// 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 }