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-rw-r--r--painting_grid_three_colours.c78
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+// 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
+}