HackerRank Sales by Match problem solution YASH PAL, 31 July 20249 August 2024 In this HackerRank Sales by Match problem in the Interview preparation kit, you need to Complete the sockMerchant function. it has the following parameter(s): int n: the number of socks in the pile int ar[n]: the colors of each sock Hackerrank sales by match solution Problem solution in Python programming. #!/bin/python3 import math import os import random import re import sys # Complete the sockMerchant function below. def sockMerchant(n, ar): num = 0 for i in range(0,n): gum = 1 for j in range(i+1,n): if ar[i] == None: continue if ar[i] == ar[j] and gum ==1: num = num + 1 gum = gum + 1 ar[j] = None return num if __name__ == '__main__': fptr = open(os.environ['OUTPUT_PATH'], 'w') n = int(input()) ar = list(map(int, input().rstrip().split())) result = sockMerchant(n, ar) fptr.write(str(result) + 'n') fptr.close() Problem solution in Java8 Programming. import java.io.*; import java.util.*; public class Solution { public static void main(String[] args) { /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */ Scanner s = new Scanner(System.in); int[] freq = new int[101]; int n = s.nextInt(); for(int i = 0; i < n; i++){ int x = s.nextInt(); freq[x]++; } int total = 0; for(int i = 1; i < 101; i++){ total+=freq[i]/2; } System.out.println(total); } } Problem solution in C++ programming. #include <bits/stdc++.h> using namespace std; using LINT = long long int; using PII = pair<int,int>; #define PB push_back #define FI first #define SE second #define REP(i,n) for(int i=0;i<(n);++i) #define FOR(i, a, b) for(int i=(a);i<(b);++i) int socks[107]; int main() { int n; cin >> n; REP(i,n){ int x; cin>>x; socks[x]++; } int ans = 0; REP(i,107)ans+=socks[i]/2; cout<<ans<<endl; return 0; } Problem solution in C programming. #include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> int main() { int n; scanf("%d",&n); int arr[101]; for(int i=0;i<101;i++){ arr[i]=0; } for(int i=0;i<n;i++){ int j; scanf("%d",&j); arr[j] += 1; } int total=0; for(int i=0;i<101;i++){ if(arr[i]> 1){ if(arr[i] % 2 == 0) total += (arr[i] / 2); else total += ((arr[i]-1) / 2); } } printf("%d",total); return 0; } Problem solution in JavaScript programming. function processData(input) { //Enter your code here var pInput = input.split('n'); var n = pInput[0]; var pInput = pInput[1].split(' '); var oddItems = {}; pInput.map(function (i) { if(oddItems[i]) { delete oddItems[i]; } else { oddItems[i] = true; } }) console.log((n - Object.keys(oddItems).length)/2); } process.stdin.resume(); process.stdin.setEncoding("ascii"); _input = ""; process.stdin.on("data", function (input) { _input += input; }); process.stdin.on("end", function () { processData(_input); }); coding problems interview prepration kit AlgorithmsHackerRank
another approach for python: def sockMerchant(n, ar): # Write your code here unique_colors = set(ar) total_pairs = 0 for color in unique_colors: total_pairs += (int(ar.count(color)/2)) return total_pairs