10000 new solution · dev-ik/leetcode-typescript@f30ea45 · GitHub
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‎018. 4Sum/index.test.ts

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import { fourSum } from './index';
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describe('fourSum', () => {
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const testCases = [
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{
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name: 'Case 1',
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input: { nums: [1, 0, -1, 0, -2, 2], target: 0 },
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expected: [
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[-2, -1, 1, 2],
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[-2, 0, 0, 2],
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[-1, 0, 0, 1],
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],
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},
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{
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name: 'Case 2',
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input: { nums: [2, 2, 2, 2, 2], target: 8 },
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expected: [[2, 2, 2, 2]],
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},
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{
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name: 'Case 3',
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input: { nums: [-2, -1, -1, 1, 1, 2, 2], target: 0 },
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expected: [
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[-2, -1, 1, 2],
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[-1, -1, 1, 1],
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],
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},
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];
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for (const testCase of testCases) {
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test(testCase.name, () => {
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expect(fourSum(testCase.input.nums, testCase.input.target)).toEqual(
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testCase.expected,
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);
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});
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}
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});

‎018. 4Sum/index.ts

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// Description:
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// Given an array nums of n integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]] such that:
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//
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// 0 <= a, b, c, d < n
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// a, b, c, and d are distinct.
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// nums[a] + nums[b] + nums[c] + nums[d] == target
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//
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// You may return the answer in any order.
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// Constraints:
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// 1 <= nums.length <= 200
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// -109 <= nums[i] <= 109
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// -109 <= target <= 109
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export const fourSum = (nums: number[], target: number): number[][] => {
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nums.sort((a, b) => a - b);
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const l = nums.length;
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const res: number[][] = [];
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for (let i = 0; i < l - 3; i++) {
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if (i > 0 && nums[i] === nums[i - 1]) {
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continue;
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}
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for (let j = i + 1; j < l - 2; j++) {
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if (j > i + 1 && nums[j] === nums[j - 1]) {
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continue;
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}
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let left = j + 1;
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let right = l - 1;
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while (left < right) {
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const sum = nums[i] + nums[j] + nums[left] + nums[right];
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if (sum === target) {
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res.push([nums[i], nums[j], nums[left], nums[right]]);
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left++;
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right--;
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while (nums[left] === nums[left - 1]) {
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left++;
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}
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while (nums[right] === nums[right + 1]) {
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right--;
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}
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} else if (sum < target) {
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left++;
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} else {
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right--;
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}
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}
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}
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}
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return res;
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};

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