Sales by Match


Problem Statement :


There is a large pile of socks that must be paired by color. Given an array of integers representing the color of each sock, determine how many pairs of socks with matching colors there are.



Function Description

Complete the sockMerchant function in the editor below.

sockMerchant has the following parameter(s):

int n: the number of socks in the pile
int ar[n]: the colors of each sock
Returns

int: the number of pairs
Input Format

The first line contains an integer n, the number of socks represented in ar.
The second line contains n space-separated integers, ar[ i ], the colors of the socks in the pile.


Constraints

1   <=   n  <=  100
1  <=  ar[ i ]  <=  100, where 0  <=  i  < n


Solution :



title-img 


                            Solution in C :

In    C   :







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

int main() {

   int n,i,t;
    scanf("%d",&n);
    int c[101];
    memset(c,0,sizeof(c));
    for(i=0;i<n;i++)
        {
        scanf("%d",&t);
        c[t]++;
    }
    long int ans=0;
    for(i=1;i<=100;i++)
        {
       ans+=(c[i]/2);
    }
    printf("%ld",ans);
    return 0;
}
                        


                        Solution in C++ :

In   C++  :







#include <bits/stdc++.h>

#define FI(i,a,b) for(int i=(a);i<=(b);i++)
#define FD(i,a,b) for(int i=(a);i>=(b);i--)

#define LL long long
#define Ldouble long double
#define PI 3.1415926535897932384626

#define PII pair<int,int>
#define PLL pair<LL,LL>
#define mp make_pair
#define fi first
#define se second

using namespace std;

int n, c[105], x;

int main(){
	scanf("%d", &n);
	FI(i, 1, n){
		scanf("%d", &x);
		c[x]++;
	}
	
	int ans = 0;
	
	FI(i, 1, 100) ans += c[i] / 2;
	printf("%d\n", ans);
	return 0;
}
                    


                        Solution in Java :

In   Java :





import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int pairs = 0;
        Set<Integer> unmatched = new HashSet<>();
        for (int i = 0; i < n; ++i) {
            int ci = in.nextInt();
            if (unmatched.contains(ci)) {
                unmatched.remove(ci);
                ++pairs;
            }
            else {
                unmatched.add(ci);
            }
        }
        System.out.println(pairs);
    }
}
                    


                        Solution in Python : 
                            
In  Python3  :






#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the sockMerchant function below.
from collections import Counter
def sockMerchant(n, ar):
    sum = 0
    for val in Counter(ar).values():
        sum += val//2
    return sum
    

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()


View More Similar Problems

Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

View Solution →

Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

View Solution →

Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

View Solution →

Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

View Solution →

Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t

View Solution →

Binary Search Tree : Lowest Common Ancestor

You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b

View Solution →