Minimum distance between any two equal elements in an Array

15 Jun, 2021
Given an array arr, the task is to find the minimum distance between any two same elements in the array. If no such element is found, return -1.
Examples:  
Input: arr = {1, 2, 3, 2, 1} Output: 2 Explanation: There are two matching pairs of values: 1 and 2 in this array. Minimum Distance between two 1’s = 4 Minimum Distance between two 2’s = 2 Therefore, Minimum distance between any two equal elements in the Array = 2
Input: arr = {3, 5, 4, 6, 5, 3} Output: 3 Explanation: There are two matching pairs of values: 3 and 5 in this array. Minimum Distance between two 3’s = 5 Minimum Distance between two 5’s = 3 Therefore, Minimum distance between any two equal elements in the Array = 3  
Naive Approach: The simplest approach is using two nested for loops to form each and every combination. If the elements are equal, find the minimum distance. Time complexity: O(N2)
Efficient Approach: An efficient approach for this problem is to use map to store array elements as a key and their index as the values. 
Below is the step by step algorithm:  
Traverse the array one by one.Check if this element is in the map or not. If the map does not contain this element, insert it as (element, current index) pair.If the array element present in the map, fetch the previous index of this element from the map.Find the difference between the previous index and the current indexCompare each difference and find the minimum distance.If no such element found, return -1.
Traverse the array one by one.
Check if this element is in the map or not. If the map does not contain this element, insert it as (element, current index) pair.If the array element present in the map, fetch the previous index of this element from the map.
If the map does not contain this element, insert it as (element, current index) pair.
If the array element present in the map, fetch the previous index of this element from the map.
Find the difference between the previous index and the current index
Compare each difference and find the minimum distance.
If no such element found, return -1.
Below is the implementation of the above approach.  
C++
Java
Python3
C#
Javascript
// C++ program to find the minimum distance// between two occurrences of the same element#include<bits/stdc++.h>using namespace std; // Function to find the minimum// distance between the same elementsint minimumDistance(int a[], int n){     // Create a HashMap to    // store (key, values) pair.    map<int,int> hmap;     int minDistance = INT_MAX;     // Initialize previousIndex    // and currentIndex as 0    int previousIndex = 0, currentIndex = 0;     // Traverse the array and    // find the minimum distance    // between the same elements with map     for (int i = 0; i < n; i++) {         if (hmap.find(a[i])!=hmap.end()) {            currentIndex = i;             // Fetch the previous index from map.            previousIndex = hmap[a[i]];             // Find the minimum distance.            minDistance = min((currentIndex -                        previousIndex),minDistance);        }         // Update the map.        hmap[a[i]] = i;    }     // return minimum distance,    // if no such elements found, return -1    return (minDistance == INT_MAX ? -1 : minDistance);} // Driver codeint main(){     // Test Case 1:    int a1[] = { 1, 2, 3, 2, 1 };    int n = sizeof(a1)/sizeof(a1[0]);     cout << minimumDistance(a1, n) << endl;     // Test Case 2:    int a2[] = { 3, 5, 4, 6, 5, 3 };    n = sizeof(a2)/sizeof(a2[0]);    cout << minimumDistance(a2, n) << endl;     // Test Case 3:    int a3[] = { 1, 2, 1, 4, 1 };    n = sizeof(a3)/sizeof(a3[0]);     cout << minimumDistance(a3, n) << endl;} // This code is contributed by Sanjit_Prasad
// Java program to find the minimum distance// between two occurrences of the same element import java.util.*;import java.math.*; class GFG {     // Function to find the minimum    // distance between the same elements    static int minimumDistance(int[] a)    {         // Create a HashMap to        // store (key, values) pair.        HashMap<Integer, Integer> hmap            = new HashMap<>();        int minDistance = Integer.MAX_VALUE;         // Initialize previousIndex        // and currentIndex as 0        int previousIndex = 0, currentIndex = 0;         // Traverse the array and        // find the minimum distance        // between the same elements with map        for (int i = 0; i < a.length; i++) {             if (hmap.containsKey(a[i])) {                currentIndex = i;                 // Fetch the previous index from map.                previousIndex = hmap.get(a[i]);                 // Find the minimum distance.                minDistance                    = Math.min(                        (currentIndex - previousIndex),                        minDistance);            }             // Update the map.            hmap.put(a[i], i);        }         // return minimum distance,        // if no such elements found, return -1        return (            minDistance == Integer.MAX_VALUE                ? -1                : minDistance);    }     // Driver code    public static void main(String args[])    {         // Test Case 1:        int a1[] = { 1, 2, 3, 2, 1 };        System.out.println(minimumDistance(a1));         // Test Case 2:        int a2[] = { 3, 5, 4, 6, 5, 3 };        System.out.println(minimumDistance(a2));         // Test Case 3:        int a3[] = { 1, 2, 1, 4, 1 };        System.out.println(minimumDistance(a3));    }}
