3Sum Smaller

Problem Summary

• Given an array of unsorted numbers, count how many triplets can be formed where the sum is smaller than given target

Key Observations

• Similar to 3Sum, 3Sum Closest

Why This Works

• result += r - l -> key piece if arr[l] + arr[r] is valid then all elements from left to right are valid because they are in sorted order

Main Concepts Used

• Two Pointers

Time & Space Complexity

• Time: $O(n^2)$ → Reason: For each number we have to visit every other number
• Space: $O(n)$ → Reason: Sorting

Code

class Solution:
    def searchTriplets(self, arr, target):
        arr.sort()
        result = 0
        for i, firstNum in enumerate(arr):
            l, r = i + 1, len(arr) - 1
            while l < r:
                threeSum = firstNum + arr[l] + arr[r]
                if threeSum < target:
                    result += r - l
                    l += 1
                else:
                    r -= 1
        return result