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918.3sum-smaller.cpp
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60 lines (58 loc) · 1.51 KB
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// Tag: Opposite Direction Two Pointers, Enumerate, Two Pointers, Array
// Time: O(N^2)
// Space: O(1)
// Ref: Leetcode-259
// Note: -
// Given an array of `n` integers nums and a `target`, find the number of index triplets `i, j, k` with `0 <= i < j < k < n` that satisfy the condition `nums[i] + nums[j] + nums[k] < target`.
//
// **Example1**
//
// ```
// Input: nums = [-2,0,1,3], target = 2
// Output: 2
// Explanation:
// Because there are two triplets which sums are less than 2:
// [-2, 0, 1]
// [-2, 0, 3]
// ```
//
// **Example2**
//
// ```
// Input: nums = [-2,0,-1,3], target = 2
// Output: 3
// Explanation:
// Because there are three triplets which sums are less than 2:
// [-2, 0, -1]
// [-2, 0, 3]
// [-2, -1, 3]
// ```
//
//
class Solution {
public:
/**
* @param nums: an array of n integers
* @param target: a target
* @return: the number of index triplets satisfy the condition nums[i] + nums[j] + nums[k] < target
*/
int threeSumSmaller(vector<int> &nums, int target) {
// Write your code here
int n = nums.size();
sort(nums.begin(), nums.end());
int res = 0;
for (int i = 0; i < n - 2; i++) {
int l = i + 1;
int r = n - 1;
while (l < r) {
if (nums[i] + nums[l] + nums[r] < target) {
res += r - l;
l += 1;
} else {
r -= 1;
}
}
}
return res;
}
};