HackerRank Js-Week3 4. Sales by Match

    AllenLi
    HackerRank

    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.

    有一大堆襪子必須按顏色搭配。 給定一個表示每隻襪子顏色的整數陣列，確定有多少雙顏色可以匹配的襪子。

    Example 1:

    n = 7
ar = [1, 2, 1, 2, 1, 3, 2]
There is one pair of color 1 and one of color 2. There are three odd socks left, one of each color. The number of pairs is 2.

    solution:
    建立一個 hashTable 物件，將所有 ar 中的種類與次數放入 key of value。
    根據這個物件表將 value 的值大於 2 的次數做加總。
    ex.奇數的部分，如無法成對，則省略不計。

    Code 1: BigO(m + n)

    'use strict';

const fs = require('fs');

process.stdin.resume();
process.stdin.setEncoding('utf-8');

let inputString = '';
let currentLine = 0;

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

process.stdin.on('end', function() {
    inputString = inputString.split('\n');

    main();
});

function readLine() {
    return inputString[currentLine++];
}

/*
 * Complete the 'sockMerchant' function below.
 *
 * The function is expected to return an INTEGER.
 * The function accepts following parameters:
 *  1. INTEGER n
 *  2. INTEGER_ARRAY ar
 */

function sockMerchant(n, ar) {
    // Write your code here
    let hashTable = {},
        count = 0
    
    for (let i = 0; i < n; i++) {
        hashTable[ar[i]] = (hashTable[ar[i]] || 0) + 1
    }
    
    for (let num in hashTable) {
        if (hashTable[num] >= 2) {
            count = count + parseInt(hashTable[num] / 2)
        }
    }
    
    return count;
}

function main() {
    const ws = fs.createWriteStream(process.env.OUTPUT_PATH);

    const n = parseInt(readLine().trim(), 10);

    const ar = readLine().replace(/\s+$/g, '').split(' ').map(arTemp => parseInt(arTemp, 10));

    const result = sockMerchant(n, ar);

    ws.write(result + '\n');

    ws.end();
}

    FlowChart:
    Example 1

    ar = [1, 2, 1, 2, 1, 3, 2]

hashTable = {
    "1": 3,
    "2": 3,
    "3": 1
}

"1" => 3 / 2 = 1.5, parseInt(1.5) = 1 //count += 1
"2" => 3 / 2 = 1.5, parseInt(1.5) = 1 //count += 1
"3" => 1 / 2 = 0.5, parseInt(0.5) = 0 //count
return count; //2

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