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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the maximum distance// between two unequal elementsint 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 codeint main(){ int arr[] = { 4, 4, 1, 2, 1, 4 }; int n = sizeof(arr) / sizeof(arr[0]); cout << maxDistance(arr, n); return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG {// Function to return the maximum distance// between two unequal elementsstatic 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 codepublic 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.. |
Python3
# Python implementation of the approach# Function to return the maximum distance# between two unequal elementsdef 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 codearr = [4, 4, 1, 2, 1, 4];n = len(arr);print(maxDistance(arr, n));# This code has been contributed by 29AjayKumar |
C#
// C# implementation of the approachusing System;class GFG{ // Function to return the maximum distance// between two unequal elementsstatic 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 codestatic 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.. |
Javascript
<script>// Javascript implementation of the approach// Function to return the maximum distance// between two unequal elementsfunction maxDistance(arr, n){ // If first and last elements are unequal // they are maximum distance apart if (arr[0] != arr[n - 1]) return (n - 1); var 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 var 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 var distLast = (i == n - 1) ? -1 : (n - 1 - i); // Maximum possible distance var maxDist = Math.max(distFirst, distLast); return maxDist;}// Driver codevar arr = [ 4, 4, 1, 2, 1, 4 ];var n = arr.length;document.write( maxDistance(arr, n));</script> |
Output:
4
Time Complexity: O(n)
Auxiliary Space: O(1)
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