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

  • Difficulty Level : Easy
  • Last Updated : 21 May, 2021

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




// 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;
}

Java




// 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

Python3




# 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

C#




// 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

PHP




<?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
?>

Javascript




<script>
// javascript implementation of the approach   
// Function to return the maximum difference
    function findMaximumDiff(a , n) {
        var ind1 = 0;
 
        // Iteratively check from back
        for (i = n - 1; i > 0; i--) {
 
            // Different numbers
            if (a[0] != a[i]) {
                ind1 = i;
                break;
            }
        }
 
        var 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 Math.max(ind1, ind2);
    }
 
    // Driver code
     
        var a = [ 1, 2, 3, 2, 3 ];
        var n = a.length;
        document.write(findMaximumDiff(a, n));
 
// This code is contributed by todaysgaurav
</script>
Output: 
4

 




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