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leet-two-sum.ts
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leet-two-sum.ts
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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]
// Explanation: 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?
//My solution 1 : weak performance
// const nums = [3, 2, 4];
// function twoSum(nums: number[], target: number) {
// let result: number[] = [];
// for (let i = 0; i < nums.length; i++) {
// for (let j = i + 1; j < nums.length; j++) {
// if (nums[i] >= Math.pow(10, 9) || Math.pow(-10, 9) >= nums[i]) throw new Error("is too long or too short");
// if (target >= Math.pow(10, 9) || target <= Math.pow(-10, 9)) throw new Error("is too long or too short");
// if (nums[i] + nums[j] === target) {
// result.push(i, j);
// return result;
// }
// }
// }
// }
// console.log(twoSum(nums, 6));
//My solution 2 : little improved performance
function twoSum(nums: number[], target: number) {
if (nums.length < 2) {
throw new Error("Array must contain at least two numbers");
}
const numMap: { [key: number]: number } = {};
for (let i = 0; i < nums.length; i++) {
const num = nums[i];
if (num < -Math.pow(10, 9) || num > Math.pow(10, 9)) {
throw new Error("Number is too long or too short");
}
const complement = target - num;
if (numMap.hasOwnProperty(complement)) {
return [numMap[complement], i];
}
numMap[num] = i;
}
throw new Error("No solution found");
}