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Posts tagged as “prefix sum”

花花酱 LeetCode 1563. Stone Game V

There are several stones arranged in a row, and each stone has an associated value which is an integer given in the array stoneValue.

In each round of the game, Alice divides the row into two non-empty rows (i.e. left row and right row), then Bob calculates the value of each row which is the sum of the values of all the stones in this row. Bob throws away the row which has the maximum value, and Alice’s score increases by the value of the remaining row. If the value of the two rows are equal, Bob lets Alice decide which row will be thrown away. The next round starts with the remaining row.

The game ends when there is only one stone remaining. Alice’s is initially zero.

Return the maximum score that Alice can obtain.

Example 1:

Input: stoneValue = [6,2,3,4,5,5]
Output: 18
Explanation: In the first round, Alice divides the row to [6,2,3], [4,5,5]. The left row has the value 11 and the right row has value 14. Bob throws away the right row and Alice's score is now 11.
In the second round Alice divides the row to [6], [2,3]. This time Bob throws away the left row and Alice's score becomes 16 (11 + 5).
The last round Alice has only one choice to divide the row which is [2], [3]. Bob throws away the right row and Alice's score is now 18 (16 + 2). The game ends because only one stone is remaining in the row.

Example 2:

Input: stoneValue = [7,7,7,7,7,7,7]
Output: 28

Example 3:

Input: stoneValue = [4]
Output: 0

Constraints:

  • 1 <= stoneValue.length <= 500
  • 1 <= stoneValue[i] <= 10^6

Solution: Range DP + Prefix Sum

dp[l][r] := max store Alice can get from range [l, r]
sum_l = sum(l, k), sum_r = sum(k + 1, r)
dp[l][r] = max{
dp[l][k] + sum_l if sum_l < sum_r
dp[k+1][r] + sum_r if sum_r < sum_l
max(dp[l][k], dp[k+1][r])) + sum_l if sum_l == sum_r)
} for k in [l, r)

Time complexity: O(n^3)
Space complexity: O(n^2)

C++

花花酱 LeetCode 1513. Number of Substrings With Only 1s

Given a binary string s (a string consisting only of ‘0’ and ‘1’s).

Return the number of substrings with all characters 1’s.

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

Example 1:

Input: s = "0110111"
Output: 9
Explanation: There are 9 substring in total with only 1's characters.
"1" -> 5 times.
"11" -> 3 times.
"111" -> 1 time.

Example 2:

Input: s = "101"
Output: 2
Explanation: Substring "1" is shown 2 times in s.

Example 3:

Input: s = "111111"
Output: 21
Explanation: Each substring contains only 1's characters.

Example 4:

Input: s = "000"
Output: 0

Constraints:

  • s[i] == '0' or s[i] == '1'
  • 1 <= s.length <= 10^5

Solution: DP / Prefix Sum

dp[i] := # of all 1 subarrays end with s[i].
dp[i] = dp[i-1] if s[i] == ‘1‘ else 0
ans = sum(dp)
s=1101
dp[0] = 1 // 1
dp[1] = 2 // 11, *1
dp[2] = 0 // None
dp[3] = 1 // ***1
ans = 1 + 2 + 1 = 5

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

C++

dp[i] only depends on dp[i-1], we can reduce the space complexity to O(1)

C++

Java

Python3


花花酱 LeetCode 1524. Number of Sub-arrays With Odd Sum

Given an array of integers arr. Return the number of sub-arrays with odd sum.

As the answer may grow large, the answer must be computed modulo 10^9 + 7.

Example 1:

Input: arr = [1,3,5]
Output: 4
Explanation: All sub-arrays are [[1],[1,3],[1,3,5],[3],[3,5],[5]]
All sub-arrays sum are [1,4,9,3,8,5].
Odd sums are [1,9,3,5] so the answer is 4.

Example 2:

Input: arr = [2,4,6]
Output: 0
Explanation: All sub-arrays are [[2],[2,4],[2,4,6],[4],[4,6],[6]]
All sub-arrays sum are [2,6,12,4,10,6].
All sub-arrays have even sum and the answer is 0.

Example 3:

Input: arr = [1,2,3,4,5,6,7]
Output: 16

Example 4:

Input: arr = [100,100,99,99]
Output: 4

Example 5:

Input: arr = [7]
Output: 1

Constraints:

  • 1 <= arr.length <= 10^5
  • 1 <= arr[i] <= 100

Solution: DP

We would like to know how many subarrays end with arr[i] have odd or even sums.

dp[i][0] := # end with arr[i] has even sum
dp[i][1] := # end with arr[i] has even sum

if arr[i] is even:

  dp[i][0]=dp[i-1][0] + 1, dp[i][1]=dp[i-1][1]

else:

  dp[i][1]=dp[i-1][0], dp[i][0]=dp[i-1][0] + 1

ans = sum(dp[i][1])

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

C++

Java

Python3

花花酱 LeetCode 1508. Range Sum of Sorted Subarray Sums

Given the array nums consisting of n positive integers. You computed the sum of all non-empty continous subarrays from the array and then sort them in non-decreasing order, creating a new array of n * (n + 1) / 2 numbers.

Return the sum of the numbers from index left to index right (indexed from 1), inclusive, in the new array. Since the answer can be a huge number return it modulo 10^9 + 7.

Example 1:

Input: nums = [1,2,3,4], n = 4, left = 1, right = 5
Output: 13 
Explanation: All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 + 2 + 3 + 3 + 4 = 13. 

Example 2:

Input: nums = [1,2,3,4], n = 4, left = 3, right = 4
Output: 6
Explanation: The given array is the same as example 1. We have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 3 to ri = 4 is 3 + 3 = 6.

Example 3:

Input: nums = [1,2,3,4], n = 4, left = 1, right = 10
Output: 50

Constraints:

  • 1 <= nums.length <= 10^3
  • nums.length == n
  • 1 <= nums[i] <= 100
  • 1 <= left <= right <= n * (n + 1) / 2

Solution 1: Brute Force

Find sums of all the subarrays and sort the values.

Time complexity: O(n^2logn)
Space complexity: O(n^2)

C++

Java

Python3

Solution 2: Priority Queue / Min Heap

For each subarray, start with one element e.g nums[i], put them into a priority queue (min heap). Each time, we have the smallest subarray sum, and extend that subarray and put the new sum back into priority queue. Thought it has the same time complexity as the brute force one in worst case, but space complexity can be reduce to O(n).

Time complexity: O(n^2logn)
Space complexity: O(n)

C++

Java

Python3

Solution 3: Binary Search + Sliding Window

Use binary search to find S s.t. that there are at least k subarrys have sum <= S.

Given S, we can use sliding window to count how many subarrays have sum <= S and their total sum.

ans = sums_of_first(right) – sums_of_first(left – 1).

Time complexity: O(n * log(sum(nums))
Space complexity: O(n)

C++

花花酱 LeetCode 1480. Running Sum of 1d Array

Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]).

Return the running sum of nums.

Example 1:

Input: nums = [1,2,3,4]
Output: [1,3,6,10]
Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].

Example 2:

Input: nums = [1,1,1,1,1]
Output: [1,2,3,4,5]
Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].

Example 3:

Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17]

Constraints:

  • 1 <= nums.length <= 1000
  • -10^6 <= nums[i] <= 10^6

Solution

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

C++