Problem Description

The problem asks us to find the maximum length of a subarray from the given array nums, such that the sum of its elements is exactly equal to the given integer k. A subarray is a contiguous part of an array. If such a subarray doesn't exist, we should return 0.

To clarify with an example, let's say our array is [1, -1, 5, -2, 3] and k is 3. The subarray [1, -1, 5, -2] sums to 3, and its length is 4, which would be the answer because there's no longer subarray that sums to 3.

To approach this problem, we need to find a way to efficiently look up the sum of elements from the beginning of the array up to a certain point, and determine if there is a previous point where the sum was exactly k less than the current sum. If we can find such a place, then the elements between these two points form a subarray that sums to k.

Intuition

The intuition behind the solution lies in the use of a running sum and a hash table (or dictionary in Python). As we iterate through the array, we keep track of the cumulative sum of elements. The hash table stores the earliest index at which each cumulative sum appears.

Here's the reasoning:

• Start with a hash table d that maps a cumulative sum to the first index where it appears. Initialize it with the pair (0: -1), which says the sum 0 is achieved before the first element.
• Initialize variables for the current sum (s) and the maximum length found (ans) to 0.
• Iterate through nums, updating the running total s by adding each element x to it.
• Check if s - k has been seen before. If it has, we've found a subarray ending at the current index with a sum of k.
• Update ans with the length of this subarray if it is longer than the maximum found so far.
• Add the current sum to the hash table with its corresponding index, but only if this sum hasn't been seen before to maintain the earliest index.

The reason we only store the earliest occurrence of a sum is because we're looking for the longest subarray; any later occurrence of the same sum would yield a shorter subarray.

Learn more about Prefix Sum patterns.

Solution Approach

The solution leverages a hashing strategy to efficiently track the sum of elements in the subarrays and their starting indices. Here's a detailed walkthrough of the implementation:

1. Hash Table Initialization: We initiate a hash table (dictionary in Python) named d with a key-value pair {0: -1}. This represents that the sum 0 is achieved before starting the iterations, essentially before the first element at index -1. This base case ensures that if the sum of elements from the beginning reaches k, the length can be calculated correctly.

2. Preparing Variables: Two variables are prepared: ans to store the maximum length of the subarray with sum k found so far and s to keep track of the running sum.

3. Iteration & Running Sum: We use a loop to iterate over the array using enumerate(nums) which gives us both the index i and the value x. We increment the running sum s by the value of x.

4. Checking for a Matching Subarray: During each iteration, after updating the running sum, we check if s - k is present in the hash table d. If it is, it means that from the earliest index d[s - k] to the current index i, the elements sum up exactly to k. We then compare (i - d[s - k]) with ans and store the larger one in ans.

5. Updating the Hash Table: The hash table is updated with the running sum only if this running sum has not been recorded before. We do this because we are interested in the first occurrence of this running sum to ensure the longest possible subarray.

6. Returning the Result: After the loop ends, ans will be holding the length of the longest subarray that sums to k. This value is returned as the final result.

This algorithm effectively uses the hashing technique to reduce the time complexity otherwise inevitable with brute force solutions. By using a hash table to record the first occurrence of each sum, the solution approaches an O(n) time complexity, since it processes each element of nums just once. The space complexity is also O(n) in the worst case, as it might store each sum occurring in the array.

Example Walkthrough

Let's use the solution approach to walk through a small example. Consider the array nums = [3, 4, -3, 2, 1] and the target sum k = 3.

1. Hash Table Initialization: Initialize a hash table d with {0: -1}. This signifies that before we start, the cumulative sum of 0 is reached at an index before the array starts.

2. Preparing Variables: The variables ans (maximum subarray length) and s (running sum) are both set to 0.

3. Iteration & Running Sum:

   • For the first element 3, the running sum s becomes 3. Since s - k = 3 - 3 = 0 and the hash table contains 0 with index -1, we update ans to 1 - (-1) = 2.
   • For the second element 4, s becomes 7. We look for s - k = 4 in the hash table, but it's not there, so no change to ans.
   • For the third element -3, s is now 4. We look for s - k = 1 which is not in the hash table, non update to ans.
   • For the fourth element 2, s becomes 6. We look for s - k = 3, and since it's not present, no update to ans.
   • For the last element 1, s increases to 7. We look for s - k = 4. It has appeared before when s was 7 for the second element. The index then was 1. Now, the index is 4, so the length of the subarray is 4 - 1 = 3 which is greater than the current ans, so we update ans to 3.

4. Checking for a Matching Subarray: Every iteration includes this check. We successfully find a matching sum during the first and last elements.

5. Updating the Hash Table: As we proceed, we update the hash table with the sums 3, 7, 4, and 6 at their respective first occurrences with the indices 0, 1, 2, and 3.

6. Returning the Result: At the end of the iteration, ans holds the value 3, which is the length of the longest subarray with sum k = 3. This is our final result.

The correct subarray that has the sum of 3 and the maximum length is [4, -3, 2]. The running sum at the start of this subarray was 4, and by adding the subarray elements, it became 7. The subarray has 3 elements, hence ans = 3 is returned.

Solution Implementation

Time and Space Complexity

The provided Python code finds the length of the longest subarray which sums up to k. It makes use of a hashmap (dictionary in Python terms) to store the cumulative sum at each index, which helps in finding subarrays that sum up to k in constant time.

Time Complexity

The time complexity of the code is O(n), where n is the number of elements in the input list nums. This is because the algorithm iterates through the list once, and for each element it performs a constant time operation of adding the element to the running sum, checking if (s - k) is in the dictionary d and updating the maximum length ans. All these operations are constant time operations: arithmetic operations, dictionary lookup and dictionary insertion (when s is not already in d).

Space Complexity

The space complexity of the code is O(n). In the worst case, every running sum is unique, and hence, each sum (represented by s in the code) would be stored in the dictionary d along with its corresponding index. Since there are n such running sums in the worst case, the dictionary could contain up to n key-value pairs.

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