The LeetCode 16 Solution is a great way to improve your problem-solving skills on LeetCode by tackling the 3Sum Closest problem. This challenge requires you to find three numbers in an array that sum up to a value closest to a given target. While it may seem difficult at first, there’s an efficient approach that’s both easy to understand and implement. In this blog post, we’ll dive into the problem, discuss an optimal solution, and break it down step by step.
Understanding the 3Sum Closest Problem
The 3Sum Closest problem asks you to select three integers from an array in such a way that their sum is as close as possible to a given target. The goal is to return the sum of those three integers. Importantly, you are guaranteed that there is exactly one solution, which means you don’t have to worry about ambiguous results.
Here’s an example to illustrate the problem:
- Input:
nums = [-1, 2, 1, -4],target = 1 - Output:
2
In this case, the three numbers -1 + 2 + 1 give us a sum of 2, which is the closest possible sum to the target 1.
How to Approach the Problem
To solve LeetCode 16 (3Sum Closest) efficiently, we need a method that works well with large arrays. A brute-force approach would check every possible combination of three numbers. However, this would be too slow, especially when the array has hundreds of elements. Instead, we can use a two-pointer technique along with sorting the array. This approach allows us to solve the problem much faster.
The basic idea of the solution is to:
- Sort the array, which helps us organize the elements and makes the two-pointer strategy possible.
- Iterate through the array, fixing one number and using two pointers to find the best pair of numbers that, together with the fixed number, come closest to the target.
- Continuously track the closest sum, adjusting the pointers based on whether the current sum is smaller or larger than the target.
LeetCode 16 Solution Using the Two-Pointer Technique
Sorting the array and using two pointers reduces the time complexity of the problem to (O(n^2)), making it efficient for large inputs.
Let’s break down the solution with a simple Python implementation:
def threeSumClosest(nums, target):
# Step 1: Sort the array
nums.sort()
# Initialize with a large value
closest_sum = float('inf')
# Step 2: Iterate through the array
for i in range(len(nums) - 2):
# Two pointers
left, right = i + 1, len(nums) - 1
# Step 3: Use the two-pointer technique
while left < right:
current_sum = nums[i] + nums[left] + nums[right]
if abs(current_sum - target) < abs(closest_sum - target):
# Update closest sum if found better
closest_sum = current_sum
if current_sum < target:
# Move left pointer to the right for a larger sum
left += 1
elif current_sum > target:
# Move right pointer to the left for a smaller sum
right -= 1
else:
# Exact match found, return immediately
return current_sum
# Return the closest sum found
return closest_sumHow the LeetCode 16 Solution Works
The 3Sum Closest solution begins by sorting the array. Sorting allows us to efficiently adjust our two pointers (left and right), which helps in finding the closest sum without having to try every possible combination.
For each element in the array, we fix that element and then move the two pointers—left (which starts from the element right after the fixed one) and right (which starts from the end of the array). As we calculate the sum of the fixed element and the two pointed elements, we compare how close that is to the target. If it is closer than any previously found sum, we update our result.
If the sum is less than the target, it means we need a larger sum, so we move the left pointer to the right. If the sum is greater than the target, we move the right pointer to the left to reduce the sum. This process continues until we find the best possible solution.
Edge Cases to Consider
There are a few scenarios that you need to consider when solving the 3Sum Closest problem:
- If the array has exactly three numbers, the answer will simply be the sum of those numbers.
- Arrays that contain all zeros (like
nums = [0, 0, 0]) will result in a sum of zero, no matter what the target is. - Very large or very small target values must also be handled. Since the array contains both negative and positive numbers, the algorithm is designed to handle this variation effectively.
Time and Space Complexity
The time complexity of the solution is dominated by the sorting step and the two-pointer technique. Sorting the array takes (O(n log n)), and for each element, the two-pointer scan runs in (O(n)), making the overall time complexity (O(n^2)). This is efficient enough for arrays up to the size of 500 elements, as allowed by the problem’s constraints.
In terms of space complexity, the solution only requires (O(1)) extra space aside from the input list, making it space-efficient as well.
LeetCode 16 Solution: Summary
The LeetCode 16 Solution for the 3Sum Closest problem shows how sorting and the two-pointer technique can simplify complex problems. This efficient approach helps avoid the pitfalls of brute-force solutions. It also ensures that the algorithm runs quickly, even with larger inputs.
By using this method, you’re not just solving this particular problem. You’re also learning how to handle similar problems that involve searching for specific sums in arrays. Mastering these techniques will give you the confidence to tackle many other array-based problems on LeetCode.
Happy Coding!
Looking for more LeetCode solutions? Stay tuned for upcoming guides on tackling other LeetCode challenges in Python. Whether you’re preparing for technical interviews or aiming to improve your coding skills, exploring a variety of problems will enhance your understanding and proficiency.
You can find this problem on LeedCode, click here.
