0325. Maximum Size Subarray Sum Equals k
https://leetcode.com/problems/maximum-size-subarray-sum-equals-k
Description
Given an integer array nums
and an integer k
, return the maximum length of a subarray that sums to k
. If there isn't one, return 0
instead.
Example 1:
**Input:** nums = [1,-1,5,-2,3], k = 3
**Output:** 4
**Explanation:** The subarray [1, -1, 5, -2] sums to 3 and is the longest.
Example 2:
**Input:** nums = [-2,-1,2,1], k = 1
**Output:** 2
**Explanation:** The subarray [-1, 2] sums to 1 and is the longest.
Constraints:
1 <= nums.length <= 2 * 105
-104 <= nums[i] <= 104
-109 <= k <= 109
ac
range sum, current sum - old sum = k. Why 2 pointers fail here? because cannot decide how to move the pointers.
class Solution {
public int maxSubArrayLen(int[] nums, int k) {
// edge cases
if (nums == null || nums.length == 0) return 0;
Map<Integer, Integer> map = new HashMap<>();
int sum = 0, max = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
if (sum == k) max = i + 1;
int oldSum = sum - k; // currentSum - oldSum = k
if (map.containsKey(oldSum)) max = Math.max(max, i - map.get(oldSum));
if (!map.containsKey(sum)) map.put(sum, i); // if contains before, skip
}
return max == Integer.MIN_VALUE ? 0 : max;
}
}
/*
edge cases
hashmap
*/
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