Two Pointers · Medium

3Sum

O(N^2) · O(1)

Initialize Two Pointers

Array is sorted. Set left pointer L=0 (val -4) and right pointer R=5 (val 2).

Array Elements & Pointers
L ↓
-4
L
-1
[1]
-1
[2]
0
[3]
1
[4]
R ↓
2
R
Live Variables & Invariants
target:0
left:0
right:5
sum:-2
Step 1 / 7
14%
Solution Code
1
function threeSum(nums: number[]): number[][] {
2
  nums.sort((a, b) => a - b);
3
  const res: number[][] = [];
4
  for (let i = 0; i < nums.length - 2; i++) {
5
    if (i > 0 && nums[i] === nums[i - 1]) continue;
6
    let l = i + 1, r = nums.length - 1;
7
    while (l < r) {
8
      const sum = nums[i] + nums[l] + nums[r];
9
      if (sum === 0) {
10
        res.push([nums[i], nums[l], nums[r]]);
11
        while (l < r && nums[l] === nums[l + 1]) l++;
12
        while (l < r && nums[r] === nums[r - 1]) r--;
13
        l++; r--;
14
      } else if (sum < 0) l++;
15
      else r--;
16
    }
17
  }
18
  return res;
19
}
LeetCode IDE Console

Test Cases

3 cases from Blind 75 & NeetCode 150

Ask for a hint whenever you get stuck.