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August 9, 2021 01:50
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Two Sums LeetCode
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| /* | |
| Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. | |
| You may assume that each input would have exactly one solution, and you may not use the same element twice. | |
| You can return the answer in any order. | |
| Example 1: | |
| Input: nums = [2,7,11,15], target = 9 | |
| Output: [0,1] | |
| Output: Because nums[0] + nums[1] == 9, we return [0, 1]. | |
| Example 2: | |
| Input: nums = [3,2,4], target = 6 | |
| Output: [1,2] | |
| Example 3: | |
| Input: nums = [3,3], target = 6 | |
| Output: [0,1] | |
| Constraints: | |
| 2 <= nums.length <= 104 | |
| -109 <= nums[i] <= 109 | |
| -109 <= target <= 109 | |
| Only one valid answer exists. | |
| Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity? | |
| */ | |
| const nums = [2,7,11,15]; | |
| const target = 9 | |
| const twoSums = (nums, target) => { | |
| const object = {}; | |
| for (let i = 0; i < nums.length; i++) { | |
| if (object[nums[i]] >= 0) { | |
| return [object[nums[i]], i]; | |
| } | |
| object[target - nums[i]] = i; | |
| } | |
| console.log(object); | |
| return object; | |
| }; | |
| twoSums(nums, target); |
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