# Posts tagged as “prefix sum”

Given an array of positive integers nums, remove the smallest subarray (possibly empty) such that the sum of the remaining elements is divisible by p. It is not allowed to remove the whole array.

Return the length of the smallest subarray that you need to remove, or -1 if it’s impossible.

subarray is defined as a contiguous block of elements in the array.

Example 1:

Input: nums = [3,1,4,2], p = 6
Output: 1
Explanation: The sum of the elements in nums is 10, which is not divisible by 6. We can remove the subarray [4], and the sum of the remaining elements is 6, which is divisible by 6.

Example 2:

Input: nums = [6,3,5,2], p = 9
Output: 2
Explanation: We cannot remove a single element to get a sum divisible by 9. The best way is to remove the subarray [5,2], leaving us with [6,3] with sum 9.

Example 3:

Input: nums = [1,2,3], p = 3
Output: 0
Explanation: Here the sum is 6. which is already divisible by 3. Thus we do not need to remove anything.

Example 4:

Input: nums = [1,2,3], p = 7
Output: -1
Explanation: There is no way to remove a subarray in order to get a sum divisible by 7.

Example 5:

Input: nums = [1000000000,1000000000,1000000000], p = 3
Output: 0

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 109
• 1 <= p <= 109

## Solution: HashTable + Prefix Sum

Very similar to subarray target sum.

Basically, we are trying to find a shortest subarray that has sum % p equals to r = sum(arr) % p.

We use a hashtable to store the last index of the prefix sum % p and check whether (prefix_sum + p – r) % p exists or not.

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

## Python3

Given an array of positive integers arr, calculate the sum of all possible odd-length subarrays.

A subarray is a contiguous subsequence of the array.

Return the sum of all odd-length subarrays of arr.

Example 1:

Input: arr = [1,4,2,5,3]
Output: 58
Explanation: The odd-length subarrays of arr and their sums are:
[1] = 1
[4] = 4
[2] = 2
[5] = 5
[3] = 3
[1,4,2] = 7
[4,2,5] = 11
[2,5,3] = 10
[1,4,2,5,3] = 15
If we add all these together we get 1 + 4 + 2 + 5 + 3 + 7 + 11 + 10 + 15 = 58

Example 2:

Input: arr = [1,2]
Output: 3
Explanation: There are only 2 subarrays of odd length, [1] and [2]. Their sum is 3.

Example 3:

Input: arr = [10,11,12]
Output: 66

Constraints:

• 1 <= arr.length <= 100
• 1 <= arr[i] <= 1000

## Solution 0: Brute Force

Enumerate all odd length subarrys: O(n^2), each take O(n) to compute the sum.

Total time complexity: O(n^3)
Space complexity: O(1)

## Solution 1: Running Prefix Sum

Reduce the time complexity to O(n^2)

## Solution 2: Math

Count how many times arr[i] can be in of an odd length subarray
we chose the start, which can be 0, 1, 2, … i, i + 1 choices
we chose the end, which can be i, i + 1, … n – 1, n – i choices
Among those 1/2 are odd length.
So there will be upper((i + 1) * (n – i) / 2) odd length subarrays contain arr[i]

ans = sum(((i + 1) * (n – i) + 1) / 2 * arr[i] for in range(n))

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

## C++

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++

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)

## Python3

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)