# Posts tagged as “subarray”

Given an array of integers, return the maximum sum for a non-empty subarray (contiguous elements) with at most one element deletion. In other words, you want to choose a subarray and optionally delete one element from it so that there is still at least one element left and the sum of the remaining elements is maximum possible.

Note that the subarray needs to be non-empty after deleting one element.

Example 1:

Input: arr = [1,-2,0,3]
Output: 4
Explanation: Because we can choose [1, -2, 0, 3] and drop -2, thus the subarray [1, 0, 3] becomes the maximum value.

Example 2:

Input: arr = [1,-2,-2,3]
Output: 3
Explanation: We just choose  and it's the maximum sum.


Example 3:

Input: arr = [-1,-1,-1,-1]
Output: -1
Explanation: The final subarray needs to be non-empty. You can't choose [-1] and delete -1 from it, then get an empty subarray to make the sum equals to 0.


Constraints:

• 1 <= arr.length <= 10^5
• -10^4 <= arr[i] <= 10^4

Solution: DP

First, handle the special case: all numbers are negative, return the max one.

s0 = max subarray sum ends with a[i]
s1 = max subarray sum ends with a[i] with at most one deletion

whenever s0 or s1 becomes negative, reset them to 0.

Time complexity: O(n)
Space complexity: O(1)

## C++

iven two integer arrays A and B, return the maximum length of an subarray that appears in both arrays.

Example 1:

Input:
A: [1,2,3,2,1]
B: [3,2,1,4,7]
Output: 3
Explanation:
The repeated subarray with maximum length is [3, 2, 1].


Note:

1. 1 <= len(A), len(B) <= 1000
2. 0 <= A[i], B[i] < 100

## Solution: DP

dp[i][j] := max length of (A[0:i], B[0:j])

dp[i][j] = dp[i – 1][j – 1] + 1 if A[i-1] == B[j-1] else 0

Time complexity: O(m*n)
Space complexity: O(m*n) -> O(n)

## C++ S:O(min(m,n))

Given an array of n positive integers and a positive integer s, find the minimal length of a contiguoussubarray of which the sum ≥ s. If there isn’t one, return 0 instead.

Example:

Input: s = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: the subarray [4,3] has the minimal length under the problem constraint.

Follow up:If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).

Solution 1: Two Pointers (Sliding Window)

Maintain a sliding window [l, r) such that sum(nums[l:r)) >= s, then move l to l + 1, and move r accordingly to make the window valid.

Time complexity: O(n)
Space complexity: O(1)

## C++

Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K.

Example 1:

Input: A = [4,5,0,-2,-3,1], K = 5 Output: 7 Explanation: There are 7 subarrays with a sum divisible by K = 5: [4, 5, 0, -2, -3, 1], , [5, 0], [5, 0, -2, -3], , [0, -2, -3], [-2, -3]

Note:

1. 1 <= A.length <= 30000
2. -10000 <= A[i] <= 10000
3. 2 <= K <= 10000

Solution: Count prefix sums

let c[i] denotes the counts of prefix_sum % K init: c = 1
Whenever we end up with the same prefix sum (after modulo), which means there are subarrys end with current element that is divisible by K (0 modulo).

e.g. A = [4,5,0,-2,-3,1], K = 5
[4,5] has prefix sum of 4, which happens at index 0 , and index 1, [4,5]
[4,5,0] also has a prefix sum of 4, which means [4, {5,0}], [4,5, {0}] are divisible by K.

ans += (c[prefix_sum] – 1)
i = 0, prefix_sum = 0, c[(0+4)%5] = c = 1, ans = 0
i = 1, prefix_sum = 4+5, c[(4+5)%5] = c = 2, ans = 0+2-1=0 => 
i = 2, prefix_sum = 4+0, c[(4+0)%5] = c = 3, ans = 1+3-1=3 => , [5,0], 
i = 3, prefix_sum = 4-2, c[(4-2)%5] = c = 1, ans = 3
i = 4, prefix_sum = 2-3, c[(2-3+5)%5] = c = 4, ans = 3+4-1=6 => ,[5,0],,[5,0,-2,-3], [0,-2,-3],[-2,-3]
i = 5, prefix_sum = 4+1, c[(4+1)%5] = c = 2, ans = 6 + 2 – 1 =>
,[5,0],,[5,0,-2,-3], [0,-2,-3],[-2,-3], [4,5,0,-2,-3,1]

Time complexity: O(n)
Space complexity: O(n)

# Problem

Given an array of integers A, find the sum of min(B), where B ranges over every (contiguous) subarray of A.

Since the answer may be large, return the answer modulo 10^9 + 7.

Example 1:

Input: [3,1,2,4]
Output: 17
Explanation: Subarrays are , , , , [3,1], [1,2], [2,4], [3,1,2], [1,2,4], [3,1,2,4].
Minimums are 3, 1, 2, 4, 1, 1, 2, 1, 1, 1.  Sum is 17.

Note:

1. 1 <= A.length <= 30000
2. 1 <= A[i] <= 30000

# Idea

1. order matters, unlike 花花酱 LeetCode 898. Bitwise ORs of Subarrays we can not sort the numbers in this problem.
1. e.g. sumSubarrayMins([3, 1, 2, 4]) !=sumSubarrayMins([1, 2, 3, 4]) since the first one will not have a subarray of [3,4].
2. For A[i], assuming there are L numbers that are greater than A[i] in range A ~ A[i – 1], and there are R numbers that are greater or equal than A[i] in the range of A[i+1] ~ A[n – 1]. Thus A[i] will be the min of (L + 1) * (R + 1) subsequences.
1. e.g. A = [3,1,2,4], A = 1, L = 1, R = 2, there are (1 + 1) * (2 + 1) = 6 subsequences are 1 is the min number. [3,1], [3,1,2], [3,1,2,4], , [1,2], [1,2,4]

# Solution 1: Brute Force

Time complexity: O(n^2)

Space complexity: O(1)

# Solution2 : Monotonic Stack

Time complexity: O(n)

Space complexity: O(n)

We can use a monotonic stack to compute left[i] and right[i] similar to 花花酱 LeetCode 901. Online Stock Span

## Python3 V2

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