Problem Description

The LeetCode problem asks us to find three numbers within an array nums such that their sum is closest to a given target value. The array nums has a length n, and it's guaranteed that there is exactly one solution for each input.

To solve this problem, we must search for triplets in the array whose sum has the smallest absolute difference from the target. The final result is not the triplet itself, but rather the sum of its components.

Intuition

The intuition behind the provided solution leverages a sorted array and a two-pointer technique for efficient searching. Here's a step-by-step breakdown of the approach:

1. Sorting: We first sort the nums array. Sorting is crucial because it allows us to efficiently sweep through the array using two pointers and easily adjust the sum of the current triplet to get closer to the target.

2. Iterating with Three Pointers: We use a for-loop to iterate over the array with an index i representing the first number of the triplet. The other two pointers, j and k, are initialized to the next element after i and the last element of the array, respectively. The three numbers represented by i, j, and k are our current candidates for the closest sum.

3. Evaluating Sums and Moving Pointers: In the while-loop, we calculate the sum of the triplet (t) and compare it to the target. If the sum exactly matches the target, we immediately return it.

   If the sum doesn't match, we compare the absolute difference of t and target with the current closest sum (ans), and if it's smaller, we update ans.

   To search through different sums, we move pointers j and k depending on whether the current sum is greater than or less than target. If it's greater, we decrement k to reduce the sum. If it's less, we increment j to increase the sum. This is possible without missing the potential answers because the array is sorted.

By analyzing the difference between t and target and adjusting j and k accordingly, we can efficiently find the triplet with the sum that has the smallest absolute difference to the target.

4. Returning the Answer: Once we've finished iterating through all possible triplets, we return the closest sum recorded as ans.

Learn more about Two Pointers and Sorting patterns.

Solution Approach

The algorithm implementation utilizes several important concepts and patterns:

1. Sorting: The array is initially sorted to simplify the searching process. By having the array in ascending order, we can make use of the two-pointer technique effectively.

2. Two-Pointer Technique: After fixing one number of the potential triplet using the first for-loop, the two other numbers are controlled by two pointers. One starts right after the fixed element (j), while the other starts from the end of the array (k). These pointers move closer to each other as they iterate.

3. Avoiding Redundancy: Since each number in nums is used as a starting point for a triplet, the solution avoids re-examining numbers that are identical to the previous starting point (i position) by skipping duplicate combinations. (This is implicit and can be added to optimization in the code if necessary by checking for the same numbers when incrementing i).

4. Closest Sum Calculation: As the pointers j and k move, the solution calculates the sum of the three numbers, compares it with the target, and keeps track of the closest sum encountered by comparing the absolute differences with the current best answer (ans).

5. Conditional Pointer Movement: Based on whether the current sum is greater than or less than target, k is decremented or j is incremented respectively. This allows the solution to narrow down the closest sum without checking all possible triplet combinations, which would result in higher computational complexity.

6. Early Termination: If at any point the sum equals the target, the loop breaks and returns the sum immediately since it cannot get any closer than an exact match.

7. Return Statement: After going through all possible iterations, the algorithm returns the sum stored in ans, which is the closest sum to target that the solution could find during the iteration process.

The code uses a simple inf value to initialize ans so that any other sum found will be closer compared to infinity. Utilizing this approach, the data structure (a single array) is kept simple, and the algorithm achieves a time complexity of O(n^2), since it only involves iterating over the array of length n and adjusting the pointers without nested full iterations.

Example Walkthrough

Let's consider a small example to illustrate the solution approach using the strategy detailed in the content provided.

Suppose we have the following nums array and target:

nums = [-1, 2, 1, -4]
target = 1

Firstly, according to our approach, we need to sort the nums:

sorted_nums = [-4, -1, 1, 2]

Now we iterate through sorted_nums with our three pointers. For simplicity, I'll walk through the first complete iteration:

1. Set i = 0, which is the value -4. This is our first number of the potential triplet.
2. Initialize two other pointers, j = i + 1 => j = 1 (value -1) and k = n - 1 => k = 3 (value 2).
3. We then enter a while loop with j < k and calculate the sum t using sorted_nums[i], sorted_nums[j], and sorted_nums[k]. So, our first sum is t = (-4) + (-1) + (2) = -3.
4. Since our goal is to get the sum close to target (1), we check the absolute difference between t and target. Abs(-3 - 1) = Abs(-4) = 4.
5. We initialize our answer ans with infinity. Our first comparison will set ans = -3 as it's the closest sum we've encountered.
6. t is less than target, we increment j to increase the sum. Now, j = 2 (value 1).
7. Calculate a new sum: t = (-4) + (1) + (2) = -1.
8. Compare the new sum's absolute difference with target. Abs(-1 - 1) = 2, which is closer to the target than our previous best of 4. Update ans to -1.
9. Since -1 is still less than the target, we increment j once again. Now, j = 3 which is equal to k, so we break out of the while-loop.
10. The loop for the index i continues, but for the sake of brevity, let's assume the remaining iterations do not find a sum closer to the target than the sum when ans was -1.

With these steps completed, we assert that the closest sum to the target we have found using this method is -1. As per our algorithm's implementation, -1 will be the final answer returned.

Note that for a real implementation, the process would involve iterating over all valid i, j, and k combinations until the end of the array is reached, continuously updating ans if closer sums are found, and potentially breaking early if an exact match is found. However, this simple example serves to convey the essentials of the solution approach.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given function depends on a few distinct steps:

1. Sorting the array: The sort() method in Python uses the Timsort algorithm, which has a time complexity of O(n log n), where n is the length of the list being sorted.

2. The for-loop: The loop runs for each element in the sorted array, resulting in a complexity of O(n) iterations.

3. The while-loop inside the for-loop: In the worst case, for each iteration of the for-loop, the while-loop can perform nearly O(n) operations since it might iterate from i+1 to the n-1 index.

Combining these complexities, the first step is dominant if n is large. However, since the nested loop inside the for-loop could potentially run n times for each n iterations of the for-loop, the resulting time complexity is O(n^2), since n(n-1)/2 simplifies to O(n^2). This nested loop is the most influential factor for large n.

So, the overall time complexity of the algorithm is O(n log n) + O(n^2), which simplifies to O(n^2) since the O(n^2) term is dominant for large n.

Space Complexity

The space complexity is O(1) if we ignore the space used for input and output since the sorting is done in-place and only a fixed number of variables are used, which does not scale with the size of the input array.

Learn more about how to find time and space complexity quickly using problem constraints.