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Two sum#1226

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nosecreek wants to merge7 commits intoTheAlgorithms:master
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1 change: 1 addition & 0 deletionsDIRECTORY.md
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Expand Up@@ -57,6 +57,7 @@
* [LocalMaximomPoint](Data-Structures/Array/LocalMaximomPoint.js)
* [NumberOfLocalMaximumPoints](Data-Structures/Array/NumberOfLocalMaximumPoints.js)
* [QuickSelect](Data-Structures/Array/QuickSelect.js)
* [TwoSum](Data-Structures/Array/TwoSum.js)
* **Graph**
* [Graph](Data-Structures/Graph/Graph.js)
* [Graph2](Data-Structures/Graph/Graph2.js)
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20 changes: 20 additions & 0 deletionsData-Structures/Array/TwoSum.js
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/**
* @function TwoSum
* @see https://leetcode.com/problems/two-sum/
* @description Given an array of integers, returns indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. This is a basic brute force approach.
* @param {Array} nums - array of integers.
* @param {number} target - target integer.
* @returns {Array} Array of the indices of the two numbers whose sum equals the target
* @example Given nums = [2, 7, 11, 15], target = 9; return [0, 1] because nums[0] + nums[1] = 2 + 7 = 9
* @complexity: O(n^2)
*/
const TwoSum = (nums, target) => {
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That's a pretty naive implementation with a runtime of O(n²). Please document this in the JSDoc comment.

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Please seethe docs and use the proper@annotations.

for (let i = 0; i < nums.length; i++) {
for (let j = i + 1; j < nums.length; j++) {
if (nums[i] + nums[j] === target) {
return [i, j]
}
}
}
}
export { TwoSum }
9 changes: 9 additions & 0 deletionsData-Structures/Array/test/TwoSum.test.js
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import { TwoSum } from '../TwoSum'

test('TwoSum tests', () => {
expect(TwoSum([2, 7, 11, 15], 9)).toEqual([0, 1])
expect(TwoSum([2, 7, 11, 15, 6], 8)).toEqual([0, 4])
expect(TwoSum([1, 0, 5, 7, 3, 4], 6)).toEqual([0, 2])
expect(TwoSum([0, 8, 3, 1, 2, 7, 3], 6)).toEqual([2, 6])
expect(TwoSum([0, 5, 4, 2, 6, 7, 9, 1], 3)).toEqual([3, 7])
})

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