Leetcode – 16. 3sum closest (yeah… threesum again)
Problems
Given an integer array nums of length n and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers.
You may assume that each input would have exactly one solution.
Solution
After solving the 3sum problem, I think that this problem is pretty easy. We will use two pointer technique to solve it.
1. sort the array of integers
2. loop through the array with i start from 0, left = i + 1, right = length – 1.
3. check the sum of the three numbers
– sum > target –> reduce the right
– sum < target –> increase the left
– get the difference between sum and target and store it if it’s the closest
– if the difference is 0, break the loop and return.
/**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
var threeSumClosest = function(nums, target) {
let diff = 999999;
let result = undefined;
nums.sort((a, b) => a - b);
for (let i = 0; i < nums.length - 2; i++) {
// ignore duplicated cases
if (i >= 1 && nums[i] === nums[i - 1]) continue;
let l = i + 1;
let r = nums.length - 1;
while (l < r) {
const sum = nums[i] + nums[l] + nums[r];
let d = Math.abs(sum - target);
if (d < diff) {
diff = d;
result = sum;
}
if (sum > target) {
r--;
} else if (sum < target) {
l++;
} else {
break;
}
}
if (diff === 0) {
break;
}
}
return result;
};