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Count Pairs Whose Sum is Less than Target

Problem 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
Explination: 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
Explination: 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

Approach​

1. Sort the List:

  • First, sort the list nums. Sorting allows us to use the two-pointer technique effectively.

2. Initialize Variables:

  • Initialize count to keep track of the number of valid pairs.
  • Initialize two pointers, left at the start (index 0) and right at the end (index nums.size() - 1) of the list.

3. Two-Pointer Technique:

  • Use a while loop to iterate as long as left is less than right.

  • Within the loop:

    • If the sum of the elements at the left and right pointers (nums.get(left) + nums.get(right)) is less than target, it means all pairs (left, left + 1), (left, left + 2), ..., (left, right) are valid. This is because the array is sorted, and any number between left and right added to nums[left] will still be less than target.
    • Increment count by right - left to account for all these pairs, and move the left pointer to the right (left++).
    • If the sum is greater than or equal to target, decrement the right pointer (right--) to try smaller numbers with the current nums[left].

    Python3​

class Solution:
  def count_pairs(nums, target):
    nums.sort()
    count = 0
    left = 0
    right = len(nums) - 1

    while left < right:
        if nums[left] + nums[right] < target:
            count += right - left
            left += 1
        else:
            right -= 1

    return count

Java​

class Solution {
     public int countPairs(List<Integer> nums, int target) {
        Collections.sort(nums);
        int count = 0;
        int left = 0;
        int right = nums.size() - 1;

        while (left < right) {
            if (nums.get(left) + nums.get(right) < target) {
                count += right - left;
                left++;
            } else {
                right--;
            }
        }

        return count;
    }

}

C++​

class Solution {
public:
    int countPairs(std::vector<int>& nums, int target) {
        std::sort(nums.begin(), nums.end());
        int count = 0;
        int left = 0;
        int right = nums.size() - 1;

        while (left < right) {
            if (nums[left] + nums[right] < target) {
                count += right - left;
                left++;
            } else {
                right--;
            }
        }

        return count;
    }
};

JavaScript​

    countPairs(nums, target) {
        nums.sort((a, b) => a - b);
        let count = 0;
        let left = 0;
        let right = nums.length - 1;

        while (left < right) {
            if (nums[left] + nums[right] < target) {
                count += right - left;
                left++;
            } else {
                right--;
            }
        }

        return count;
    }

TypeScript​

 countPairs(nums: number[], target: number): number {
        nums.sort((a, b) => a - b);
        let count = 0;
        let left = 0;
        let right = nums.length - 1;

        while (left < right) {
            if (nums[left] + nums[right] < target) {
                count += right - left;
                left++;
            } else {
                right--;
            }
        }

        return count;
    }

References