int arr[], int n, int pos)
{
    int num = arr[pos];
  
    // Start from arr[pos - 1]
    int i = pos - 1;
    while (i >= 0) {
        if (arr[i] != num)
            break;
        i--;
    }
  
    // All elements are equal
    // from arr[i + 1] to arr[pos]
    return i + 1;
}
  
// Driver code
int main()
{
    int arr[] = { 2, 1, 1, 1, 5, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int pos = 4;
  
    cout << minIndex(arr, n, pos);
    return 0;
}

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Java














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// Java implementation of the approach 


class GFG 


{ 


      


// Function to return the minimum required index 


static int minIndex(int arr[], int n, int pos) 


{ 


    int num = arr[pos]; 


  


    // Start from arr[pos - 1] 


    int i = pos - 1; 


    while (i >= 0)  


    { 


        if (arr[i] != num) 


            break; 


        i--; 


    } 


  


    // All elements are equal 


    // from arr[i + 1] to arr[pos] 


    return i + 1; 


} 


  


// Driver code 


public static void main(String[] args) 


{ 


    int arr[] = { 2, 1, 1, 1, 5, 2 }; 


    int n = arr.length; 


    int pos = 4; 


  


    System.out.println(minIndex(arr, n, pos)); 


} 


} 


  


// This code is contributed by Code_Mech. 



















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Python3














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# Python3 implementation of the approach 


  


# Function to return the minimum 


# required index 


def minIndex(arr, n, pos): 


  


    num = arr[pos] 


  


    # Start from arr[pos - 1] 


    i = pos - 1


    while (i >= 0): 


        if (arr[i] != num): 


            break


        i -= 1


      


    # All elements are equal 


    # from arr[i + 1] to arr[pos] 


    return i + 1


  


# Driver code 


arr = [2, 1, 1, 1, 5, 2 ] 


n = len(arr) 


pos = 4


  


print(minIndex(arr, n, pos)) 


  


# This code is contributed by  


# Mohit Kumar 29 



















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C#














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// C# implementation of the approach 


using System; 


class GFG 


{ 


      


// Function to return the minimum required index 


static int minIndex(int []arr, int n, int pos) 


{ 


    int num = arr[pos]; 


  


    // Start from arr[pos - 1] 


    int i = pos - 1; 


    while (i >= 0)  


    { 


        if (arr[i] != num) 


            break; 


        i--; 


    } 


  


    // All elements are equal 


    // from arr[i + 1] to arr[pos] 


    return i + 1; 


} 


  


// Driver code 


public static void Main() 


{ 


    int []arr = { 2, 1, 1, 1, 5, 2 }; 


    int n = arr.Length; 


    int pos = 4; 


  


    Console.WriteLine(minIndex(arr, n, pos)); 


} 


} 


  


// This code is contributed  


// by Akanksha Rai 



















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PHP














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<?php 


// PHP implementation of the approach  


  


// Function to return the minimum  


// required index  


function minIndex($arr, $n, $pos)  


{  


    $num = $arr[$pos];  


  


    // Start from arr[pos - 1]  


    $i = $pos - 1;  


    while ($i >= 0)  


    {  


        if ($arr[$i] != $num)  


            break;  


        $i--;  


    }  


  


    // All elements are equal  


    // from arr[i + 1] to arr[pos]  


    return $i + 1;  


}  


  


// Driver code  


$arr = array(2, 1, 1, 1, 5, 2 );  


$n = sizeof($arr);  


$pos = 4;  


  


echo minIndex($arr, $n, $pos);  


  


// This code is contributed by Ryuga 


?> 



















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Output:


4






          
          
          
          

            

    

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