2824. Count Pairs Whose Sum is Less than Target

Given a 0-indexed integer array nums of length n and an integer target, return 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

Solution:

class Solution {
    public int countPairs(List<Integer> nums, int target) {
        int result = 0;


        int left = 0;

        for (int i = 0; i < nums.size(); i++){
            for (int j = i + 1; j < nums.size(); j++){
                if (nums.get(i) + nums.get(j) < target){
                    result++;
                }
            }
        }

        return result;
    }
}

// TC: O(n^2)
// SC: O(1)

class Solution {
    public int countPairs(List<Integer> nums, int target) {
        Collections.sort(nums); 
        /*
         0  1 2 3
        [-1,1,1,2,3]
            l
              r

        s = -1
        r = (3 -0) = 3
        1 + 2 > target

        */
        int result = 0;
        int left = 0;
        int right = nums.size() - 1;

        while(left < right){
            int curLeft = nums.get(left);
            int curRight = nums.get(right);

            int curSum = curLeft + curRight;

            if (curSum < target){
                result = result + (right - left);
                left++;
            }else{
                right--;
            }
        }
        return result;
    }
}
// TC: O(nlogn)
// SC: O(1)