# Posts published in “Recursion”

There is a tournament where n players are participating. The players are standing in a single row and are numbered from 1 to n based on their initial standing position (player 1 is the first player in the row, player 2 is the second player in the row, etc.).

The tournament consists of multiple rounds (starting from round number 1). In each round, the ith player from the front of the row competes against the ith player from the end of the row, and the winner advances to the next round. When the number of players is odd for the current round, the player in the middle automatically advances to the next round.

• For example, if the row consists of players 1, 2, 4, 6, 7
• Player 1 competes against player 7.
• Player 2 competes against player 6.
• Player 4 automatically advances to the next round.

After each round is over, the winners are lined back up in the row based on the original ordering assigned to them initially (ascending order).

The players numbered firstPlayer and secondPlayer are the best in the tournament. They can win against any other player before they compete against each other. If any two other players compete against each other, either of them might win, and thus you may choose the outcome of this round.

Given the integers nfirstPlayer, and secondPlayer, return an integer array containing two values, the earliest possible round number and the latest possible round number in which these two players will compete against each other, respectively.

Example 1:

Input: n = 11, firstPlayer = 2, secondPlayer = 4
Output: [3,4]
Explanation:
One possible scenario which leads to the earliest round number:
First round: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
Second round: 2, 3, 4, 5, 6, 11
Third round: 2, 3, 4
One possible scenario which leads to the latest round number:
First round: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
Second round: 1, 2, 3, 4, 5, 6
Third round: 1, 2, 4
Fourth round: 2, 4


Example 2:

Input: n = 5, firstPlayer = 1, secondPlayer = 5
Output: [1,1]
Explanation: The players numbered 1 and 5 compete in the first round.
There is no way to make them compete in any other round.


Constraints:

• 2 <= n <= 28
• 1 <= firstPlayer < secondPlayer <= n

## Solution 1: Simulation using recursion

All possible paths,
Time complexity: O(n2*2n)
Space complexity: O(logn)

dfs(s, i, j, d) := let i battle with j at round d, given s (binary mask of dead players).

## C++

You are given a valid boolean expression as a string expression consisting of the characters '1','0','&' (bitwise AND operator),'|' (bitwise OR operator),'(', and ')'.

• For example, "()1|1" and "(1)&()" are not valid while "1""(((1))|(0))", and "1|(0&(1))" are valid expressions.

Return the minimum cost to change the final value of the expression.

• For example, if expression = "1|1|(0&0)&1", its value is 1|1|(0&0)&1 = 1|1|0&1 = 1|0&1 = 1&1 = 1. We want to apply operations so that the new expression evaluates to 0.

The cost of changing the final value of an expression is the number of operations performed on the expression. The types of operations are described as follows:

• Turn a '1' into a '0'.
• Turn a '0' into a '1'.
• Turn a '&' into a '|'.
• Turn a '|' into a '&'.

Note: '&' does not take precedence over '|' in the order of calculation. Evaluate parentheses first, then in left-to-right order.

Example 1:

Input: expression = "1&(0|1)"
Output: 1
Explanation: We can turn "1&(0|1)" into "1&(0&1)" by changing the '|' to a '&' using 1 operation.
The new expression evaluates to 0.


Example 2:

Example 3:

Input: expression = "(0|(1|0&1))"
Output: 1
Explanation: We can turn "(0|(1|0&1))" into "(0|(0|0&1))" using 1 operation.
The new expression evaluates to 0.

Constraints:

• 1 <= expression.length <= 105
• expression only contains '1','0','&','|','(', and ')'
• All parentheses are properly matched.
• There will be no empty parentheses (i.e: "()" is not a substring of expression).

## Solution: DP, Recursion / Simulation w/ Stack

For each expression, stores the min cost to change value to 0 and 1.

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

## C++

You are given an integer n, the number of teams in a tournament that has strange rules:

• If the current number of teams is even, each team gets paired with another team. A total of n / 2 matches are played, and n / 2 teams advance to the next round.
• If the current number of teams is odd, one team randomly advances in the tournament, and the rest gets paired. A total of (n - 1) / 2 matches are played, and (n - 1) / 2 + 1 teams advance to the next round.

Return the number of matches played in the tournament until a winner is decided.

Example 1:

Input: n = 7
Output: 6
Explanation: Details of the tournament:
- 1st Round: Teams = 7, Matches = 3, and 4 teams advance.
- 2nd Round: Teams = 4, Matches = 2, and 2 teams advance.
- 3rd Round: Teams = 2, Matches = 1, and 1 team is declared the winner.
Total number of matches = 3 + 2 + 1 = 6.


Example 2:

Input: n = 14
Output: 13
Explanation: Details of the tournament:
- 1st Round: Teams = 14, Matches = 7, and 7 teams advance.
- 2nd Round: Teams = 7, Matches = 3, and 4 teams advance.
- 3rd Round: Teams = 4, Matches = 2, and 2 teams advance.
- 4th Round: Teams = 2, Matches = 1, and 1 team is declared the winner.
Total number of matches = 7 + 3 + 2 + 1 = 13.


Constraints:

• 1 <= n <= 200

## Solution: Simulation / Recursion

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

## C++

A kingdom consists of a king, his children, his grandchildren, and so on. Every once in a while, someone in the family dies or a child is born.

The kingdom has a well-defined order of inheritance that consists of the king as the first member. Let’s define the recursive function Successor(x, curOrder), which given a person x and the inheritance order so far, returns who should be the next person after x in the order of inheritance.

