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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
}
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