# Python3 program to find the minimum distance# between two occurrences of the same element # Function to find the minimum# distance between the same elementsdef minimumDistance(a):     # Create a HashMap to    # store (key, values) pair.    hmap = dict()    minDistance = 10**9     # Initialize previousIndex    # and currentIndex as 0    previousIndex = 0    currentIndex = 0     # Traverse the array and    # find the minimum distance    # between the same elements with map    for i in range(len(a)):         if a[i] in hmap:            currentIndex = i             # Fetch the previous index from map.            previousIndex = hmap[a[i]]             # Find the minimum distance.            minDistance = min((currentIndex -                        previousIndex), minDistance)         # Update the map.        hmap[a[i]] = i     # return minimum distance,    # if no such elements found, return -1    if minDistance == 10**9:        return -1    return minDistance # Driver codeif __name__ == '__main__':         # Test Case 1:    a1 = [1, 2, 3, 2, 1 ]    print(minimumDistance(a1))     # Test Case 2:    a2 = [3, 5, 4, 6, 5,3]    print(minimumDistance(a2))     # Test Case 3:    a3 = [1, 2, 1, 4, 1 ]    print(minimumDistance(a3))     # This code is contributed by mohit kumar 29   
// C# program to find the minimum distance// between two occurrences of the same elementusing System;using System.Collections.Generic; class GFG{     // Function to find the minimum// distance between the same elementsstatic int minimumDistance(int[] a){         // Create a HashMap to    // store (key, values) pair.    Dictionary<int,               int> hmap = new Dictionary<int,                                          int>();    int minDistance = Int32.MaxValue;     // Initialize previousIndex    // and currentIndex as 0    int previousIndex = 0, currentIndex = 0;     // Traverse the array and    // find the minimum distance    // between the same elements with map    for(int i = 0; i < a.Length; i++)    {        if (hmap.ContainsKey(a[i]))        {            currentIndex = i;                         // Fetch the previous index from map.            previousIndex = hmap[a[i]];             // Find the minimum distance.            minDistance = Math.Min((currentIndex -                                    previousIndex),                                    minDistance);        }         // Update the map.        if (!hmap.ContainsKey(a[i]))            hmap.Add(a[i], i);        else            hmap[a[i]] = i;    }     // Return minimum distance,    // if no such elements found, return -1    return(minDistance == Int32.MaxValue ?                    -1 : minDistance);} // Driver codestatic public void Main(){         // Test Case 1:    int[] a1 = { 1, 2, 3, 2, 1 };    Console.WriteLine(minimumDistance(a1));         // Test Case 2:    int[] a2 = { 3, 5, 4, 6, 5, 3 };    Console.WriteLine(minimumDistance(a2));         // Test Case 3:    int[] a3 = { 1, 2, 1, 4, 1 };    Console.WriteLine(minimumDistance(a3));}} // This code is contributed by unknown2108
<script> // Javascript program to find the minimum distance// between two occurrences of the same element // Function to find the minimum// distance between the same elementsfunction minimumDistance(a, n){     // Create a HashMap to    // store (key, values) pair.    var hmap = new Map();     var minDistance = 1000000000;     // Initialize previousIndex    // and currentIndex as 0    var previousIndex = 0, currentIndex = 0;     // Traverse the array and    // find the minimum distance    // between the same elements with map     for (var i = 0; i < n; i++) {         if (hmap.has(a[i])) {            currentIndex = i;             // Fetch the previous index from map.            previousIndex = hmap.get(a[i]);             // Find the minimum distance.            minDistance = Math.min((currentIndex -                        previousIndex),minDistance);        }         // Update the map.        hmap.set(a[i], i);    }     // return minimum distance,    // if no such elements found, return -1    return (minDistance == 1000000000 ? -1 : minDistance);} // Driver code// Test Case 1:var a1 = [1, 2, 3, 2, 1];var n = a1.length;document.write( minimumDistance(a1, n) + "<br>"); // Test Case 2:var a2 = [3, 5, 4, 6, 5, 3];n = a2.length;document.write( minimumDistance(a2, n) + "<br>"); // Test Case 3:var a3 = [1, 2, 1, 4, 1];n = a3.length;document.write( minimumDistance(a3, n)); // This code is contributed by famously.</script>
2
3
2
 
Time complexity: O(N) 
mohit kumar 29
Sanjit_Prasad
famously
unknown2108
cpp-map
Java-HashMap
java-map
Arrays
Arrays
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