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 4int dp[MAX][MAXP + 1][MAXK][MAXPREV + 1];// Visited array to keep track// of which columns were paintedbool vis[MAX];// Recursive Function to compute the// number of ways to paint the K cells// of the 3 x N gridint 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 gridint 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 Codeint main(){ int N = 2, K = 2, P = 2; cout << solve(N, P, K) << endl; return 0;} |
Java
// Java implementation to find the// number of ways to paint K cells of// 3 x N grid such that No two adjacent// cells are paintedimport java.util.*;class GFG{static int mod = (int)(1e9 + 7);static final int MAX = 301;static final int MAXP = 3;static final int MAXK = 600;static final int MAXPREV = 4;static int [][][][]dp = new int[MAX][MAXP + 1][MAXK][MAXPREV + 1];// Visited array to keep track// of which columns were paintedstatic boolean []vis = new boolean[MAX];// Recursive Function to compute the// number of ways to paint the K cells// of the 3 x N gridstatic 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 = Math.max( maxContinuousCol, continuousCol); continuousCol = 0; } } maxContinuousCol = Math.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 gridstatic int solve(int n, int p, int K){ // Set all values // of dp to -1; for(int i = 0; i < MAX; i++) for(int j = 0; j < MAXP + 1; j++) for(int k = 0; k < MAXK; k++) for(int l = 0; l < MAXPREV + 1; l++) dp[i][j][k][l] = -1; // Set all values of Visited // array to false Arrays.fill(vis, false); return helper(0, 0, 0, 0, n, p, K);}// Driver Codepublic static void main(String[] args){ int N = 2, K = 2, P = 2; System.out.print(solve(N, P, K) + "\n");}}// This code is contributed by Amit Katiyar |
Python3
# Python 3 implementation to find the# number of ways to paint K cells of# 3 x N grid such that No two adjacent# cells are paintedmod = 1e9 + 7MAX = 301MAXP = 3MAXK = 600MAXPREV = 4dp = [[[[-1 for x in range(MAXPREV + 1)]for y in range(MAXK)] for z in range(MAXP + 1)]for k in range(MAX)]# Visited array to keep track# of which columns were paintedvis = [False] * MAX# Recursive Function to compute the# number of ways to paint the K cells# of the 3 x N griddef helper(col, prevCol, painted, prev, N, P, K): # Condition to check if total # cells painted are K if (painted >= K): continuousCol = 0 maxContinuousCol = 0 # Check if any P continuous # columns were left unpainted for i in range(N): if (vis[i] == False): continuousCol += 1 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] 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 elif (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 elif (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 elif (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 dp[col][prevCol][painted][prev] = res % mod return dp[col][prevCol][painted][prev]# Function to find the number of# ways to paint 3 x N griddef solve(n, p, k): # Set all values # of dp to -1; global dp # Set all values of Visited # array to false global vis return helper(0, 0, 0, 0, n, p, k)# Driver Codeif __name__ == "__main__": N = 2 K = 2 P = 2 print(int(solve(N, P, K))) # This code is contributed by ukasp. |
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 using System;class GFG{ static int mod = (int)(1e9 + 7); static readonly int MAX = 301; static readonly int MAXP = 3; static readonly int MAXK = 600; static readonly int MAXPREV = 4; static int [,,,]dp = new int[MAX, MAXP + 1, MAXK, MAXPREV + 1]; // Visited array to keep track // of which columns were painted static bool []vis = new bool[MAX]; // Recursive Function to compute the // number of ways to paint the K cells // of the 3 x N grid static 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 = Math.Max( maxContinuousCol, continuousCol); continuousCol = 0; } } maxContinuousCol = Math.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 static int solve(int n, int p, int K) { // Set all values // of dp to -1; for(int i = 0; i < MAX; i++) for(int j = 0; j < MAXP + 1; j++) for(int k = 0; k < MAXK; k++) for(int l = 0; l < MAXPREV + 1; l++) dp[i, j, k, l] = -1; // Set all values of Visited // array to false for(int i = 0; i < vis.Length; i++) vis[i] = false; return helper(0, 0, 0, 0, n, p, K); } // Driver Code public static void Main(String[] args) { int N = 2, K = 2, P = 2; Console.Write(solve(N, P, K) + "\n"); } } // This code is contributed by Rohit_ranjan |
Javascript
<script>// Javascript implementation to find the// number of ways to paint K cells of// 3 x N grid such that No two adjacent// cells are paintedlet mod = (1e9 + 7);let MAX = 301;let MAXP = 3;let MAXK = 600;let MAXPREV = 4;let dp = new Array(MAX);for(let i = 0; i < MAX; i++){ dp[i] = new Array(MAXP + 1); for(let j = 0; j < (MAXP + 1); j++) { dp[i][j] = new Array(MAXK); for(let k = 0; k < MAXK; k++) { dp[i][j][k] = new Array(MAXPREV + 1); for(let l = 0; l < (MAXPREV + 1); l++) { dp[i][j][k][l] = -1; } } }}// Visited array to keep track// of which columns were paintedlet vis = new Array(MAX);for(let i = 0; i < MAX; i++){ vis[i] = false;}// Recursive Function to compute the// number of ways to paint the K cells// of the 3 x N gridfunction helper(col, prevCol, painted, prev, N, P, K){ // Condition to check if total // cells painted are K if (painted >= K) { let continuousCol = 0; let maxContinuousCol = 0; // Check if any P continuous // columns were left unpainted for(let i = 0; i < N; i++) { if (vis[i] == false) continuousCol++; else { maxContinuousCol = Math.max( maxContinuousCol, continuousCol); continuousCol = 0; } } maxContinuousCol = Math.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]; let 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 gridfunction solve(n,p,k){ return helper(0, 0, 0, 0, n, p, K);}// Driver Codelet N = 2, K = 2, P = 2;document.write(solve(N, P, K) + "<br>");// This code is contributed by avanitrachhadiya2155</script> |
Output:
8
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