Program to find the maximum difference between the index of any two different numbers

Given an array of N integers. The task is to find the maximum difference between the index of any two different numbers. Note that there is a minimum of two different numbers.

Examples:

Input: a[] = {1, 2, 3, 2, 3}
Output: 4
The difference between 1 and last 3.



Input: a[] = {1, 1, 3, 1, 1, 1}
Output: 3
The difference between the index of 3 and last 1.

Approach: Initially, check the first number which is different from a[0] starting from the end, store the difference of their index as ind1. Also, check for the first number which is different from a[n – 1] from the beginning, store the difference of their index as ind2. The answer will be max(ind1, ind2).

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the maximum difference
int findMaximumDiff(int a[], int n)
{
    int ind1 = 0;
  
    // Iteratively check from back
    for (int i = n - 1; i > 0; i--) {
  
        // Different numbers
        if (a[0] != a[i]) {
            ind1 = i;
            break;
        }
    }
  
    int ind2 = 0;
  
    // Iteratively check from
    // the beginning
    for (int i = 0; i < n - 1; i++) {
  
        // Different numbers
        if (a[n - 1] != a[i]) {
            ind2 = (n - 1 - i);
            break;
        }
    }
  
    return max(ind1, ind2);
}
  
// Driver code
int main()
{
    int a[] = { 1, 2, 3, 2, 3 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << findMaximumDiff(a, n);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
class GFG
{
      
// Function to return the maximum difference
static int findMaximumDiff(int []a, int n)
{
    int ind1 = 0;
  
    // Iteratively check from back
    for (int i = n - 1; i > 0; i--) 
    {
  
        // Different numbers
        if (a[0] != a[i]) 
        {
            ind1 = i;
            break;
        }
    }
  
    int ind2 = 0;
  
    // Iteratively check from
    // the beginning
    for (int i = 0; i < n - 1; i++)
    {
  
        // Different numbers
        if (a[n - 1] != a[i]) 
        {
            ind2 = (n - 1 - i);
            break;
        }
    }
  
    return Math.max(ind1, ind2);
}
  
// Driver code
public static void main(String args[])
{
    int []a = { 1, 2, 3, 2, 3 };
    int n = a.length;
    System.out.println(findMaximumDiff(a, n));
}
}
  
// This code is contributed by Akanksha_Rai

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
  
# Function to return the maximum difference
def findMaximumDiff(a, n):
    ind1 = 0
  
    # Iteratively check from back
    for i in range(n - 1, -1, -1):
  
        # Different numbers
        if (a[0] != a[i]):
            ind1 = i
            break
  
    ind2 = 0
  
    # Iteratively check from
    # the beginning
    for i in range(n - 1):
  
        # Different numbers
        if (a[n - 1] != a[i]):
            ind2 = (n - 1 - i)
            break
  
    return max(ind1, ind2)
  
# Driver code
a = [1, 2, 3, 2, 3]
n = len(a)
print(findMaximumDiff(a, n))
  
# This code is contributed by mohit kumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
  
class GFG
{
      
// Function to return the maximum difference
static int findMaximumDiff(int []a, int n)
{
    int ind1 = 0;
  
    // Iteratively check from back
    for (int i = n - 1; i > 0; i--) 
    {
  
        // Different numbers
        if (a[0] != a[i]) 
        {
            ind1 = i;
            break;
        }
    }
  
    int ind2 = 0;
  
    // Iteratively check from
    // the beginning
    for (int i = 0; i < n - 1; i++)
    {
  
        // Different numbers
        if (a[n - 1] != a[i]) 
        {
            ind2 = (n - 1 - i);
            break;
        }
    }
  
    return Math.Max(ind1, ind2);
}
  
// Driver code
static void Main()
{
    int []a = { 1, 2, 3, 2, 3 };
    int n = a.Length;
    Console.WriteLine(findMaximumDiff(a, n));
}
}
  
// This code is contributed by mits

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of the approach 
  
// Function to return the maximum difference 
function findMaximumDiff($a, $n
    $ind1 = 0; 
  
    // Iteratively check from back 
    for ($i = $n - 1; $i > 0; $i--) 
    
  
        // Different numbers 
        if ($a[0] != $a[$i])
        
            $ind1 = $i
            break
        
    
  
    $ind2 = 0; 
  
    // Iteratively check from 
    // the beginning 
    for ($i = 0; $i < $n - 1; $i++) 
    
  
        // Different numbers 
        if ($a[$n - 1] != $a[$i]) 
        
            $ind2 = ($n - 1 - $i); 
            break
        
    
  
    return max($ind1, $ind2); 
  
// Driver code 
$a = array( 1, 2, 3, 2, 3 ); 
$n = count($a);
  
echo findMaximumDiff($a, $n); 
  
// This code is contributed by Ryuga
?>

chevron_right


Output:

4


My Personal Notes arrow_drop_up

Striver(underscore)79 at Codechef and codeforces D

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.