Let's begin with set 6,

**13)Given an array of integer nums and an integer target, return indices of the two numbers such that they add up to the target.**

**Assumptions:** Each input would have exactly one solution, and you may not use the same element twice.

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]

**Solution:**

const twoSum = (nums, target) => {

const map = {};

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

const another = target - nums[i];

if (another in map) {

return [map[another], i];

}

map[nums[i]] = i;

}

return null;

**};**

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**14)Given an array of numbers containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.**

**Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?**

**Solution:**

`var missingNumber = function (nums) { var sum = 0; for (let i = 0; i < nums.length; i++) { sum += i + 1 - nums[i]; }`

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