4Sum

• Difficulty: Medium
• Primary pattern: Arrays & Strings
• Tags: Array, Two Pointers, Sorting
• Time taken: 10:14
• LeetCode Link

Key Idea

• Lock i and j, then move Left and Right
• Left and Right are dynamic, dont try and manage 4 pointers at ones, always lock to only 2 dynamic

Solution

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        # Inputs: List[int], int
        # Outputs: List[List[int]]
        # Description:
            # all unique quadruplets that nums[a] + nums[b] + nums[c] + nums[d] == target
            # 0 <= a, b, c, d < n
            # a, b, c, and d are distinct.
        # Sort according to size
        nums.sort()
        
        # Initialize
        output_list = []
        n = len(nums)

        # Iterate
        # Fix i
        for i in range(n-3):
            # Skip duplicate
            if i > 0 and nums[i] == nums[i-1]:
                continue
            # Fix j
            for j in range(i + 1, n -2):
                # Skip duplicate
                if j > i + 1 and nums[j] == nums[j - 1]:
                    continue
                # [ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 ]
                #   ^   ^   ^                       ^
                #   i   j   L                       R
                left = j + 1
                right = n - 1
                while left < right:
                    current_sum = nums[i] + nums[j] + nums[left] + nums[right]
                    if current_sum == target:
                        output_list.append([nums[i], nums[j], nums[left], nums[right]])
                        left += 1
                        right -= 1
                        # Move if it is duplicate digit
                        while left < right and nums[left] == nums[left - 1]:
                            left += 1
                        while left < right and nums[right] == nums[right + 1]:
                            right -= 1
                    elif current_sum < target:
                        left += 1
                    else:
                        right -= 1
        return output_list

Complexity

• Time: O(n^3) ``` first loop i -> O(n) second loop j -> O(n) two-pointer scan -> O(n)

Total: O(n × n × n) = O(n^3) Why not O(n^4)? Because after fixing i and j, left and right do one linear scan, not a nested loop. ```

• Space: O(1)