Maximum distance between two occurrences of same element in array

Given an array with repeated elements, the task is to find the maximum distance between two occurrences of an element.

Examples: 

Input : arr[] = {3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2}
Output: 10
// maximum distance for 2 is 11-1 = 10 
// maximum distance for 1 is 4-2 = 2 
// maximum distance for 4 is 10-5 = 5  

A simple solution for this problem is to, one by one, pick each element from the array and find its first and last occurrence in the array and take the difference between the first and last occurrence for maximum distance. The time complexity for this approach is O(n²).

Below is the implementations of the idea.

C++

// C++ program to find the maximum distance between
// two equal elements
#include <bits/stdc++.h>

// function to find the maximum distance
int maxDistance(int arr[], int n)
{

    // initialize the maxD to -1
    int maxD = -1;
    for (int i = 0; i < n - 1; i++)
        for (int j = i + 1; j < n; j++)

            // check if two elements are equal
            if (arr[i] == arr[j]) {
                // if yes then calculate the distance and
                // update maxD
                int temp = abs(j - i);
                maxD = maxD > temp ? maxD : temp;
            }
    // return maximum distance
    return maxD;
}
// Driver code
int main()
{
    int Arr[] = { 1, 2, 4, 1, 3, 4, 2, 5, 6, 5 };
    printf("Maximum distance between two occurrences of "
           "same element in array:%d",
           maxDistance(Arr, 10));
    return 0;
}

Java

import java.util.*;

public class Main {
    // function to find the maximum distance
    public static int maxDistance(int[] arr, int n)
    {
        // initialize the maxD to -1
        int maxD = -1;
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                // check if two elements are equal
                if (arr[i] == arr[j]) {
                    // if yes then calculate the distance
                    // and update maxD
                    int temp = Math.abs(j - i);
                    maxD = maxD > temp ? maxD : temp;
                }
            }
        }
        // return maximum distance
        return maxD;
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] Arr = { 1, 2, 4, 1, 3, 4, 2, 5, 6, 5 };
        System.out.printf(
            "Maximum distance between two occurrences of same element in array:%d",
            maxDistance(Arr, 10));
    }
}

Output

Maximum distance between two occurrences of same element in array:5

Time complexity : O(n2)
Auxiliary Space : O(n)

An efficient solution to this problem is to use hashing. The idea is to traverse the input array and store the index of the first occurrence in a hash map. For every other occurrence, find the difference between the index and the first index stored in the hash map. If the difference is more than the result so far, then update the result.

Below are implementations of the idea. The implementation uses unordered_map in. 

C++

// C++ program to find maximum distance between two
// same occurrences of a number.
#include<bits/stdc++.h>
using namespace std;

// Function to find maximum distance between equal elements
int maxDistance(int arr[], int n)
{
    // Used to store element to first index mapping
    unordered_map<int, int> mp;

    // Traverse elements and find maximum distance between
    // same occurrences with the help of map.
    int max_dist = 0;
    for (int i=0; i<n; i++)
    {
        // If this is first occurrence of element, insert its
        // index in map
        if (mp.find(arr[i]) == mp.end())
            mp[arr[i]] = i;

        // Else update max distance
        else
            max_dist = max(max_dist, i - mp[arr[i]]);
    }

    return max_dist;
}

// Driver program to run the case
int main()
{
    int arr[] = {3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << maxDistance(arr, n);
    return 0;
}

Java

// Java program to find maximum distance between two
// same occurrences of a number.
import java.io.*;
import java.util.*;

class GFG
{

    // Function to find maximum distance between equal elements
    static int maxDistance(int[] arr, int n)
    {
        // Used to store element to first index mapping
        HashMap<Integer, Integer> map = new HashMap<>();

        // Traverse elements and find maximum distance between
        // same occurrences with the help of map.
        int max_dist = 0;

        for (int i = 0; i < n; i++)
        {
            // If this is first occurrence of element, insert its
            // index in map
            if (!map.containsKey(arr[i]))
                map.put(arr[i], i);

            // Else update max distance
            else
                max_dist = Math.max(max_dist, i - map.get(arr[i]));
        }

        return max_dist;
}

// Driver code
public static void main(String args[])
{
    int[] arr = {3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2};
    int n = arr.length;
    System.out.println(maxDistance(arr, n));
}
}

// This code is contributed by rachana soma

Python3

# Python program to find maximum distance between two
# same occurrences of a number.

# Function to find maximum distance between equal elements
def maxDistance(arr, n):

    # Used to store element to first index mapping
    mp = {}

    # Traverse elements and find maximum distance between
    # same occurrences with the help of map.
    maxDict = 0
    for i in range(n):

        # If this is first occurrence of element, insert its
        # index in map
        if arr[i] not in mp.keys():
            mp[arr[i]] = i

        # Else update max distance
        else:
            maxDict = max(maxDict, i-mp[arr[i]])

    return maxDict

# Driver Program
if __name__=='__main__':
    arr = [3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2]
    n = len(arr)
    print (maxDistance(arr, n))

# Contributed By: Harshit Sidhwa

C#

// C# program to find maximum distance between two
// same occurrences of a number.

using System;
using System.Collections.Generic;

class GFG
{

    // Function to find maximum distance between equal elements
    static int maxDistance(int[] arr, int n)
    {
        // Used to store element to first index mapping
        Dictionary<int, int> map = new Dictionary<int, int>();

        // Traverse elements and find maximum distance between
        // same occurrences with the help of map.
        int max_dist = 0;

        for (int i = 0; i < n; i++)
        {
            // If this is first occurrence of element, insert its
            // index in map
            if (!map.ContainsKey(arr[i]))
                map.Add(arr[i], i);

            // Else update max distance
            else
                max_dist = Math.Max(max_dist, i - map[arr[i]]);
        }

        return max_dist;
}

// Driver code
public static void Main(String []args)
{
    int[] arr = {3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2};
    int n = arr.Length;
    Console.WriteLine(maxDistance(arr, n));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// Javascript program to find maximum distance between two
// same occurrences of a number.

    // Function to find maximum distance between equal elements
    function maxDistance(arr, n)
    {
        // Used to store element to first index mapping
        let map = new Map();

        // Traverse elements and find maximum distance between
        // same occurrences with the help of map.
        let max_dist = 0;

        for (let i = 0; i < n; i++)
        {
            // If this is first occurrence of element, insert its
            // index in map
            if (!map.has(arr[i]))
                map.set(arr[i], i);

            // Else update max distance
            else
                max_dist = Math.max(max_dist, i - map.get(arr[i]));
        }

        return max_dist;
}

// Driver program

    let arr = [3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2];
    let n = arr.length;
    document.write(maxDistance(arr, n));

</script>

Output

10

Time complexity : O(n) under the assumption that unordered_map’s search and insert operations take O(1) time.
Auxiliary Space : O(n).

This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.