Pyramid Transition Matrix
Pyramid Transition Matrix
LWC 65: 756. Pyramid Transition Matrix
Problem:
We are stacking blocks to form a pyramid. Each block has a color which is a one letter string, like
'Z'. For every block of colorCwe place not in the bottom row, we are placing it on top of a left block of colorAand right block of colorB. We are allowed to place the block there only if(A, B, C)is an allowed triple. We start with a bottom row of bottom, represented as a single string. We also start with a list of allowed triples allowed. Each allowed triple is represented as a string of length 3. Return true if we can build the pyramid all the way to the top, otherwise false.
Example 1:
Input: bottom = “XYZ”, allowed = [“XYD”, “YZE”, “DEA”, “FFF”] Output: true Explanation: We can stack the pyramid like this: A / \ D E / \ / \ X Y Z This works because (‘X’, ‘Y’, ‘D’), (‘Y’, ‘Z’, ‘E’), and (‘D’, ‘E’, ‘A’) are allowed triples.
Example 1:
Input: bottom = “XXYX”, allowed = [“XXX”, “XXY”, “XYX”, “XYY”, “YXZ”] Output: false Explanation: We can’t stack the pyramid to the top. Note that there could be allowed triples (A, B, C) and (A, B, D) with C != D.
Note:
- bottom will be a string with length in range [2, 12].
- allowed will have length in range [0, 343].
- Letters in all strings will be chosen from the set {‘A’, ‘B’, ‘C’, ‘D’, ‘E’, ‘F’, ‘G’}.
思路： 注意allowed中的三元组是可以重复利用的，这样我们定义dp：
dp[i][j][k]: 表示第i层上，第j个元素为k根据bottom，可以初始化每个位置j上含有的字符bottom[j]， dp更新式如下：
dp[i][j][k] = true if dp[i + 1][j][l] = true && dp[i + 1][j + 1][r] = true && lrk组成的字符串在allowed中出现过。Java版本：
public boolean pyramidTransition(String bottom, List<String> allowed) {
Map<String, List<String>> mem = new HashMap<>();
boolean[][] dp = new boolean[20][7];
int n = bottom.length();
for (String allow : allowed) {
mem.computeIfAbsent(allow.substring(0, 2), k -> new ArrayList<>()).add(allow.substring(2));
}
for (int i = 0; i < n; ++i) {
dp[i][bottom.charAt(i) - 'A'] = true;
}
for (int i = n - 1; i >= 1; --i) {
boolean[][] ndp = new boolean[20][7];
for (int j = 0; j < i; ++j) {
for (int l = 0; l < 7; ++l) {
for (int r = 0; r < 7; ++r) {
if (dp[j][l] && dp[j + 1][r]) {
if (mem.containsKey((char)(l + 'A') + "" + (char)(r + 'A'))) {
for (String s : mem.get((char)(l + 'A') + "" + (char)(r + 'A'))) {
ndp[j][s.charAt(0) - 'A'] = true;
}
}
}
}
}
}
dp = ndp;
}
for (int i = 0; i < 7; ++i) {
if (dp[0][i]) return true;
}
return false;
}Python版本：
class Solution(object):
def pyramidTransition(self, bottom, allowed):
from collections import defaultdict
"""
:type bottom: str
:type allowed: List[str]
:rtype: bool
"""
mem = defaultdict(list)
for allow in allowed:
mem[allow[0:2]].append(allow[2])
dp = [[False] * 10 for i in range(20)]
n = len(bottom)
for i in range(n):
dp[i][ord(bottom[i]) - ord('A')] = True
for i in range(n - 1, 0, -1):
ndp = [[False] * 10 for i in range(20)]
for j in range(i):
for l in range(7):
for r in range(7):
if (dp[j][l] and dp[j + 1][r]):
if str(chr(65 + l) + "" + chr(65 + r)) in mem:
for c in mem[chr(65 + l) + "" + chr(65 + r)]:
ndp[j][ord(c) - ord('A')] = True
dp = ndp
for i in range(7):
if (dp[0][i]): return True
return False