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Removed "Two Sum" for testing purposes
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@@ -1,14 +0,0 @@
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/**
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* @param {number[]} nums
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* @param {number} target
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* @return {number[]}
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*/
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var twoSum = function(nums, target) {
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for(let i = 0; i < nums.length; i++) {
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for(let j = i + 1; j < nums.length; j++) {
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if(nums[i] + nums[j] == target) {
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return [i, j];
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}
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}
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}
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};
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@@ -1,11 +0,0 @@
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class Solution(object):
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def twoSum(self, nums, target):
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"""
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:type nums: List[int]
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:type target: int
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:rtype: List[int]
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"""
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for i in range(len(nums)):
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for j in range(i + 1, len(nums)):
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if nums[i] + nums[j] == target:
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return [i, j]
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@@ -1 +0,0 @@
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@@ -1,38 +0,0 @@
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<h2><a href="https://leetcode.com/problems/two-sum/">1. Two Sum</a></h2><h3>Easy</h3><hr><div><p>Given an array of integers <code>nums</code> and an integer <code>target</code>, return <em>indices of the two numbers such that they add up to <code>target</code></em>.</p>
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<p>You may assume that each input would have <strong><em>exactly</em> one solution</strong>, and you may not use the <em>same</em> element twice.</p>
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<p>You can return the answer in any order.</p>
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<p> </p>
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<p><strong>Example 1:</strong></p>
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<pre><strong>Input:</strong> nums = [2,7,11,15], target = 9
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<strong>Output:</strong> [0,1]
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<strong>Explanation:</strong> Because nums[0] + nums[1] == 9, we return [0, 1].
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre><strong>Input:</strong> nums = [3,2,4], target = 6
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<strong>Output:</strong> [1,2]
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre><strong>Input:</strong> nums = [3,3], target = 6
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<strong>Output:</strong> [0,1]
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>2 <= nums.length <= 10<sup>4</sup></code></li>
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<li><code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code></li>
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<li><code>-10<sup>9</sup> <= target <= 10<sup>9</sup></code></li>
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<li><strong>Only one valid answer exists.</strong></li>
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</ul>
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<p> </p>
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<strong>Follow-up: </strong>Can you come up with an algorithm that is less than <code>O(n<sup>2</sup>) </code>time complexity?</div>
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