2824

2824. Count Pairs Whose Sum is Less than Target (Easy)

0-indexed nums n target the number of pairs (i, j) where 0 <= i < j < n and nums[i] + nums[j] < target

 

Example 1:

Input: nums = [-1,1,2,3,1], target = 2
Output: 3
Explanation: There are 3 pairs of indices that satisfy the conditions in the statement:
- (0, 1) since 0 < 1 and nums[0] + nums[1] = 0 < target
- (0, 2) since 0 < 2 and nums[0] + nums[2] = 1 < target 
- (0, 4) since 0 < 4 and nums[0] + nums[4] = 0 < target
Note that (0, 3) is not counted since nums[0] + nums[3] is not strictly less than the target.

Example 2:

Input: nums = [-6,2,5,-2,-7,-1,3], target = -2
Output: 10
Explanation: There are 10 pairs of indices that satisfy the conditions in the statement:
- (0, 1) since 0 < 1 and nums[0] + nums[1] = -4 < target
- (0, 3) since 0 < 3 and nums[0] + nums[3] = -8 < target
- (0, 4) since 0 < 4 and nums[0] + nums[4] = -13 < target
- (0, 5) since 0 < 5 and nums[0] + nums[5] = -7 < target
- (0, 6) since 0 < 6 and nums[0] + nums[6] = -3 < target
- (1, 4) since 1 < 4 and nums[1] + nums[4] = -5 < target
- (3, 4) since 3 < 4 and nums[3] + nums[4] = -9 < target
- (3, 5) since 3 < 5 and nums[3] + nums[5] = -3 < target
- (4, 5) since 4 < 5 and nums[4] + nums[5] = -8 < target
- (4, 6) since 4 < 6 and nums[4] + nums[6] = -4 < target

 

Constraints:

• 1 <= nums.length == n <= 50
• -50 <= nums[i], target <= 50

Companies: Facebook

Related Topics:
Array, Two Pointers, Sorting

Similar Questions:

• Two Sum (Easy)
• Count the Number of Fair Pairs (Medium)

Solution 1. Two Pointers

// OJ: https://leetcode.com/problems/count-pairs-whose-sum-is-less-than-target
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int countPairs(vector<int>& A, int target) {
        sort(begin(A), end(A));
        int ans = 0, N = A.size();
        for (int i = 0, j = N - 1; i < j;) {
            if (A[i] + A[j] < target) {
                ans += j - i;
                ++i;
            } else --j;
        }
        return ans;
    }
};