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

  • Difficulty Level : Hard
  • Last Updated : 11 Aug, 2021

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:
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:
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 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 color 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;
}

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




<?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
?>

Javascript




<script>
 
    // JavaScript Program to Paint N boxes using M
    // colors such that K boxes have color
    // different from color of box on its left
     
    let m = 1001;
    let MOD = 998244353;
   
    let dp = new Array(m);
    for(let i = 0; i < m; i++)
    {
        dp[i] = new Array(m);
        for(let j = 0; j < m; j++)
        {
            dp[i][j] = 0;
        }
    }
   
    // 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)
    {
        // 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
        let 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];
    }
     
    let N = 3, M = 3, K = 0;
    for(let i = 0; i <= M; i++)
      for(let 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
    document.write((M * solve(2, 0, N, M, K)));
     
</script>
Output: 
3

 

Time Complexity: O(M*M)
Auxiliary Space: O(M*M)




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