2824. Count Pairs Whose Sum is Less than Target

Method 1

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

Method 1

【O(nlogn) time | O(1) space】

package Leetcode.TwoPointer;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;

/**
 * @author zhengxingxing
 * @date 2025/02/22
 */
public class CountPairsWhoseSumIsLessThanTarget {
    
    public static int countPairs(List<Integer> nums, int target) {
        // Sort the list to enable two pointer approach
        Collections.sort(nums);

        int count = 0;  // Counter for valid pairs
        int left = 0;   // Left pointer starting from beginning
        int right = nums.size() - 1;  // Right pointer starting from end

        while (left < right) {
            // Calculate current sum of elements at left and right pointers
            int currentSum = nums.get(left) + nums.get(right);

            if (currentSum < target) {
                // If current sum is less than target:
                // All pairs between left and right will also be valid
                // Because array is sorted, nums[left] + nums[right-1] will be even smaller
                // We add (right - left) pairs to our count
                count += right - left;
                // Move left pointer to find more pairs
                left++;
            } else {
                // If current sum is >= target:
                // Need to decrease sum, so move right pointer left
                right--;
            }
        }

        return count;
    }

    public static void main(String[] args) {
        // Test Case 1: Normal case with positive and negative numbers
        List<Integer> nums1 = new ArrayList<>(Arrays.asList(-1, 1, 2, 3, 1));
        int target1 = 2;
        System.out.println("Test Case 1 Result: " + countPairs(nums1, target1)); // Expected output: 3

        // Test Case 2: Case with more negative numbers
        List<Integer> nums2 = new ArrayList<>(Arrays.asList(-6, 2, 5, -2, -7, -1, 3));
        int target2 = -2;
        System.out.println("Test Case 2 Result: " + countPairs(nums2, target2)); // Expected output: 10

        // Test Case 3: Edge case - empty list
        List<Integer> nums3 = new ArrayList<>();
        int target3 = 1;
        System.out.println("Test Case 3 Result: " + countPairs(nums3, target3)); // Expected output: 0

        // Test Case 4: Edge case - single element list
        List<Integer> nums4 = new ArrayList<>(Arrays.asList(1));
        int target4 = 1;
        System.out.println("Test Case 4 Result: " + countPairs(nums4, target4)); // Expected output: 0
    }
}

Method 1

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