HackerRank Minimum Distances Problem Solution

Yashwant Parihar,

In this post, we will solve HackerRank Minimum Distances Problem Solution.

The distance between two array values is the number of indices between them. Given a find the minimum distance between any pair of equal elements in the array. If no such value exists, return -1.
Example
a = [3, 2, 1, 2, 3]
There are two matching pairs of values: 3 and 2. The indices of the 3’s are i=0 and j = 4, so their distance is d[i, j] = |j — i| = 4. The indices of the 2’s are i = 1 and j = 3, so their distance is d[i, j] = |ji| = 2. The minimum distance is 2.

Function Description

Complete the minimumDistances function in the editor below.

minimumDistances has the following parameter(s):

  • int a[n]: an array of integers

Returns

  • int: the minimum distance found or -1 if there are no matching elements

Input Format
The first line contains an integer n, the size of array a. The second line contains n space-separated integers a[i].

Output Format
Print a single integer denoting the minimum d[i, j] in a. If no such value exists, print -1.

Sample Input

STDIN           Function
-----           --------
6               arr[] size n = 6
7 1 3 4 1 7     arr = [7, 1, 3, 4, 1, 7]

Sample Output

3

Explanation
There are two pairs to consider:

  • a[1] and @[4] are both 1, so d[1, 4] = |1 — 4 = 3.
  • a[0] and @[5] are both 7, so d[0, 5] = 10-5 = 5.

The answer is min(3,5) = 3.

HackerRank Minimum Distances Problem Solution
HackerRank Minimum Distances Problem Solution

Minimum Distances C Solution

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
    int n; 
    scanf("%d",&n);
    int *A = malloc(sizeof(int) * n);
    for(int i = 0; i < n; i++){
       scanf("%d",&A[i]);
    }
    
    int i, j;
    int distance = n;
    
    for(i=0; i<n-1; i++)
        for(j=i+1; j<n; j++)
        {
            if(A[i]==A[j])
            {
                distance = (distance<(j-i))?distance:(j-i);
            }
        }
    if(distance == n)
        distance = -1;
    printf("%d\n", distance);
    return 0;
}

Minimum Distances C++ Solution

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;


int main(){
    int n;
    cin >> n;
    vector<int> A(n);
    for(int i = 0;i < n;i++){
        cin >> A[i];
    }
    int result = -1;
    for(int i = 0; i < n - 1; i++){
        for(int j = i + 1; j < n; j++){
            if(A[j] == A[i]){
                int dist = j - i;
                if(result == -1 || dist < result){
                    result = dist;
                }
                break;
            }
        }
    }
    cout << result << endl;
    return 0;
}

Minimum Distances C Sharp Solution

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
class Solution {

    static void Main(String[] args) {
        int n = Convert.ToInt32(Console.ReadLine());
        string[] A_temp = Console.ReadLine().Split(' ');
        int[] A = Array.ConvertAll(A_temp,Int32.Parse);
        
        	Console.WriteLine(minDistance(n, A));
    }
    public static object minDistance(int n, int[] arr)
	{
		if ((n < 1 | n > 1000))
			return -1;

		int dist = 0;
		List<int> distances = new List<int>();

		for (int i = 0; i <= n - 1; i++) {
			  int cur = arr[i];
			if ((cur < 1 | cur > 100000))
				return -1;

			int curPos = i;
			for (int j = i + 1; j <= n - 1; j++) {
				if (arr[j] == cur) {
					dist = j - curPos;

					distances.Add(dist);
				}
			}
		}

		if (distances.Count == 0)
			return -1;
		  return distances.Min();

	}
}

Minimum Distances Java Solution

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

class Result {

    /*
     * Complete the 'minimumDistances' function below.
     *
     * The function is expected to return an INTEGER.
     * The function accepts INTEGER_ARRAY a as parameter.
     */

    public static int minimumDistances(List<Integer> a) {
             HashMap<Integer, Integer> nums =  new  HashMap<Integer, Integer>();

        nums.put(a.get(0), 0);
        int min = Integer.MAX_VALUE;
        boolean f =false;
        for(int i = 1; i < a.size(); i++){
            if(nums.containsKey(a.get(i))){
                nums.put(a.get(i), i - nums.get(a.get(i)));
                f = true;
                if( nums.get(a.get(i)) < min){
                    min = nums.get(a.get(i));
                }
            }else{
                nums.put(a.get(i), i);
            }
        }
        if(!f)
            return -1;
        return min;
    }

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int n = Integer.parseInt(bufferedReader.readLine().trim());

        List<Integer> a = Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
            .map(Integer::parseInt)
            .collect(toList());

        int result = Result.minimumDistances(a);

        bufferedWriter.write(String.valueOf(result));
        bufferedWriter.newLine();

        bufferedReader.close();
        bufferedWriter.close();
    }
}

Minimum Distances JavaScript Solution

process.stdin.resume();
process.stdin.setEncoding('ascii');

var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;

process.stdin.on('data', function (data) {
    input_stdin += data;
});

process.stdin.on('end', function () {
    input_stdin_array = input_stdin.split("\n");
    main();    
});

function readLine() {
    return input_stdin_array[input_currentline++];
}

/////////////// ignore above this line ////////////////////

function main() {
    var n = parseInt(readLine());
    A = readLine().split(' ');
    A = A.map(Number);
    
    var B = A.map(function (val, index) {
        return {value: val, index: index};
    });
    
    B.sort(function(obj1, obj2) {
        var diff = obj1.value - obj2.value;
        if (diff != 0) {
            return diff;
        }
        return obj1.index - obj2.index;
    });
    
    var minDistance = -1;
    
    for (var i = 1; i < B.length; i++) {
        if (B[i].value == B[i - 1].value) {
            var distance = B[i].index - B[i - 1].index;
            if (minDistance == -1) {
                minDistance = distance;
            } else if (distance < minDistance) {
                minDistance = distance;
            }
        }
    }
    
    console.log(minDistance);
    
    //B.forEach(function(obj) { console.log(obj.value + " " + obj.index)} );

}

Minimum Distances Python Solution

n = int(input())
a = [int(_) for _ in input().split()]
b = {}
ans = n
for i in range(n):
    if a[i] in b:
        ans = min(ans, i - b[a[i]])
    b[a[i]] = i
    
if ans == n:
    print(-1)
else:
    print(ans)

Other Solutions

c C# C++ HackerRank Solutions java javascript python

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