Maximum distance between two occurrences of same element in array

06 Jul, 2022
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).
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++
Java
Python3
C#
Javascript
// 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 elementsint 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 caseint 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 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 codepublic 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
# Python program to find maximum distance between two# same occurrences of a number. # Function to find maximum distance between equal elementsdef 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 Programif __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# 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 codepublic 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
<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>
10
Time complexity : O(n) under the assumption that unordered_map’s search and insert operations take O(1) time.Auxiliary Space : O(n).
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