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

**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 **1 ^{st}**, 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 <bits/stdc++.h> ` `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; ` `} ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## PHP

`<?php ` `// PHP 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; ` ` ` `$dp` `= ` `array_fill` `(0, ` `$M` `, ` ` ` `array_fill` `(0, ` `$M` `, -1)); ` ` ` `// This function returns the required number ` `// of ways where idx is the current index ` `// and diff is number of boxes having ` `// different color from box on its left ` `function` `solve(` `$idx` `, ` `$diff` `, ` `$N` `, ` `$M` `, ` `$K` `) ` `{ ` ` ` `global` `$dp` `, ` `$MOD` `; ` ` ` ` ` `// 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 ` ` ` `$ans` `= solve(` `$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 ` `?> ` |

*chevron_right*

*filter_none*

**Output:**

3

## Recommended Posts:

- Ways to fill N positions using M colors such that there are exactly K pairs of adjacent different colors
- Ways to color a skewed tree such that parent and child have different colors
- Color all boxes in line such that every M consecutive boxes are unique
- Minimum colors required such that edges forming cycle do not have same color
- Ways to paint N paintings such that adjacent paintings don't have same colors
- Minimize number of boxes by putting small box inside bigger one
- Count of Array elements greater than all elements on its left and at least K elements on its right
- Count of Array elements greater than all elements on its left and next K elements on its right
- Total number of different staircase that can made from N boxes
- Maximum area of triangle having different vertex colors
- Maximize the number of indices such that element is greater than element to its left
- Minimum steps to color the tree with given colors
- Check if quantities of 3 distinct colors can be converted to a single color by given merge-pair operations
- Find N numbers such that a number and its reverse are divisible by sum of its digits
- Delete linked list nodes which have a greater value on left side
- Delete linked list nodes which have a Lesser Value on Left Side
- Find the number of points that have atleast 1 point above, below, left or right of it
- Color a grid such that all same color cells are connected either horizontally or vertically
- Minimum number of stacks possible using boxes of given capacities
- Querying the number of distinct colors in a subtree of a colored tree using BIT

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.