There are N rooms and you start in room 0. Each room has a distinct number in 0, 1, 2, ..., N-1, and each room may have some keys to access the next room.
Formally, each room i has a list of keys rooms[i], and each key rooms[i][j] is an integer in [0, 1, ..., N-1] where N = rooms.length. A key rooms[i][j] = v opens the room with number v.
Initially, all the rooms start locked (except for room 0).
You can walk back and forth between rooms freely.
Return true if and only if you can enter every room.
Example 1:
Input: [[1],[2],[3],[]]
Output: true
Explanation:
We start in room 0, and pick up key 1.
We then go to room 1, and pick up key 2.
We then go to room 2, and pick up key 3.
We then go to room 3. Since we were able to go to every room, we return true.
Example 2:
Input: [[1,3],[3,0,1],[2],[0]]
Output: false
Explanation: We can't enter the room with number 2.
Note:
1 <= rooms.length <= 1000
0 <= rooms[i].length <= 1000
The number of keys in all rooms combined is at most 3000.
Given an integer matrix, find the length of the longest increasing path.
From each cell, you can either move to four directions: left, right, up or down. You may NOT move diagonally or move outside of the boundary (i.e. wrap-around is not allowed).
Example 1:
Input: nums =
[
[9,9,4],
[6,6,8],
[2,1,1]
]
Output: 4
Explanation: The longest increasing path is [1, 2, 6, 9].
Example 2:
Input: nums =
[
[3,4,5],
[3,2,6],
[2,2,1]
]
Output: 4
Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.
In a given integer array A, we must move every element of A to either list B or list C. (B and C initially start empty.)
Return true if and only if after such a move, it is possible that the average value of B is equal to the average value of C, and B and C are both non-empty.
Example :Input:
[1,2,3,4,5,6,7,8]
Output: true
Explanation: We can split the array into [1,4,5,8] and [2,3,6,7], and both of them have the average of 4.5.
Note:
The length of A will be in the range [1, 30].
A[i] will be in the range of [0, 10000].
Solution: Search
Time complexity: O(2^n)
Space complexity: O(n)
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// Author: Huahua
// Running time: 771 ms
classSolution{
public:
boolsplitArraySameAverage(vector<int>& A) {
std::sort(A.begin(), A.end());
sum=std::accumulate(A.begin(),A.end(),0);
n=A.size();
returndfs(A,1,0,0);
}
private:
intsum;
intn;
booldfs(constvector<int>& A, int c, int s, int cur) {
We have a grid of 1s and 0s; the 1s in a cell represent bricks. A brick will not drop if and only if it is directly connected to the top of the grid, or at least one of its (4-way) adjacent bricks will not drop.
We will do some erasures sequentially. Each time we want to do the erasure at the location (i, j), the brick (if it exists) on that location will disappear, and then some other bricks may drop because of that erasure.
Return an array representing the number of bricks that will drop after each erasure in sequence.
Example 1:Input:
grid = [[1,0,0,0],[1,1,1,0]]
hits = [[1,0]]
Output: [2]
Explanation:
If we erase the brick at (1, 0), the brick at (1, 1) and (1, 2) will drop. So we should return 2.
Example 2:Input:
grid = [[1,0,0,0],[1,1,0,0]]
hits = [[1,1],[1,0]]
Output: [0,0]
Explanation:
When we erase the brick at (1, 0), the brick at (1, 1) has already disappeared due to the last move. So each erasure will cause no bricks dropping. Note that the erased brick (1, 0) will not be counted as a dropped brick.
Idea
For each day, hit and clear the specified brick.
Find all connected components (CCs) using DFS.
For each CC, if there is no brick that is on the first row that the entire cc will drop. Clear those CCs.
