LeetCode 2824: Count Pairs Whose Sum is Less than Target Solution

Master LeetCode problem 2824 (Count Pairs Whose Sum is Less than Target), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2824. Count Pairs Whose Sum is Less than Target

Easy

Array Two Pointers Binary Search Sorting

Problem Explanation

Explanation:

To solve this problem, we can iterate through all pairs of indices (i, j) where i < j and calculate the sum of nums[i] + nums[j]. If the sum is less than the target, we increment a counter to keep track of the valid pairs.

Algorithmic Idea:

Initialize a counter variable to keep track of the number of valid pairs.
Iterate through all pairs of indices (i, j) where i < j.
Calculate the sum of nums[i] + nums[j].
If the sum is less than the target, increment the counter.
Return the counter as the result.

Time Complexity:

The time complexity of this approach is O(n^2), where n is the length of the input array nums.

Space Complexity:

The space complexity is O(1) as we are using only a constant amount of extra space.

Solution Code

class Solution {
    public int countPairs(int[] nums, int target) {
        int n = nums.length;
        int count = 0;
        
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (nums[i] + nums[j] < target) {
                    count++;
                }
            }
        }
        
        return count;
    }
}

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What is the time complexity of LeetCode 2824 (Count Pairs Whose Sum is Less than Target)?

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