[LeetCode] 325. Maximum Size Subarray Sum Equals k

Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn't one, return 0 instead.

Note:
The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range.

Example 1:

Given nums = [1, -1, 5, -2, 3], k = 3,
return 4. (because the subarray [1, -1, 5, -2] sums to 3 and is the longest)

Example 2:

Given nums = [-2, -1, 2, 1], k = 1,
return 2. (because the subarray [-1, 2] sums to 1 and is the longest)

Follow Up:
Can you do it in O(n) time?

Thought process:

Get the prefix sums of the array. Then iterate the prefix sums using a decrementing length variable.

Solution 1 (Brute force):

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public class Solution { public int maxSubArrayLen(int[] nums, int k) { if (nums.length == 0) { return 0; } int[] prefixSum = new int[nums.length + 1]; for (int i = 1; i <= nums.length; i++) { prefixSum[i] = prefixSum[i - 1] + nums[i - 1]; } for (int i = nums.length; i >= 0; i--) { for (int j = prefixSum.length - 1; j - i >= 0; j--) { if (prefixSum[j] - prefixSum[j - i] == k) { return i; } } } return 0; } }

Time complexity: O(n^2).

Follow-up:
Iterate through the array. Calculate prefix sums as I iterate. Use a hash map to keep track of prefix sum -> index. For each prefix sum, check if map contains a key that equals prefix sum - k. If so, the length of the sub-array is index - map.get(prefixSum - k). Put current prefix sum -> index into the map only if prefix sum doesn't exist in the map yet. Initialize the hash map with an entry 0 -> -1.

Further thoughts: because I'm not using all of the prefix sums, I only need a variable to keep track of the previous sum.

Solution 2 (Hash table):

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class Solution { public int maxSubArrayLen(int[] nums, int k) { int max = 0; int sum = 0; Map<Integer, Integer> map = new HashMap<>(); map.put(0, -1); for (int i = 0; i < nums.length; i++) { sum += nums[i]; if (map.containsKey(sum - k)) { max = Math.max(max, i - map.get(sum - k)); } if (!map.containsKey(sum)) { map.put(sum, i); } } return max; } }

Time complexity: O(n).