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Posts tagged as “subarray”

花花酱 LeetCode 1343. Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold

Given an array of integers arr and two integers k and threshold.

Return the number of sub-arrays of size k and average greater than or equal to threshold.

Example 1:

Input: arr = [2,2,2,2,5,5,5,8], k = 3, threshold = 4
Output: 3
Explanation: Sub-arrays [2,5,5],[5,5,5] and [5,5,8] have averages 4, 5 and 6 respectively. All other sub-arrays of size 3 have averages less than 4 (the threshold).

Example 2:

Input: arr = [1,1,1,1,1], k = 1, threshold = 0
Output: 5

Example 3:

Input: arr = [11,13,17,23,29,31,7,5,2,3], k = 3, threshold = 5
Output: 6
Explanation: The first 6 sub-arrays of size 3 have averages greater than 5. Note that averages are not integers.

Example 4:

Input: arr = [7,7,7,7,7,7,7], k = 7, threshold = 7
Output: 1

Example 5:

Input: arr = [4,4,4,4], k = 4, threshold = 1
Output: 1

Constraints:

  • 1 <= arr.length <= 10^5
  • 1 <= arr[i] <= 10^4
  • 1 <= k <= arr.length
  • 0 <= threshold <= 10^4

Solution: Sliding Window

  1. Window size = k
  2. Maintain the sum of the window
  3. Check sum >= threshold * k

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

C++

花花酱 LeetCode 1186. Maximum Subarray Sum with One Deletion

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 [3] 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++

花花酱 LeetCode 718. Maximum Length of Repeated Subarray

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(mn)

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

花花酱 LeetCode 209. Minimum Size Subarray Sum

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++

花花酱 LeetCode 974. Subarray Sums Divisible by K

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

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

C++

Python3