Given an undirected tree consisting of n vertices numbered from 1 to n. A frog starts jumping from the vertex 1. In one second, the frog jumps from its current vertex to another unvisited vertex if they are directly connected. The frog can not jump back to a visited vertex. In case the frog can jump to several vertices it jumps randomly to one of them with the same probability, otherwise, when the frog can not jump to any unvisited vertex it jumps forever on the same vertex.
The edges of the undirected tree are given in the array edges, where edges[i] = [fromi, toi] means that exists an edge connecting directly the vertices fromi and toi.
Return the probability that after t seconds the frog is on the vertex target.
Example 1:
Input: n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 2, target = 4
Output: 0.16666666666666666
Explanation: The figure above shows the given graph. The frog starts at vertex 1, jumping with 1/3 probability to the vertex 2 after second 1 and then jumping with 1/2 probability to vertex 4 after second 2. Thus the probability for the frog is on the vertex 4 after 2 seconds is 1/3 * 1/2 = 1/6 = 0.16666666666666666.
Example 2:
Input: n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 1, target = 7
Output: 0.3333333333333333
Explanation: The figure above shows the given graph. The frog starts at vertex 1, jumping with 1/3 = 0.3333333333333333 probability to the vertex 7 after second 1.
Example 3:
Input: n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 20, target = 6
Output: 0.16666666666666666
Constraints:
1 <= n <= 100
edges.length == n-1
edges[i].length == 2
1 <= edges[i][0], edges[i][1] <= n
1 <= t <= 50
1 <= target <= n
Answers within 10^-5 of the actual value will be accepted as correct.
Solution: BFS
key: if a node has children, the fog jumps to to children so the probability at current node will become 0.
A company has n employees with a unique ID for each employee from 0 to n - 1. The head of the company has is the one with headID.
Each employee has one direct manager given in the manager array where manager[i] is the direct manager of the i-th employee, manager[headID] = -1. Also it’s guaranteed that the subordination relationships have a tree structure.
The head of the company wants to inform all the employees of the company of an urgent piece of news. He will inform his direct subordinates and they will inform their subordinates and so on until all employees know about the urgent news.
The i-th employee needs informTime[i] minutes to inform all of his direct subordinates (i.e After informTime[i] minutes, all his direct subordinates can start spreading the news).
Return the number of minutes needed to inform all the employees about the urgent news.
Example 1:
Input: n = 1, headID = 0, manager = [-1], informTime = [0]
Output: 0
Explanation: The head of the company is the only employee in the company.
Example 2:
Input: n = 6, headID = 2, manager = [2,2,-1,2,2,2], informTime = [0,0,1,0,0,0]
Output: 1
Explanation: The head of the company with id = 2 is the direct manager of all the employees in the company and needs 1 minute to inform them all.
The tree structure of the employees in the company is shown.
Example 3:
Input: n = 7, headID = 6, manager = [1,2,3,4,5,6,-1], informTime = [0,6,5,4,3,2,1]
Output: 21
Explanation: The head has id = 6. He will inform employee with id = 5 in 1 minute.
The employee with id = 5 will inform the employee with id = 4 in 2 minutes.
The employee with id = 4 will inform the employee with id = 3 in 3 minutes.
The employee with id = 3 will inform the employee with id = 2 in 4 minutes.
The employee with id = 2 will inform the employee with id = 1 in 5 minutes.
The employee with id = 1 will inform the employee with id = 0 in 6 minutes.
Needed time = 1 + 2 + 3 + 4 + 5 + 6 = 21.
Example 4:
Input: n = 15, headID = 0, manager = [-1,0,0,1,1,2,2,3,3,4,4,5,5,6,6], informTime = [1,1,1,1,1,1,1,0,0,0,0,0,0,0,0]
Output: 3
Explanation: The first minute the head will inform employees 1 and 2.
The second minute they will inform employees 3, 4, 5 and 6.
The third minute they will inform the rest of employees.
There is a room with n bulbs, numbered from 1 to n, arranged in a row from left to right. Initially, all the bulbs are turned off.
At moment k (for k from 0 to n - 1), we turn on the light[k] bulb. A bulb change color to blue only if it is on and all the previous bulbs (to the left) are turned on too.
Return the number of moments in which all turned on bulbs are blue.
Example 1:
Input: light = [2,1,3,5,4]
Output: 3
Explanation: All bulbs turned on, are blue at the moment 1, 2 and 4.
Example 2:
Input: light = [3,2,4,1,5]
Output: 2
Explanation: All bulbs turned on, are blue at the moment 3, and 4 (index-0).
Example 3:
Input: light = [4,1,2,3]
Output: 1
Explanation: All bulbs turned on, are blue at the moment 3 (index-0).
Bulb 4th changes to blue at the moment 3.
Example 4:
Input: light = [2,1,4,3,6,5]
Output: 3
Example 5:
Input: light = [1,2,3,4,5,6]
Output: 6
Constraints:
n == light.length
1 <= n <= 5 * 10^4
light is a permutation of [1, 2, ..., n]
Solution
Track the right most light l_k, all turned-on lights are blue if and only if the right most one is k, and there are exact k lights on right now.
Given an integer n, return a string with n characters such that each character in such string occurs an odd number of times.
The returned string must contain only lowercase English letters. If there are multiples valid strings, return any of them.
Example 1:
Input: n = 4
Output: "pppz"
Explanation: "pppz" is a valid string since the character 'p' occurs three times and the character 'z' occurs once. Note that there are many other valid strings such as "ohhh" and "love".
Example 2:
Input: n = 2
Output: "xy"
Explanation: "xy" is a valid string since the characters 'x' and 'y' occur once. Note that there are many other valid strings such as "ag" and "ur".
Example 3:
Input: n = 7
Output: "holasss"
Constraints:
1 <= n <= 500
Solution: Greedy
if n is odd, return n ‘a’s. otherwise, return n -1 ‘a’s and 1 ‘b’
Time complexity: O(n) Space complexity: O(n) or O(1)
