You are given a 0-indexed integer array nums. In one operation, select any non-negative integer x and an index i, then update nums[i] to be equal to nums[i] AND (nums[i] XOR x).

Note that AND is the bitwise AND operation and XOR is the bitwise XOR operation.

Return the maximum possible bitwise XOR of all elements of nums after applying the operation any number of times.

Example 1:

Input: nums = [3,2,4,6]
Output: 7
Explanation: Apply the operation with x = 4 and i = 3, num[3] = 6 AND (6 XOR 4) = 6 AND 2 = 2.
Now, nums = [3, 2, 4, 2] and the bitwise XOR of all the elements = 3 XOR 2 XOR 4 XOR 2 = 7.
It can be shown that 7 is the maximum possible bitwise XOR.
Note that other operations may be used to achieve a bitwise XOR of 7.

Example 2:

Input: nums = [1,2,3,9,2]
Output: 11
Explanation: Apply the operation zero times.
The bitwise XOR of all the elements = 1 XOR 2 XOR 3 XOR 9 XOR 2 = 11.
It can be shown that 11 is the maximum possible bitwise XOR.

Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i] <= 108

## Solution: Bitwise OR

The maximum possible number MAX = nums[0] | nums[1] | … | nums[n – 1].

We need to prove:
1) MAX is achievable.
2) MAX is the largest number we can get.

nums[i] AND (nums[i] XOR x) means that we can turn any 1 bits to 0 for nums[i].

1) If the i-th bit of MAX is 1, which means there are at least one number with i-th bit equals to 1, however, for XOR, if there are even numbers with i-th bit equal to one, the final results will be 0 for i-th bit, we get a smaller number. By using the operation, we can choose one of them and flip the bit.

**1** XOR **1** XOR **1** XOR **1** = **0** =>
**0** XOR **1** XOR **1** XOR **1** = **1**

2) If the i-th bit of MAX is 0, which means the i-th bit of all the numbers is 0, there is nothing we can do with the operation, and the XOR will be 0 as well.
e.g. **0** XOR **0** XOR **0** XOR **0** = **0**

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

## C++

You are given an integer n. There is an undirected graph with n nodes, numbered from 0 to n - 1. You are given a 2D integer array edges where edges[i] = [ai, bi] denotes that there exists an undirected edge connecting nodes ai and bi.

Return the number of pairs of different nodes that are unreachable from each other.

Example 1:

Input: n = 3, edges = [[0,1],[0,2],[1,2]]
Output: 0
Explanation: There are no pairs of nodes that are unreachable from each other. Therefore, we return 0.


Example 2:

Input: n = 7, edges = [[0,2],[0,5],[2,4],[1,6],[5,4]]
Output: 14
Explanation: There are 14 pairs of nodes that are unreachable from each other:
[[0,1],[0,3],[0,6],[1,2],[1,3],[1,4],[1,5],[2,3],[2,6],[3,4],[3,5],[3,6],[4,6],[5,6]].
Therefore, we return 14.


Constraints:

• 1 <= n <= 105
• 0 <= edges.length <= 2 * 105
• edges[i].length == 2
• 0 <= ai, bi < n
• ai != bi
• There are no repeated edges.

## Solution 1: DFS

Use DFS to find all CCs

Time complexity: O(V+E)
Space complexity: O(V+E)

## Solution 2: Union Find

Time complexity: O(V+E)
Space complexity: O(V)

## C++

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.