Max Distance Between Two Occurrences

Last Updated : 27 Dec, 2024

Given an array arr[], the task is to find the maximum distance between two occurrences of any element. If no element occurs twice, return 0.

Examples:

Input: arr = [1, 1, 2, 2, 2, 1]
Output: 5
Explanation: distance for 1 is: 5-0 = 5, distance for 2 is: 4-2 = 2, So max distance is 5.
Input : arr[] = [3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2]
Output: 10
Explanation : Max distance for 2 is 11-1 = 10, max distance for 1 is 4-2 = 2 and max distance for 4 is 10-5 = 5
Input: arr[] = [1, 2, 3, 6, 5, 4]
Output: 0
Explanation: No element has two occurrence, so maximum distance = 0.

Table of Content

[Naive Approach] Exploring all pairs - O(n^2) Time and O(1) Space

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.

C++

// C++ Program to find max distance between two occurrences
// in array by exploring all pairs

#include <iostream>
#include <vector>
using namespace std;

// function to find the maximum distance
int maxDistance(vector<int>& arr) {
    int res = 0;
    for (int i = 0; i < arr.size() - 1; i++) {
        for (int j = i + 1; j < arr.size(); j++) {
          
            // check if two elements are equal
            if (arr[i] == arr[j]) {
          
              // calculate the distance and update res
               res = max(res, j - i);
            }
        }
    }
    return res;
}

// Driver code
int main() {
    vector<int> arr = {1, 2, 4, 1, 3, 4, 2, 5, 6, 5};
    cout << maxDistance(arr) << endl;
    return 0;
}

// C Program to find max distance between two occurrences
// in array by exploring all pairs

#include <stdio.h>
#include <stdlib.h>

// function to find the maximum distance
int maxDistance(int arr[], int n) {
    int res = 0;
    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]) {
          
              // calculate the distance and update res
                res = (res > (j - i)) ? res : (j - i);
            }
        }
    }
    return res;
}

int main() {
    int arr[] = {1, 2, 4, 1, 3, 4, 2, 5, 6, 5};
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("%d\n", maxDistance(arr, n));
    return 0;
}

Java

// Java Program to find max distance between two occurrences
// in array by exploring all pairs

import java.util.Arrays;

class GfG {
  
    // function to find the maximum distance
    static int maxDistance(int[] arr) {
        int res = 0;
        for (int i = 0; i < arr.length - 1; i++) {
            for (int j = i + 1; j < arr.length; j++) {
              
                // check if two elements are equal
                if (arr[i] == arr[j]) {
                  
                    // calculate the distance and update res
                    res = Math.max(res, j - i);
                }
            }
        }
        return res;
    }

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

Python

# Python Program to find max distance between two occurrences
# in array by exploring all pairs

def maxDistance(arr):
    res = 0
    for i in range(len(arr) - 1):
        for j in range(i + 1, len(arr)):
          
            # check if two elements are equal
            if arr[i] == arr[j]:
          
                # calculate the distance and update res
                res = max(res, j - i)
    return res

if __name__ == "__main__":
    arr = [1, 2, 4, 1, 3, 4, 2, 5, 6, 5]
    print(maxDistance(arr))

// C# Program to find max distance between two occurrences
// in array by exploring all pairs

using System;

class GfG {
  
    // function to find the maximum distance
    static int maxDistance(int[] arr) {
        int res = 0;
        for (int i = 0; i < arr.Length - 1; i++) {
            for (int j = i + 1; j < arr.Length; j++) {
              
                // check if two elements are equal
                if (arr[i] == arr[j]) {
              
                    // calculate the distance and update res
                    res = Math.Max(res, j - i);
                }
            }
        }
        return res;
    }

    static void Main() {
        int[] arr = new int[]{1, 2, 4, 1, 3, 4, 2, 5, 6, 5};
        Console.WriteLine(maxDistance(arr));
    }
}

JavaScript

// JavaScript Program to find max distance between two occurrences
// in array by exploring all pairs

// function to find the maximum distance
function maxDistance(arr) {
    let res = 0;
    for (let i = 0; i < arr.length - 1; i++) {
        for (let j = i + 1; j < arr.length; j++) {
        
            // check if two elements are equal
            if (arr[i] === arr[j]) {
            
                // calculate the distance and update res
                res = Math.max(res, j - i);
            }
        }
    }
    return res;
}

// Driver code
const arr = [1, 2, 4, 1, 3, 4, 2, 5, 6, 5];
console.log(maxDistance(arr));

Output

[Expected Approach] Using Hash Map or Dictionary - O(n) Time and O(n) Space

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.

