# Number of ways to paint K cells in 3 x N grid such that no P continuous columns are left unpainted

Given three integers **N**, **P** and **K**, the task is to find the number of ways of painting K cells of 3 x N grid such that no adjacent cells are painted and also no continuous P columns are left unpainted.

**Note**: Diagonal cells are not considered as adjacent cells.

**Examples:**

Input:N = 1, P = 3, K = 1

Output:3

There are 3 ways to paint 1 cell in a 3 x 1 grid.

Input:N = 2, P = 2, K = 2

Output:8

There are 8 ways to paint 2 cells in a 3×2 grid.

Combinations of cells those are painted is shown below –

1) (0, 0) and (1, 1)

2) (0, 0) and (2, 1)

3) (0, 0) and (2, 0)

4) (1, 0) and (0, 1)

5) (1, 0) and (2, 1)

6) (2, 0) and (0, 1)

7) (2, 0) and (1, 1)

8) (0, 1) and (2, 1)

**Approach:** The idea is to use Dynamic Programming to solve this problem. As we know from the problem that column can be painted only when column is not painted. If column is not painted then we have following five cases –

- Paint the first Row.
- Paint the second row.
- Paint the third row.
- Paint first and third row.
- Leave the current column if atleast one column is painted.

Therefore, using this fact we can solve this problem easily.

Below is the implementation of the above approach:

