# 2824. Count Pairs Whose Sum is Less than Target

## Description

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

## Solutions

Solution 1: Sorting + Binary Search

First, we sort the array $nums$. Then, for each $j$, we use binary search in the range $[0, j)$ to find the first index $i$ that is greater than or equal to $target - nums[j]$. All indices $k$ in the range $[0, i)$ meet the condition, so the answer increases by $i$.

After the traversal, we return the answer.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $nums$.

• class Solution {
public int countPairs(List<Integer> nums, int target) {
Collections.sort(nums);
int ans = 0;
for (int j = 0; j < nums.size(); ++j) {
int x = nums.get(j);
int i = search(nums, target - x, j);
ans += i;
}
return ans;
}

private int search(List<Integer> nums, int x, int r) {
int l = 0;
while (l < r) {
int mid = (l + r) >> 1;
if (nums.get(mid) >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
}

• class Solution {
public:
int countPairs(vector<int>& nums, int target) {
sort(nums.begin(), nums.end());
int ans = 0;
for (int j = 0; j < nums.size(); ++j) {
int i = lower_bound(nums.begin(), nums.begin() + j, target - nums[j]) - nums.begin();
ans += i;
}
return ans;
}
};

• class Solution:
def countPairs(self, nums: List[int], target: int) -> int:
nums.sort()
ans = 0
for j, x in enumerate(nums):
i = bisect_left(nums, target - x, hi=j)
ans += i
return ans


• func countPairs(nums []int, target int) (ans int) {
sort.Ints(nums)
for j, x := range nums {
i := sort.SearchInts(nums[:j], target-x)
ans += i
}
return
}

• function countPairs(nums: number[], target: number): number {
nums.sort((a, b) => a - b);
let ans = 0;
const search = (x: number, r: number): number => {
let l = 0;
while (l < r) {
const mid = (l + r) >> 1;
if (nums[mid] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
};
for (let j = 0; j < nums.length; ++j) {
const i = search(target - nums[j], j);
ans += i;
}
return ans;
}