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## Maximum Size subarray problem

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)

## Algorithm

The brute force solution is straightforward, try all possible combinations:

``````for index i ~ (0, n):
for index j ~ (i, n):
if sum(i, j) == k
minLen = min(minLen, j - i + 1);
``````

This algorithm takes O(n ^ 2) time, which is too slow.

The second solution I thought of was a `sliding window algorithm`.

``````while (right < boundary):
sum += nums[right++]
while (sum == target):
minLen = min(minLen, right - left)
sum -= nums[left++]
``````

Sliding window does not work for this problem because:

• Skipped part of the window may also contribute to a maxSize subarray.
• Move right does not make the sum bigger, move left does not make the sum smaller. because number can be positive or negative.

A better solution is based on the following idea:

• sum(i, j) = sum(0, j) - sum(i)
• sum(0, i), sum(0, j) can be computed incrementally

The following diagram better illustrate the idea:

So we have the following algorithm:

``````for i ~ (0, n):
sum[i] = sum[i - 1] + num[i]
gap = sum[i] - target
if gap in map:
maxWindow = Math.max(maxWindow, i - map(gap))

if sum[i] not in map:
map.put(sum[i], i)
``````

## Implementation

The following code implemented the algorithm above:

``````public int maxSubArrayLen(int[] nums, int k) {
if (nums == null || nums.length == 0) {
return 0;
}

int sum = 0, maxLen = 0;
Map<Integer, Integer> sumIndex = new HashMap<Integer, Integer>();
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
if (sum == k) {
maxLen = i + 1;
} else if (sumIndex.containsKey(sum - k)) {
maxLen = Math.max(maxLen, i - sumIndex.get(sum - k));
}
if (!sumIndex.containsKey(sum)) {
sumIndex.put(sum, i);
} // not update index because we want max sum
}
return maxLen;
}``````
• Runtime: O(n)