Efficiently Counting Non-Common Element Pairs in JavaScript

Efficiently Counting Non-Common Element Pairs in JavaScript | CodeSignal Learn

Introduction

Hello, coding enthusiast! In our journey to master coding and problem-solving, we've arrived at an interesting challenge today. We're going to focus heavily on combinatorial problems . Specifically, we're examining combinatorial problems that involve working with large datasets and multiple pairs of numbers. We'll learn how to solve significant problems efficiently by implementing smart use of data structures like objects and sidestepping expensive operations like iterating over large arrays. Are you ready? Let's dive in!

Task Statement

i \ne j

Understanding the Solution: The Idea

n \cdot (n - 1) / 2

Join the 1M+ learners on CodeSignal

Be a part of our community of 1M+ users who develop and demonstrate their skills on CodeSignal

Initialize Data Structures and Calculate Total Pairs

n \cdot (n - 1) / 2

Populate Indices Object with Pair Data

With the first step completed, our next move is to populate the object by iterating over the array of pairs. For each pair, we'll examine its two elements and either append the current index to the list of indices for this number (if it’s already in the object) or start a new list for it (if it isn't). Here's how we modify our function to carry out this operation: JavaScript function nonCommonPairs(arr) {let indices = {}; let totalPairs = arr.length * (arr.length - 1) / 2; arr.forEach((pair, idx) => {pair.forEach(num => {if (!indices[num]) indices[num] = []; indices[num].push(idx);});}); // (The rest of the code will be added in the next step) return totalPairs; // Temporary return to avoid error} console.log(nonCommonPairs([])); // Example usage, to be updated later function nonCommonPairs(arr) {let indices = {}; let totalPairs = arr.length * (arr.length - 1) / 2; arr.forEach((pair, idx) => {pair.forEach(num => {if (!indices[num]) indices[num] = []; indices[num].push(idx);});}); // (The rest of the code will be added in the next step) return totalPairs; // Temporary return to avoid error} console.log(nonCommonPairs([])); // Example usage, to be updated later

Calculate and Subtract Common Pairs

Finally, with all the data in place, we arrive at our final step of calculation. We need to calculate the total pairs of indices that share at least one common element. For that, we'll consider each number in the array and count the number of times those numbers occur in different pairs. We'll use the same formula as before. Finally, we subtract these common pairs from the total pairs to get our answer — the count of pairs without a common number. Adding this last part to our function gives us the solution: JavaScript function nonCommonPairs(arr) {let indices = {}; let totalPairs = arr.length * (arr.length - 1) / 2; arr.forEach((pair, idx) => {pair.forEach(num => {if (!indices[num]) indices[num] = []; indices[num].push(idx);});}); let commonPairs = 0; Object.keys(indices).forEach(num => {let size = indices[num].length; commonPairs += size * (size - 1) / 2;}); return totalPairs - commonPairs;} const testArr = [[2, 5], [1, 6], [3, 2], [4, 2], [5, 1], [6, 3]]; console.log("Count of non-common pairs: " + nonCommonPairs(testArr)); // Output: Count of non-common pairs: 8 function nonCommonPairs(arr) {let indices = {}; let totalPairs = arr.length * (arr.length - 1) / 2; arr.forEach((pair, idx) => {pair.forEach(num => {if (!indices[num]) indices[num] = []; indices[num].push(idx);});}); let commonPairs = 0; Object.keys(indices).forEach(num => {let size = indices[num].length; commonPairs += size * (size - 1) / 2;}); return totalPairs - commonPairs;} const testArr = [[2, 5], [1, 6], [3, 2], [4, 2], [5, 1], [6, 3]]; console.log("Count of non-common pairs: " + nonCommonPairs(testArr)); // Output: Count of non-common pairs: 8

Let's begin with the initial steps of our solution. The first thing we need is a convenient place to store the occurrence of each number in the pairs. Here, JavaScript's object shines. It enables us to efficiently track the number and its corresponding occurrences.

Next, we calculate the total number of pairs using the formula $n \cdot (n - 1) / 2$ . We'll need this for our final calculation.

Let's initialize an empty object and calculate the total pairs.

JavaScript

function nonCommonPairs(arr) {
    let indices = {};
    let totalPairs = arr.length * (arr.length - 1) / 2;

    // (The rest of the code will be added in subsequent steps)
    return totalPairs; // Temporary return to avoid error
}

console.log(nonCommonPairs([])); // Example usage, to be updated later