Patterns

For each item, look up its complement (target - item) in the map

When to useTwo Sum and variants — pair-finding where you know what to look for

iterative

Hash Map

two_sum

Hashmap Intro

hashmap_intro

Two Sum - Unique Pairs

two_sum_unique_pairs

Pairs of Songs With Total Durations Divisible by 60

pairs_of_songs

Prefix Sum

subarray_sum

INITIALIZE

empty Map of value → index

const seen = new Map();

const counts = {};

const complement = new Set();
const seen = new Set();

const counts = new Array(60).fill(0);
let pairs = 0;

const prefixSums = new Map([[0, 0]]);
let curSum = 0;

TRAVERSE

for each item

for (let i = 0; i < arr.length; i++) {

for (const x of arr) {

for (const num of nums) {

for (const duration of durations) {

for (let i = 0; i < arr.length; i++) {

[COMPLEMENT]

compute complement; check map

  const complement = target - arr[i];
  if (seen.has(complement)) return [seen.get(complement), i];

  counts[x] = (counts[x] || 0) + 1;

  if (complement.has(target - num)) {
    const pair = num < target - num ? `${num},${target - num}` : `${target - num},${num}`;
    seen.add(pair);
  }

  const remainder = duration % 60;
  const complement = (60 - remainder) % 60;
  pairs += counts[complement];

  curSum += arr[i];
  const complement = curSum - target;
  if (prefixSums.has(complement)) return [prefixSums.get(complement), i + 1];

store

if not found, store current value

  seen.set(arr[i], i);

—

  complement.add(num);

  counts[remainder] += 1;

  prefixSums.set(curSum, i + 1);

RETURN

result or null

return null;

return counts;

return seen.size;

return pairs;

return [];