Maximum distance between two unequal elements

Given an array arr[], the task is to find the maximum distance between two unequal elements of the given array.

Examples:

Input: arr[] = {3, 2, 1, 2, 1}
Output: 4
The maximum distance is between the first and the last element.



Input: arr[] = {3, 3, 1, 3, 3}
Output: 2

Naive approach: Traverse the whole array for every single element and find the longest distance of element which is unequal.

Efficient approach: By using the fact that the pair of unequal elements must include either first or last element or both, calculate the longest distance between unequal element by traversing the array either by fixing the first element or fixing the last element.

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 maximum distance
// between two unequal elements
int maxDistance(int arr[], int n)
{
    // If first and last elements are unequal
    // they are maximum distance apart
    if (arr[0] != arr[n - 1])
        return (n - 1);
  
    int i = n - 1;
  
    // Fix first element as one of the elements
    // and start traversing from the right
    while (i > 0) {
  
        // Break for the first unequal element
        if (arr[i] != arr[0])
            break;
        i--;
    }
  
    // To store the distance from the first element
    int distFirst = (i == 0) ? -1 : i;
  
    i = 0;
  
    // Fix last element as one of the elements
    // and start traversing from the left
    while (i < n - 1) {
  
        // Break for the first unequal element
        if (arr[i] != arr[n - 1])
            break;
        i++;
    }
  
    // To store the distance from the last element
    int distLast = (i == n - 1) ? -1 : (n - 1 - i);
  
    // Maximum possible distance
    int maxDist = max(distFirst, distLast);
    return maxDist;
}
  
// Driver code
int main()
{
    int arr[] = { 4, 4, 1, 2, 1, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << maxDistance(arr, n);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.io.*;
  
class GFG 
{
  
// Function to return the maximum distance
// between two unequal elements
static int maxDistance(int arr[], int n)
{
    // If first and last elements are unequal
    // they are maximum distance apart
    if (arr[0] != arr[n - 1])
        return (n - 1);
  
    int i = n - 1;
  
    // Fix first element as one of the elements
    // and start traversing from the right
    while (i > 0
    {
  
        // Break for the first unequal element
        if (arr[i] != arr[0])
            break;
        i--;
    }
  
    // To store the distance from the first element
    int distFirst = (i == 0) ? -1 : i;
  
    i = 0;
  
    // Fix last element as one of the elements
    // and start traversing from the left
    while (i < n - 1)
    {
  
        // Break for the first unequal element
        if (arr[i] != arr[n - 1])
            break;
        i++;
    }
  
    // To store the distance from the last element
    int distLast = (i == n - 1) ? -1 : (n - 1 - i);
  
    // Maximum possible distance
    int maxDist = Math.max(distFirst, distLast);
    return maxDist;
}
  
// Driver code
public static void main (String[] args) 
{
    int arr[] = { 4, 4, 1, 2, 1, 4 };
    int n = arr.length;
    System.out.print(maxDistance(arr, n));
}
}
  
// This code is contributed by anuj_67..

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Python3

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# Python implementation of the approach
  
# Function to return the maximum distance
# between two unequal elements
def maxDistance(arr, n):
      
    # If first and last elements are unequal
    # they are maximum distance apart
    if (arr[0] != arr[n - 1]):
        return (n - 1);
  
    i = n - 1;
  
    # Fix first element as one of the elements
    # and start traversing from the right
    while (i > 0):
  
        # Break for the first unequal element
        if (arr[i] != arr[0]):
            break;
        i-=1;
  
    # To store the distance from the first element
    distFirst = -1 if(i == 0) else i;
  
    i = 0;
  
    # Fix last element as one of the elements
    # and start traversing from the left
    while (i < n - 1):
  
        # Break for the first unequal element
        if (arr[i] != arr[n - 1]):
            break;
        i+=1;
  
    # To store the distance from the last element
    distLast = -1 if(i == n - 1) else (n - 1 - i);
  
    # Maximum possible distance
    maxDist = max(distFirst, distLast);
    return maxDist;
  
# Driver code
arr = [4, 4, 1, 2, 1, 4];
n = len(arr);
print(maxDistance(arr, n));
  
# This code has been contributed by 29AjayKumar

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
      
// Function to return the maximum distance
// between two unequal elements
static int maxDistance(int []arr, int n)
{
    // If first and last elements are unequal
    // they are maximum distance apart
    if (arr[0] != arr[n - 1])
        return (n - 1);
  
    int i = n - 1;
  
    // Fix first element as one of the elements
    // and start traversing from the right
    while (i > 0) 
    {
  
        // Break for the first unequal element
        if (arr[i] != arr[0])
            break;
        i--;
    }
  
    // To store the distance from the first element
    int distFirst = (i == 0) ? -1 : i;
  
    i = 0;
  
    // Fix last element as one of the elements
    // and start traversing from the left
    while (i < n - 1)
    {
  
        // Break for the first unequal element
        if (arr[i] != arr[n - 1])
            break;
        i++;
    }
  
    // To store the distance from the last element
    int distLast = (i == n - 1) ? -1 : (n - 1 - i);
  
    // Maximum possible distance
    int maxDist = Math.Max(distFirst, distLast);
    return maxDist;
}
  
// Driver code
static public void Main ()
{
    int []arr = { 4, 4, 1, 2, 1, 4 };
    int n = arr.Length;
    Console.WriteLine(maxDistance(arr, n));
}
}
  
// This code is contributed by Tushil..

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Output:

4


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Improved By : vt_m, jit_t, 29AjayKumar