## C++

`// C++ implementation to find the ` `// number of ways to paint K cells of ` `// 3 x N grid such that No two adjacent ` `// cells are painted ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `int` `mod = 1e9 + 7; ` `#define MAX 301 ` `#define MAXP 3 ` `#define MAXK 600 ` `#define MAXPREV 4 ` ` ` `int` `dp[MAX][MAXP + 1][MAXK][MAXPREV + 1]; ` ` ` `// Visited array to keep track ` `// of which columns were painted ` `bool` `vis[MAX]; ` ` ` `// Recursive Function to compute the ` `// number of ways to paint the K cells ` `// of the 3 x N grid ` `int` `helper(` `int` `col, ` `int` `prevCol, ` ` ` `int` `painted, ` `int` `prev, ` ` ` `int` `N, ` `int` `P, ` `int` `K) ` `{ ` ` ` ` ` `// Condition to check if total ` ` ` `// cells painted are K ` ` ` `if` `(painted >= K) { ` ` ` `int` `continuousCol = 0; ` ` ` `int` `maxContinuousCol = 0; ` ` ` ` ` `// Check if any P continuous ` ` ` `// columns were left unpainted ` ` ` `for` `(` `int` `i = 0; i < N; i++) { ` ` ` ` ` `if` `(vis[i] == ` `false` `) ` ` ` `continuousCol++; ` ` ` `else` `{ ` ` ` `maxContinuousCol ` ` ` `= max(maxContinuousCol, ` ` ` `continuousCol); ` ` ` `continuousCol = 0; ` ` ` `} ` ` ` `} ` ` ` ` ` `maxContinuousCol = max( ` ` ` `maxContinuousCol, ` ` ` `continuousCol); ` ` ` ` ` `// Condition to check if no P ` ` ` `// continuous columns were ` ` ` `// left unpainted ` ` ` `if` `(maxContinuousCol < P) ` ` ` `return` `1; ` ` ` ` ` `// return 0 if there are P ` ` ` `// continuous columns are ` ` ` `// left unpainted ` ` ` `return` `0; ` ` ` `} ` ` ` ` ` `// Condition to check if No ` ` ` `// further cells can be ` ` ` `// painted, so return 0 ` ` ` `if` `(col >= N) ` ` ` `return` `0; ` ` ` ` ` `// if already calculated the value ` ` ` `// return the val instead ` ` ` `// of calculating again ` ` ` `if` `(dp[col][prevCol][painted][prev] != -1) ` ` ` `return` `dp[col][prevCol][painted][prev]; ` ` ` ` ` `int` `res = 0; ` ` ` ` ` `// Previous column was not painted ` ` ` `if` `(prev == 0) { ` ` ` ` ` `// Column is painted so, ` ` ` `// make vis[col]=true ` ` ` `vis[col] = ` `true` `; ` ` ` `res += (helper( ` ` ` `col + 1, 0, painted + 1, ` ` ` `1, N, P, K)) ` ` ` `% mod; ` ` ` ` ` `res += (helper( ` ` ` `col + 1, 0, painted + 1, ` ` ` `2, N, P, K)) ` ` ` `% mod; ` ` ` ` ` `res += (helper( ` ` ` `col + 1, 0, painted + 1, ` ` ` `3, N, P, K)) ` ` ` `% mod; ` ` ` ` ` `// Condition to check if the number ` ` ` `// of cells to be painted is equal ` ` ` `// to or more than 2, then we can ` ` ` `// paint first and third row ` ` ` `if` `(painted + 2 <= K) { ` ` ` `res ` ` ` `+= (helper( ` ` ` `col + 1, 0, painted + 2, ` ` ` `4, N, P, K)) ` ` ` `% mod; ` ` ` `} ` ` ` `vis[col] = ` `false` `; ` ` ` ` ` `// Condition to check if number of ` ` ` `// previous continuous columns left ` ` ` `// unpainted is less than P ` ` ` `if` `(prevCol + 1 < P) { ` ` ` `res ` ` ` `+= (helper( ` ` ` `col + 1, prevCol + 1, ` ` ` `painted, 0, N, P, K)) ` ` ` `% mod; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Condition to check if first row ` ` ` `// was painted in previous column ` ` ` `else` `if` `(prev == 1) { ` ` ` `vis[col] = ` `true` `; ` ` ` `res += (helper( ` ` ` `col + 1, 0, painted + 1, ` ` ` `2, N, P, K)) ` ` ` `% mod; ` ` ` `res += (helper( ` ` ` `col + 1, 0, painted + 1, ` ` ` `3, N, P, K)) ` ` ` `% mod; ` ` ` `vis[col] = ` `false` `; ` ` ` `if` `(prevCol + 1 < P) { ` ` ` `res += (helper( ` ` ` `col + 1, prevCol + 1, ` ` ` `painted, 0, N, P, K)) ` ` ` `% mod; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Condition to check if second row ` ` ` `// was painted in previous column ` ` ` `else` `if` `(prev == 2) { ` ` ` `vis[col] = ` `true` `; ` ` ` `res += (helper( ` ` ` `col + 1, 0, painted + 1, ` ` ` `1, N, P, K)) ` ` ` `% mod; ` ` ` `res += (helper( ` ` ` `col + 1, 0, painted + 1, ` ` ` `3, N, P, K)) ` ` ` `% mod; ` ` ` ` ` `// Condition to check if the number ` ` ` `// of cells to be painted is equal to ` ` ` `// or more than 2, then we can ` ` ` `// paint first and third row ` ` ` `if` `(painted + 2 <= K) { ` ` ` `res ` ` ` `+= (helper( ` ` ` `col + 1, 0, painted + 2, ` ` ` `4, N, P, K)) ` ` ` `% mod; ` ` ` `} ` ` ` `vis[col] = ` `false` `; ` ` ` `if` `(prevCol + 1 < P) { ` ` ` `res ` ` ` `+= (helper( ` ` ` `col + 1, prevCol + 1, ` ` ` `painted, 0, N, P, K)) ` ` ` `% mod; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Condition to check if third row ` ` ` `// was painted in previous column ` ` ` `else` `if` `(prev == 3) { ` ` ` `vis[col] = ` `true` `; ` ` ` `res += (helper( ` ` ` `col + 1, 0, painted + 1, ` ` ` `1, N, P, K)) ` ` ` `% mod; ` ` ` `res += (helper( ` ` ` `col + 1, 0, painted + 1, ` ` ` `2, N, P, K)) ` ` ` `% mod; ` ` ` `vis[col] = ` `false` `; ` ` ` `if` `(prevCol + 1 < P) { ` ` ` `res += (helper( ` ` ` `col + 1, prevCol + 1, ` ` ` `painted, 0, N, P, K)) ` ` ` `% mod; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Condition to check if first and ` ` ` `// third row were painted ` ` ` `// in previous column ` ` ` `else` `{ ` ` ` `vis[col] = ` `true` `; ` ` ` `res += (helper( ` ` ` `col + 1, 0, painted + 1, ` ` ` `2, N, P, K)) ` ` ` `% mod; ` ` ` `vis[col] = ` `false` `; ` ` ` `if` `(prevCol + 1 < P) { ` ` ` `res += (helper( ` ` ` `col + 1, prevCol + 1, ` ` ` `painted, 0, N, P, K)) ` ` ` `% mod; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Memoize the data and return the ` ` ` `// Computed value ` ` ` `return` `dp[col][prevCol][painted][prev] ` ` ` `= res % mod; ` `} ` ` ` `// Function to find the number of ` `// ways to paint 3 x N grid ` `int` `solve(` `int` `n, ` `int` `p, ` `int` `k) ` `{ ` ` ` `// Set all values ` ` ` `// of dp to -1; ` ` ` `memset` `(dp, -1, ` `sizeof` `(dp)); ` ` ` ` ` `// Set all values of Visited ` ` ` `// array to false ` ` ` `memset` `(vis, ` `false` `, ` `sizeof` `(vis)); ` ` ` ` ` `return` `helper(0, 0, 0, 0, n, p, k); ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` ` ` `int` `N = 2, K = 2, P = 2; ` ` ` `cout << solve(N, P, K) << endl; ` ` ` ` ` `return` `0; ` `} ` |

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*filter_none*

**Output:**

8

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