Programa para encontrar la diferencia máxima entre el índice de dos números diferentes

Dada una array de N enteros. La tarea es encontrar la diferencia máxima entre el índice de dos números diferentes cualesquiera. Tenga en cuenta que hay un mínimo de dos números diferentes. 
Ejemplos: 
 

Entrada: a[] = {1, 2, 3, 2, 3} 
Salida: 4 
La diferencia entre 1 y los últimos 3.
Entrada: a[] = {1, 1, 3, 1, 1, 1} 
Salida: 3 
La diferencia entre el índice de 3 y el último 1. 
 

Enfoque: Inicialmente, verifique el primer número que es diferente de a[0] comenzando desde el final, almacene la diferencia de su índice como ind1 . Además, verifique el primer número que es diferente de a[n – 1] desde el principio, almacene la diferencia de su índice como ind2 . La respuesta será max(ind1, ind2) . 
A continuación se muestra la implementación del enfoque anterior: 
 

// 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 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 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# 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 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
?>

<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>

4

Complejidad temporal: O(n)
Espacio auxiliar: O(1)