ProblemPermalink

Problem ID: 259

Title: 3Sum Smaller

Difficulty: Medium

Description:

ThoughtsPermalink

SolutionPermalink

General Idea

The main observation: for 2 sum smaller, if there exists two pointer such that the sum is less than the target, moving the right pointer all the way until right before the left pointer will give the smaller.

ie:

target = 9
[1, 2, 3, 5, 8]
 ↑        ↑
left    right
  • (1,5), (1,3), (1,2) are valid combinations and we can easily calculate this by right - left

ImplementationPermalink

class Solution {
public:
    int threeSumSmaller(vector<int>& nums, int target) {
        
        auto n = nums.size();
        int ret = 0;
        sort(nums.begin(), nums.end());
        
        for (auto i = size_t{0}; i < n; i++) {
            auto j = i + 1;
            auto k = n - 1;
            while (j < k) {
                int sum = nums[i] + nums[j] + nums[k];
                if (sum < target) {
                    ret += (k - j);
                    j++;
                } else {
                   k--; 
                }
            }
        }
        return ret;
    }
};