For example, assume we have a kingdom that consists of the king, his children Alice and Bob (Alice is older than Bob), and finally Alice’s son Jack.

1. In the beginning, curOrder will be ["king"].
2. Calling Successor(king, curOrder) will return Alice, so we append to curOrder to get ["king", "Alice"].
3. Calling Successor(Alice, curOrder) will return Jack, so we append to curOrder to get ["king", "Alice", "Jack"].
4. Calling Successor(Jack, curOrder) will return Bob, so we append to curOrder to get ["king", "Alice", "Jack", "Bob"].
5. Calling Successor(Bob, curOrder) will return null. Thus the order of inheritance will be ["king", "Alice", "Jack", "Bob"].

Using the above function, we can always obtain a unique order of inheritance.

Implement the ThroneInheritance class:

• ThroneInheritance(string kingName) Initializes an object of the ThroneInheritance class. The name of the king is given as part of the constructor.
• void birth(string parentName, string childName) Indicates that parentName gave birth to childName.
• void death(string name) Indicates the death of name. The death of the person doesn’t affect the Successor function nor the current inheritance order. You can treat it as just marking the person as dead.
• string[] getInheritanceOrder() Returns a list representing the current order of inheritance excluding dead people.

Example 1:

Input
["ThroneInheritance", "birth", "birth", "birth", "birth", "birth", "birth", "getInheritanceOrder", "death", "getInheritanceOrder"]
[["king"], ["king", "andy"], ["king", "bob"], ["king", "catherine"], ["andy", "matthew"], ["bob", "alex"], ["bob", "asha"], [null], ["bob"], [null]]
Output
[null, null, null, null, null, null, null, ["king", "andy", "matthew", "bob", "alex", "asha", "catherine"], null, ["king", "andy", "matthew", "alex", "asha", "catherine"]]
Explanation
ThroneInheritance t= new ThroneInheritance("king"); // order: king
t.birth("king", "andy"); // order: king > andy
t.birth("king", "bob"); // order: king > andy > bob
t.birth("king", "catherine"); // order: king > andy > bob > catherine
t.birth("andy", "matthew"); // order: king > andy > matthew > bob > catherine
t.birth("bob", "alex"); // order: king > andy > matthew > bob > alex > catherine
t.birth("bob", "asha"); // order: king > andy > matthew > bob > alex > asha > catherine
t.getInheritanceOrder(); // return ["king", "andy", "matthew", "bob", "alex", "asha", "catherine"]
t.death("bob"); // order: king > andy > matthew > bob > alex > asha > catherine
t.getInheritanceOrder(); // return ["king", "andy", "matthew", "alex", "asha", "catherine"]



Constraints:

• 1 <= kingName.length, parentName.length, childName.length, name.length <= 15
• kingNameparentNamechildName, and name consist of lowercase English letters only.
• All arguments childName and kingName are distinct.
• All name arguments of death will be passed to either the constructor or as childName to birth first.
• For each call to birth(parentName, childName), it is guaranteed that parentName is alive.
• At most 105 calls will be made to birth and death.
• At most 10 calls will be made to getInheritanceOrder.

## Solution: HashTable + DFS

Record :
1. mapping from parent to children (ordered)

Time complexity: getInheritanceOrder O(n), other O(1)
Space complexity: O(n)

## Python3

Given an array nums that represents a permutation of integers from 1 to n. We are going to construct a binary search tree (BST) by inserting the elements of nums in order into an initially empty BST. Find the number of different ways to reorder nums so that the constructed BST is identical to that formed from the original array nums.

For example, given nums = [2,1,3], we will have 2 as the root, 1 as a left child, and 3 as a right child. The array [2,3,1] also yields the same BST but [3,2,1] yields a different BST.

Return the number of ways to reorder nums such that the BST formed is identical to the original BST formed from nums.

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

Example 1:

Input: nums = [2,1,3]
Output: 1
Explanation: We can reorder nums to be [2,3,1] which will yield the same BST. There are no other ways to reorder nums which will yield the same BST.


Example 2:

Input: nums = [3,4,5,1,2]
Output: 5
Explanation: The following 5 arrays will yield the same BST:
[3,1,2,4,5]
[3,1,4,2,5]
[3,1,4,5,2]
[3,4,1,2,5]
[3,4,1,5,2]


Example 3:

Input: nums = [1,2,3]
Output: 0
Explanation: There are no other orderings of nums that will yield the same BST.


Example 4:

Input: nums = [3,1,2,5,4,6]
Output: 19


Example 5:

Input: nums = [9,4,2,1,3,6,5,7,8,14,11,10,12,13,16,15,17,18]
Output: 216212978
Explanation: The number of ways to reorder nums to get the same BST is 3216212999. Taking this number modulo 10^9 + 7 gives 216212978.


Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= nums.length
• All integers in nums are distinct.

## Solution: Recursion + Combinatorics

For a given root (first element of the array), we can split the array into left children (nums[i] < nums[0]) and right children (nums[i] > nums[0]). Assuming there are l nodes for the left and r nodes for the right. We have C(l + r, l) different ways to insert l elements into a (l + r) sized array. Within node l / r nodes, we have ways(left) / ways(right) different ways to re-arrange those nodes. So the total # of ways is:
C(l + r, l) * ways(l) * ways(r)
Don’t forget to minus one for the final answer.

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