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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 $nums$.
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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; }