# Color N boxes using M colors such that K boxes have different color from the box on its left

Given N number of boxes arranged in a row and M number of colors. The task is to find the number of ways to paint those N boxes using M colors such that there are exactly K boxes with a color different from the color of the box on its left. Print this answer modulo 998244353.

Examples:

Input: N = 3, M = 3, K = 0
Output: 3
Since the value of K is zero, no box can have a different color from color of the box on its left. Thus, all boxes should be painted with same color and since there are 3 types of colors, so there are total 3 ways.

Input: N = 3, M = 2, K = 1
Output: 4
Let’s number the colors as 1 and 2. Four possible sequences of painting 3 boxes with 1 box having different color from color of box on its left are (1 2 2), (1 1 2), (2 1 1) (2 2 1)

Prerequisites : Dynamic Programming

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: This problem can be solved using dynamic programming where dp[i][j] will denote the number of ways to paint i boxes using M colors such that there are exactly j boxes with a color different from the color of the box on its left. For every current box except 1st, either we can paint the same color as painted on its left box and solve for dp[i – 1][j] or we can paint it with remaining M – 1 colors and solve for dp[i – 1][j – 1] recursively.

Below is the implementation of the above approach:

## C++

 `// CPP Program to Paint N boxes using M ` `// colors such that K boxes have color ` `// different from color of box on its left ` `#include ` `using` `namespace` `std; ` ` `  `const` `int` `M = 1001; ` `const` `int` `MOD = 998244353; ` ` `  `int` `dp[M][M]; ` ` `  `// This function returns the required number ` `// of ways where idx is the current index and ` `// diff is number of boxes having different ` `// color from box on its left ` `int` `solve(``int` `idx, ``int` `diff, ``int` `N, ``int` `M, ``int` `K) ` `{ ` `    ``// Base Case ` `    ``if` `(idx > N) { ` `        ``if` `(diff == K) ` `            ``return` `1; ` `        ``return` `0; ` `    ``} ` ` `  `    ``// If already computed ` `    ``if` `(dp[idx][ diff] != -1) ` `        ``return` `dp[idx][ diff]; ` ` `  `    ``// Either paint with same color as ` `    ``// previous one ` `    ``int` `ans = solve(idx + 1, diff, N, M, K); ` ` `  `    ``// Or paint with remaining (M - 1) ` `    ``// colors ` `    ``ans += (M - 1) * solve(idx + 1, diff + 1, N, M, K); ` ` `  `    ``return` `dp[idx][ diff] = ans % MOD; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 3, M = 3, K = 0; ` `    ``memset``(dp, -1, ``sizeof``(dp)); ` ` `  `    ``// Multiply M since first box can be ` `    ``// painted with any of the M colors and ` `    ``// start solving from 2nd box ` `    ``cout << (M * solve(2, 0, N, M, K)) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java Program to Paint N boxes using M ` `// colors such that K boxes have color ` `// different from color of box on its left ` ` `  `class` `GFG ` `{ ` `     `  `    ``static` `int` `M = ``1001``; ` `    ``static` `int` `MOD = ``998244353``; ` ` `  `    ``static` `int``[][] dp = ``new` `int``[M][M]; ` ` `  `    ``// This function returns the required number ` `    ``// of ways where idx is the current index and ` `    ``// diff is number of boxes having different ` `    ``// color from box on its left ` `    ``static` `int` `solve(``int` `idx, ``int` `diff, ` `                        ``int` `N, ``int` `M, ``int` `K) ` `    ``{ ` `        ``// Base Case ` `        ``if` `(idx > N)  ` `        ``{ ` `            ``if` `(diff == K) ` `                ``return` `1``; ` `            ``return` `0``; ` `        ``} ` ` `  `        ``// If already computed ` `        ``if` `(dp[idx][ diff] != -``1``) ` `            ``return` `dp[idx][ diff]; ` ` `  `        ``// Either paint with same color as ` `        ``// previous one ` `        ``int` `ans = solve(idx + ``1``, diff, N, M, K); ` ` `  `        ``// Or paint with remaining (M - 1) ` `        ``// colors ` `        ``ans += (M - ``1``) * solve(idx + ``1``,  ` `                ``diff + ``1``, N, M, K); ` ` `  `        ``return` `dp[idx][ diff] = ans % MOD; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `N = ``3``, M = ``3``, K = ``0``; ` `        ``for``(``int` `i = ``0``; i <= M; i++) ` `            ``for``(``int` `j = ``0``; j <= M; j++) ` `                ``dp[i][j] = -``1``; ` `     `  `        ``// Multiply M since first box can be ` `        ``// painted with any of the M colors and ` `        ``// start solving from 2nd box ` `        ``System.out.println((M * solve(``2``, ``0``, N, M, K))); ` `    ``} ` `} ` ` `  `// This code is contributed by mits `

