You are given a string s, where every two consecutive vertical bars '|' are grouped into a pair. In other words, the 1st and 2nd '|' make a pair, the 3rd and 4th '|' make a pair, and so forth.

Return the number of '*' in sexcluding the '*' between each pair of '|'.

Note that each '|' will belong to exactly one pair.

Example 1:

Input: s = "l|*e*et|c**o|*de|"
Output: 2
Explanation: The considered characters are underlined: "l|*e*et|c**o|*de|".
The characters between the first and second '|' are excluded from the answer.
Also, the characters between the third and fourth '|' are excluded from the answer.
There are 2 asterisks considered. Therefore, we return 2.

Example 2:

Input: s = "iamprogrammer"
Output: 0
Explanation: In this example, there are no asterisks in s. Therefore, we return 0.


Example 3:

Input: s = "yo|uar|e**|b|e***au|tifu|l"
Output: 5
Explanation: The considered characters are underlined: "yo|uar|e**|b|e***au|tifu|l". There are 5 asterisks considered. Therefore, we return 5.

Constraints:

• 1 <= s.length <= 1000
• s consists of lowercase English letters, vertical bars '|', and asterisks '*'.
• s contains an even number of vertical bars '|'.

## Solution: Counting

Count the number of bars so far, and only count ‘*’ when there are even number of bars on the left.

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

## C++

You are given a 0-indexed m x n integer matrix grid consisting of distinct integers from 0 to m * n - 1. You can move in this matrix from a cell to any other cell in the next row. That is, if you are in cell (x, y) such that x < m - 1, you can move to any of the cells (x + 1, 0)(x + 1, 1), …, (x + 1, n - 1)Note that it is not possible to move from cells in the last row.

Each possible move has a cost given by a 0-indexed 2D array moveCost of size (m * n) x n, where moveCost[i][j] is the cost of moving from a cell with value i to a cell in column j of the next row. The cost of moving from cells in the last row of grid can be ignored.

The cost of a path in grid is the sum of all values of cells visited plus the sum of costs of all the moves made. Return the minimum cost of a path that starts from any cell in the first row and ends at any cell in the last row.

Example 1:

Input: grid = [[5,3],[4,0],[2,1]], moveCost = [[9,8],[1,5],[10,12],[18,6],[2,4],[14,3]]
Output: 17
Explanation: The path with the minimum possible cost is the path 5 -> 0 -> 1.
- The sum of the values of cells visited is 5 + 0 + 1 = 6.
- The cost of moving from 5 to 0 is 3.
- The cost of moving from 0 to 1 is 8.
So the total cost of the path is 6 + 3 + 8 = 17.


Example 2:

Input: grid = [[5,1,2],[4,0,3]], moveCost = [[12,10,15],[20,23,8],[21,7,1],[8,1,13],[9,10,25],[5,3,2]]
Output: 6
Explanation: The path with the minimum possible cost is the path 2 -> 3.
- The sum of the values of cells visited is 2 + 3 = 5.
- The cost of moving from 2 to 3 is 1.
So the total cost of this path is 5 + 1 = 6.


Constraints:

• m == grid.length
• n == grid[i].length
• 2 <= m, n <= 50
• grid consists of distinct integers from 0 to m * n - 1.
• moveCost.length == m * n
• moveCost[i].length == n
• 1 <= moveCost[i][j] <= 100

## Solution: DP

Let dp[i][j] := min cost to reach grid[i][j] from the first row.

dp[i][j] = min{grid[i][j] + dp[i – 1][k] + moveCost[grid[i – 1][k]][j]} 0 <= k < n

For each node, try all possible nodes from the previous row.

Time complexity: O(m*n2)
Space complexity: O(m*n) -> O(n)

## C++

You are given a 0-indexed 2D integer array brackets where brackets[i] = [upperi, percenti] means that the ith tax bracket has an upper bound of upperi and is taxed at a rate of percenti. The brackets are sorted by upper bound (i.e. upperi-1 < upperi for 0 < i < brackets.length).

Tax is calculated as follows:

• The first upper0 dollars earned are taxed at a rate of percent0.
• The next upper1 - upper0 dollars earned are taxed at a rate of percent1.
• The next upper2 - upper1 dollars earned are taxed at a rate of percent2.
• And so on.

You are given an integer income representing the amount of money you earned. Return the amount of money that you have to pay in taxes. Answers within 10-5 of the actual answer will be accepted.