C++

// C++ Program to find max distance between two occurrences
// in array using hashing

#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;

int maxDistance(vector<int>& arr) {
  
    // Stores element to first index mapping
    unordered_map<int, int> mp; 
  
    int res = 0;
  
    for (int i = 0; i < arr.size(); i++) {
      
        // If this is the first occurrence of 
        // the element, store its index
        if (mp.find(arr[i]) == mp.end())
            mp[arr[i]] = i;
      
        // Else update max distance
        else
            res = max(res, i - mp[arr[i]]);
    }

    return res;
}

int main() {
    vector<int> arr = {1, 1, 2, 2, 2, 1};
    cout << maxDistance(arr);
    return 0;
}

Java

// Java Program to find max distance between two occurrences
// in array using hashing

import java.util.*;

class GfG {
    static int maxDistance(int[] arr) {
      
        // Stores element to first index mapping
        HashMap<Integer, Integer> mp = new HashMap<>(); 
        int res = 0;

        for (int i = 0; i < arr.length; i++) {
          
            // If this is the first occurrence of the 
            // element, store its index
            if (!mp.containsKey(arr[i]))
                mp.put(arr[i], i);
          
            // Else update max distance
            else
                res = Math.max(res, i - mp.get(arr[i]));
        }

        return res;
    }

    public static void main(String[] args) {
        int[] arr = {1, 1, 2, 2, 2, 1};
        System.out.println(maxDistance(arr));
    }
}

Python

# Python Program to find max distance between two occurrences
# in array using hashing

def maxDistance(arr):
  
    # Stores element to first index mapping
    mp = {}
    res = 0

    for i in range(len(arr)):
      
        # If this is the first occurrence of the
        # element, store its index
        if arr[i] not in mp:
            mp[arr[i]] = i
            
        # Else update max distance
        else:
            res = max(res, i - mp[arr[i]])

    return res

if __name__ == "__main__":
    arr = [1, 1, 2, 2, 2, 1]
    print(maxDistance(arr))

// C# Program to find max distance between two occurrences
// in array using hashing

using System;
using System.Collections.Generic;

class GfG {
    static int maxDistance(int[] arr) {
      
        // Stores element to first index mapping
        Dictionary<int, int> mp = new Dictionary<int, int>();
        int res = 0;

        for (int i = 0; i < arr.Length; i++) {
          
            // If this is the first occurrence of the 
            // element, store its index
            if (!mp.ContainsKey(arr[i])) {
                mp[arr[i]] = i;
            } else {
              
                // Else update max distance
                res = Math.Max(res, i - mp[arr[i]]);
            }
        }

        return res;
    }

    static void Main() {
        int[] arr = {1, 1, 2, 2, 2, 1};
        Console.WriteLine(maxDistance(arr));
    }
}

JavaScript

// JavaScript Program to find max distance between two 
// occurrences in array using hashing

function maxDistance(arr) {

    // Stores element to first index mapping
    const mp = {};
    let res = 0;

    for (let i = 0; i < arr.length; i++) {
    
        // If this is the first occurrence of the
        // element, store its index
        if (!(arr[i] in mp)) {
            mp[arr[i]] = i;
        } else {
        
            // Else update max distance
            res = Math.max(res, i - mp[arr[i]]);
        }
    }

    return res;
}

// Driver Code
const arr = [1, 1, 2, 2, 2, 1];
console.log(maxDistance(arr));

Output

Most frequent element in an array

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Practice Tags :

Max Distance Between Two Occurrences

[Naive Approach] Exploring all pairs - O(n^2) Time and O(1) Space

[Expected Approach] Using Hash Map or Dictionary - O(n) Time and O(n) Space

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