## Python3

 `# Python3 Program to Paint N boxes using M  ` `# colors such that K boxes have color  ` `# different from color of box on its left  ` ` `  `M ``=` `1001``;  ` `MOD ``=` `998244353``;  ` ` `  `dp ``=` `[[``-``1``]``*` `M ] ``*` `M ` ` `  `# This function returns the required number  ` `# of ways where idx is the current index and  ` `# diff is number of boxes having different  ` `# color from box on its left  ` `def` `solve(idx, diff, N, M, K) : ` `     `  `    ``# Base Case  ` `    ``if` `(idx > N) :  ` `        ``if` `(diff ``=``=` `K) : ` `            ``return` `1`  `        ``return` `0` ` `  `    ``# If already computed  ` `    ``if` `(dp[idx][ diff] !``=` `-``1``) : ` `        ``return` `dp[idx];  ` ` `  `    ``# Either paint with same color as  ` `    ``# previous one  ` `    ``ans ``=` `solve(idx ``+` `1``, diff, N, M, K);  ` ` `  `    ``# Or paint with remaining (M - 1)  ` `    ``# colors  ` `    ``ans ``+``=` `(M ``-` `1``) ``*` `solve(idx ``+` `1``, diff ``+` `1``, N, M, K);  ` ` `  `    ``dp[idx][ diff] ``=` `ans ``%` `MOD;  ` `     `  `    ``return` `dp[idx][ diff] ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``N ``=` `3` `    ``M ``=` `3` `    ``K ``=` `0`  ` `  `    ``# Multiply M since first box can be  ` `    ``# painted with any of the M colors and  ` `    ``# start solving from 2nd box  ` `    ``print``(M ``*` `solve(``2``, ``0``, N, M, K))  ` ` `  `# This code is contributed by Ryuga `

## C#

 `// C# Program to Paint N boxes using M ` `// colors such that K boxes have color ` `// different from color of box on its left ` `using` `System; ` `class` `GFG ` `{ ` `     `  `static` `int` `M = 1001; ` `static` `int` `MOD = 998244353; ` ` `  `static` `int``[,] dp = ``new` `int``[M, M]; ` ` `  `// This function returns the required number ` `// of ways where idx is the current index and ` `// diff is number of boxes having different ` `// color from box on its left ` `static` `int` `solve(``int` `idx, ``int` `diff, ` `                 ``int` `N, ``int` `M, ``int` `K) ` `{ ` `    ``// Base Case ` `    ``if` `(idx > N)  ` `    ``{ ` `        ``if` `(diff == K) ` `            ``return` `1; ` `        ``return` `0; ` `    ``} ` ` `  `    ``// If already computed ` `    ``if` `(dp[idx, diff] != -1) ` `        ``return` `dp[idx, diff]; ` ` `  `    ``// Either paint with same color as ` `    ``// previous one ` `    ``int` `ans = solve(idx + 1, diff, N, M, K); ` ` `  `    ``// Or paint with remaining (M - 1) ` `    ``// colors ` `    ``ans += (M - 1) * solve(idx + 1,  ` `                ``diff + 1, N, M, K); ` ` `  `    ``return` `dp[idx, diff] = ans % MOD; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main ()  ` `{ ` `    ``int` `N = 3, M = 3, K = 0; ` `    ``for``(``int` `i = 0; i <= M; i++) ` `        ``for``(``int` `j = 0; j <= M; j++) ` `            ``dp[i, j] = -1; ` ` `  `    ``// Multiply M since first box can be ` `    ``// painted with any of the M colors and ` `    ``// start solving from 2nd box ` `    ``Console.WriteLine((M * solve(2, 0, N, M, K))); ` `} ` `} ` ` `  `// This code is contributed by chandan_jnu `

## PHP

 ` ``\$N``)  ` `    ``{ ` `        ``if` `(``\$diff` `== ``\$K``) ` `            ``return` `1; ` `        ``return` `0; ` `    ``} ` ` `  `    ``// If already computed ` `    ``if` `(``\$dp``[``\$idx``][``\$diff``] != -1) ` `        ``return` `\$dp``[``\$idx``][``\$diff``]; ` ` `  `    ``// Either paint with same color  ` `    ``// as previous one ` `    ``\$ans` `= solve(``\$idx` `+ 1, ``\$diff``, ``\$N``, ``\$M``, ``\$K``); ` ` `  `    ``// Or paint with remaining (M - 1) ` `    ``// colors ` `    ``\$ans` `+= (``\$M` `- 1) * solve(``\$idx` `+ 1, ` `             ``\$diff` `+ 1, ``\$N``, ``\$M``, ``\$K``); ` ` `  `    ``return` `\$dp``[``\$idx``][``\$diff``] = ``\$ans` `% ``\$MOD``; ` `} ` ` `  `// Driver code ` `\$N` `= 3; ` `\$M` `= 3; ` `\$K` `= 0; ` ` `  `// Multiply M since first box can be ` `// painted with any of the M colors and ` `// start solving from 2nd box ` `echo` `(``\$M` `* solve(2, 0, ``\$N``, ``\$M``, ``\$K``)); ` ` `  `// This code is contributed by chandan_jnu ` `?> `

Output:

```3
```

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.