Example 1:

Input: brackets = [[3,50],[7,10],[12,25]], income = 10
Output: 2.65000
Explanation:
The first 3 dollars you earn are taxed at 50%. You have to pay $3 * 50% =$1.50 dollars in taxes.
The next 7 - 3 = 4 dollars you earn are taxed at 10%. You have to pay $4 * 10% =$0.40 dollars in taxes.
The final 10 - 7 = 3 dollars you earn are taxed at 25%. You have to pay $3 * 25% =$0.75 dollars in taxes.
You have to pay a total of $1.50 +$0.40 + $0.75 =$2.65 dollars in taxes.


Example 2:

Input: brackets = [[1,0],[4,25],[5,50]], income = 2
Output: 0.25000
Explanation:
The first dollar you earn is taxed at 0%. You have to pay $1 * 0% =$0 dollars in taxes.
The second dollar you earn is taxed at 25%. You have to pay $1 * 25% =$0.25 dollars in taxes.
You have to pay a total of $0 +$0.25 = $0.25 dollars in taxes.  Example 3: Input: brackets = [[2,50]], income = 0 Output: 0.00000 Explanation: You have no income to tax, so you have to pay a total of$0 dollars in taxes.


Constraints:

• 1 <= brackets.length <= 100
• 1 <= upperi <= 1000
• 0 <= percenti <= 100
• 0 <= income <= 1000
• upperi is sorted in ascending order.
• All the values of upperi are unique.
• The upper bound of the last tax bracket is greater than or equal to income.

“Nothing is certain except death and taxes” – Benjamin Franklin

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

## C++

You are given a 0-indexed integer array nums of length n.

nums contains a valid split at index i if the following are true:

• The sum of the first i + 1 elements is greater than or equal to the sum of the last n - i - 1 elements.
• There is at least one element to the right of i. That is, 0 <= i < n - 1.

Return the number of valid splits in nums.

Example 1:

Input: nums = [10,4,-8,7]
Output: 2
Explanation:
There are three ways of splitting nums into two non-empty parts:
- Split nums at index 0. Then, the first part is , and its sum is 10. The second part is [4,-8,7], and its sum is 3. Since 10 >= 3, i = 0 is a valid split.
- Split nums at index 1. Then, the first part is [10,4], and its sum is 14. The second part is [-8,7], and its sum is -1. Since 14 >= -1, i = 1 is a valid split.
- Split nums at index 2. Then, the first part is [10,4,-8], and its sum is 6. The second part is , and its sum is 7. Since 6 < 7, i = 2 is not a valid split.
Thus, the number of valid splits in nums is 2.


Example 2:

Input: nums = [2,3,1,0]
Output: 2
Explanation:
There are two valid splits in nums:
- Split nums at index 1. Then, the first part is [2,3], and its sum is 5. The second part is [1,0], and its sum is 1. Since 5 >= 1, i = 1 is a valid split.
- Split nums at index 2. Then, the first part is [2,3,1], and its sum is 6. The second part is , and its sum is 0. Since 6 >= 0, i = 2 is a valid split.


Constraints:

• 2 <= nums.length <= 105
• -105 <= nums[i] <= 105

## Solution: Prefix/Suffix Sum

Note: sum can be greater than 2^31, use long!

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

## C++

The k-beauty of an integer num is defined as the number of substrings of num when it is read as a string that meet the following conditions:

• It has a length of k.
• It is a divisor of num.

Given integers num and k, return the k-beauty of num.

Note:

• 0 is not a divisor of any value.

substring is a contiguous sequence of characters in a string.

Example 1:

Input: num = 240, k = 2
Output: 2
Explanation: The following are the substrings of num of length k:
- "24" from "240": 24 is a divisor of 240.
- "40" from "240": 40 is a divisor of 240.
Therefore, the k-beauty is 2.


Example 2:

Input: num = 430043, k = 2
Output: 2
Explanation: The following are the substrings of num of length k:
- "43" from "430043": 43 is a divisor of 430043.
- "30" from "430043": 30 is not a divisor of 430043.
- "00" from "430043": 0 is not a divisor of 430043.
- "04" from "430043": 4 is not a divisor of 430043.
- "43" from "430043": 43 is a divisor of 430043.
Therefore, the k-beauty is 2.


Constraints:

• 1 <= num <= 109
• 1 <= k <= num.length (taking num as a string)

## Solution: Substring

Note: the substring can be 0, e.g. “00”

Time complexity: O((l-k)*k)
Space complexity: O(l + k) -> O(1)