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Short simpler solution python O(n^2)
Isabela Vlls
Nov 26, 2024
class Solution: def searchTriplets(self, arr, target): count = 0 arr.sort() for i in range(len(arr)-2): left, right = i + 1, len(arr)-1 # pointers while left < right: sum_elements = arr[i] + arr[left] + arr[right] if sum_elements < target: # if it's small then all to the left are valid count += (right - left) left += 1 else: right -= 1 return count
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