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.
Example 1:
Given nums =
return
[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 =
return
[-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?
Understand the problem:Can you do it in O(n) time?
The problem is equal to: find out a range from i to j, in which the sum (nums[i], ..., nums[j]) = k. What is the maximal range?
So we can first calculate the prefix sum of each number, so sum(i, j) = sum(j) - sum(i - 1) = k. Therefore, for each sum(j), we only need to check if there was a sum(i - 1) which equals to sum(j) - k. We can use a hash map to store the previous calculated sum.
Code (Java):
public class Solution { public int maxSubArrayLen(int[] nums, int k) { if(nums == null || nums.length == 0) { return 0; } int maxLen = 0; Map<Integer, Integer> map = new HashMap<>(); map.put(0, -1); // IMPOARTANT int sum = 0; for (int i = 0; i < nums.length; i++) { sum += nums[i]; if (!map.containsKey(sum)) { map.put(sum, i); } if (map.containsKey(sum - k)) { maxLen = Math.max(maxLen, i - map.get(sum - k)); } } return maxLen; } }
Comments:
Note the map.put(0, -1). We need to put this entry into the map before, because if the maximal range starts from 0, we need to calculate sum(j) - sum(i - 1).
Good Solution!
ReplyDeleteGenius.
ReplyDeleteHi,
ReplyDeleteif array = {1, 2, 1, 0, 1, 2, 1}
and k = 3
it should return 5 (longest sub-array is from 1 included to 5 included)
but it returns 3
indexes 3 to 5, using a zero-base (i3=0, i4=1, i5=2 -> 0+1+2 -> 3)
DeleteThis comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteDictionary map = new Dictionary();
ReplyDeleteint sum = 0, maxLength = 0;
for (int i = 0; i < nums.Length; i++)
{
sum += nums[i];
if (sum == k)
{
maxLength = i + 1;
}
if (!map.ContainsKey(sum))
{
map.Add(sum, i);
}
if (map.ContainsKey(sum - k))
{
maxLength = Math.Max(maxLength, i - map[sum - k]);
}
}
return maxLength;
What a genius!!! Thank you for your clear explanation! I am from the future 2021, and thank you a random stranger from 2016!!
ReplyDelete