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authorJack Sangdahl <[email protected]>2025-05-17 23:15:29 -0600
committerJack Sangdahl <[email protected]>2025-05-17 23:15:29 -0600
commit87badbaba07e4cd13c2281e15c96ccf2dd8b3106 (patch)
treeceb97f38b7a69211175f4bb25c4b707c29f4bd5d
parent2304d32820d403740108de84125059c623686fc9 (diff)
downloadleetcode-87badbaba07e4cd13c2281e15c96ccf2dd8b3106.tar.gz
leetcode-87badbaba07e4cd13c2281e15c96ccf2dd8b3106.zip
1931. painting a grid with three different colours
-rw-r--r--painting_grid_three_colours.c78
-rw-r--r--painting_grid_three_colours.py42
2 files changed, 120 insertions, 0 deletions
diff --git a/painting_grid_three_colours.c b/painting_grid_three_colours.c
new file mode 100644
index 0000000..82c9d53
--- /dev/null
+++ b/painting_grid_three_colours.c
@@ -0,0 +1,78 @@
+// 1931. Painting a Grid With Three Different Colors
+#include "leetcode.h"
+
+/*
+ * you are given two integers 'm' and 'n'. consider an 'm x n' grid where each
+ * cell is initially white. you can paint each cell red, green, or blue. all
+ * cells must be painted. return the number of ways to colour the grid with no
+ * two adjacent cells having the same colour. since the answer can be very
+ * large, return it modulo 10^9+7.
+ */
+
+int colorTheGrid(int m, int n) {
+ const int mod = 1e9 + 7;
+ int max_mask = 1;
+ for (int i = 0; i < m; ++i)
+ max_mask *= 3;
+ int states[48], s = 0;
+ for (int i = 0; i < max_mask; ++i) {
+ int curr = i, prev = -1;
+ bool valid = true;
+ for (int j = 0; j < m; ++j) {
+ int dir = curr % 3;
+ if (dir == prev) {
+ valid = false;
+ break;
+ }
+ prev = dir;
+ curr /= 3;
+ }
+ if (valid)
+ states[s++] = i;
+ }
+ int compat[48][48], compat_size[48];
+ for (int i = 0; i < s; ++i)
+ compat_size[i] = 0;
+ for (int i = 0; i < s; ++i)
+ for (int j = 0; j < s; ++j) {
+ int x = states[i], y = states[j];
+ bool valid = true;
+ for (int k = 0; k < m; ++k) {
+ if (x % 3 == y % 3) {
+ valid = false;
+ break;
+ }
+ x /= 3;
+ y /= 3;
+ }
+ if (valid)
+ compat[i][compat_size[i]++] = j;
+ }
+ int dp[48], copy[48];
+ for (int i = 0; i < s; ++i)
+ dp[i] = 1;
+ for (int i = 0; i < n - 1; ++i) {
+ for (int j = 0; j < s; ++j)
+ copy[j] = 0;
+ for (int j = 0; j < s; ++j) {
+ int dir = dp[j];
+ if (dir)
+ for (int k = 0; k < compat_size[j]; ++k) {
+ int new = compat[j][k];
+ copy[new] = (copy[new] + dir) % mod;
+ }
+ }
+ for (int j = 0; j < s; ++j)
+ dp[j] = copy[j];
+ }
+ long long ans = 0;
+ for (int i = 0; i < s; ++i)
+ ans = (ans + dp[i]) % mod;
+ return (int)ans;
+}
+
+int main() {
+ printf("%d\n", colorTheGrid(1, 1)); // expect: 3
+ printf("%d\n", colorTheGrid(1, 2)); // expect: 6
+ printf("%d\n", colorTheGrid(5, 5)); // expect: 580986
+}
diff --git a/painting_grid_three_colours.py b/painting_grid_three_colours.py
new file mode 100644
index 0000000..d33b1d8
--- /dev/null
+++ b/painting_grid_three_colours.py
@@ -0,0 +1,42 @@
+# 1931. Painting a Grid With Three Different Colors
+from collections import Counter
+
+"""
+you are given two integers 'm' and 'n'. consider an 'm x n' grid where each
+cell is initially white. you can paint each cell red, green, or blue. all
+cells must be painted. return the number of ways to colour the grid with no
+two adjacent cells having the same colour. since the answer can be very
+large, return it modulo 10^9+7.
+"""
+
+
+class Solution(object):
+ def colorTheGrid(self, m, n):
+ """
+ :type m: int
+ :type n: int
+ :rtype: int
+ """
+ dp = {(1 << (2 * m)) - 1: 1}
+ mod = 10**9 + 7
+ for i in range(m * n - 1, -1, -1):
+ copy = Counter()
+ x, y = divmod(i, m)
+ for j in dp:
+ valid = {0, 1, 2}
+ if y < m - 1:
+ valid.discard(j >> (2 * m - 2))
+ if x < n - 1:
+ valid.discard(j & 3)
+ for k in valid:
+ new = (j >> 2) + (k << (2 * m - 2))
+ copy[new] = (copy[new] + dp[j]) % mod
+ dp = copy
+ return sum(dp.values()) % mod
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.colorTheGrid(m=1, n=1))
+ print(obj.colorTheGrid(m=1, n=2))
+ print(obj.colorTheGrid(m=5, n=5))