Question
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
For example, given array S = {-1 2 1 -4}, and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
Stats
| Frequency | 1 |
| Diffficulty | 3 |
| Adjusted Difficulty | 4 |
| Time to use | ---------- |
Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)
Analysis
This question is a simplified version of [LeetCode 15] 3Sum. The required return is an integer instead of a list.
This makes life easier because we do not need to consider duplications (think about it, why?).
Solution
The code is 3Sum solution without duplication avoidance.
My code
public class Solution {
public int threeSumClosest(int[] num, int target) {
if (num == null || num.length < 3) {
return 0;
}
Arrays.sort(num);
int len = num.length;
int ans = num[0] + num[1] + num[2];
for (int i = 0; i < len; i++) {
// if (i != 0 && num[i - 1] == num[i]) {
// continue;
// }
// similar to 3sum question, but without dup avoidance
int left = i + 1;
int right = len - 1;
while (left < right) {
int sum = num[i] + num[left] + num[right];
// if found triplet that sums to target, return!
if (sum == target) {
return target;
}
// then update ans variable - if it's closer to target
if (Math.abs(sum - target) < Math.abs(ans - target)) {
ans = sum;
}
// now move either left or right pointer
if (sum > target) {
right--;
} else {
left++;
}
}
}
return ans;
}
}