Day 2 - Count Pairs whose sum is less than target.md

Difficulty	Source	Tags
Medium	160 Days of Problem Solving	Sorting two-pointer-algorithm Arrays

🚀 Day 2. Count Pairs whose sum is less than target 🧠

The problem can be found at the following link: Problem Link

💡 Problem Description:

You are given an array arr[] of integers and a target integer. Your task is to find the number of pairs in the array whose sum is strictly less than the given target.

🔍 Example Walkthrough:

Input:
arr[] = [7, 2, 5, 3], target = 8
Output:
2
Explanation:
There are 2 pairs with sum less than 8: (2, 5) and (2, 3).

Input:
arr[] = [5, 2, 3, 2, 4, 1], target = 5
Output:
4
Explanation:
There are 4 pairs whose sum is less than 5: (2, 2), (2, 1), (3, 1), and (2, 1).

Input:
arr[] = [2, 1, 8, 3, 4, 7, 6, 5], target = 7
Output:
6
Explanation:
There are 6 pairs whose sum is less than 7: (2, 1), (2, 3), (2, 4), (1, 3), (1, 4), and (1, 5).

Constraints:

• $1 <= arr.size() <= 10^5$
• $0 <= arr[i] <= 10^4$
• $1 <= target <= 10^4$

🎯 My Approach:

1. Sorting and Two Pointers Technique:
   We can efficiently solve this problem using a two-pointer approach. By sorting the array first, we can use two pointers: one at the start of the array (l) and one at the end (r). We then check if the sum of the elements at these pointers is less than the target.

   • If the sum of arr[l] and arr[r] is less than the target, all pairs formed by arr[l] with any element between l and r will also be valid. Hence, we can add (r - l) to our result and move the left pointer (l) to the right.
   • If the sum is greater than or equal to the target, we move the right pointer (r) to the left.

2. Steps:

   • Sort the array.
   • Use two pointers: l = 0 and r = arr.size() - 1.
   • Traverse the array while l < r and check if arr[l] + arr[r] < target.
   • Adjust the pointers accordingly and count the pairs.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n log n), where n is the number of elements in the array. This comes from the sorting step, and the two-pointer traversal takes O(n) time.

• Expected Auxiliary Space Complexity: O(1), as we only use a constant amount of extra space (for the two pointers and the result variable).

📝 Solution Code

Code (C++)

class Solution {
public:
    int countPairs(vector<int>& arr, int target) {
        sort(arr.begin(), arr.end());
        int l = 0, r = arr.size() - 1, ans = 0;
        while (l < r) (arr[l] + arr[r] < target) ? ans += (r - l), l++ : r--;
        return ans;
    }
};

Code (Java)

class Solution {
    int countPairs(int[] arr, int target) {
        Arrays.sort(arr);
        int l = 0, r = arr.length - 1, ans = 0;
        while (l < r) if (arr[l] + arr[r] < target) ans += (r - l++);
        else r--;
        return ans;
    }
}

Code (Python)

class Solution:
    def countPairs(self, arr, target):
        arr.sort()
        l, r, ans = 0, len(arr) - 1, 0
        while l < r:
            if arr[l] + arr[r] < target: ans += (r - l); l += 1
            else: r -= 1
        return ans

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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