# Ways to paint N paintings such that adjacent paintings don’t have same colors

Given two integers n and m, where n represent some paintings numbered from 1 to n and m represent some colours 1 to m with unlimited amount. The task is to find the number of ways to paint the paintings such that no two consecutive paintings have the same colors.

**Note:** Answer must be calculated in modulo 10^9 +7 as answer can be very large.

**Examples:**

Input:n = 4, m = 2Output:2Input:n = 4, m = 6Output:750

Asked in : National Instruments

**Approach:**

The total number of given color is **m** and the total paintings are from 1 to **n**. As per the condition of no two adjacent painting having the same color, first painting can be painted by anyone out of n colors and the rest of any painting can be painted by any of n-1 color except the color used for the painting just preceding that. Hence if we derive the solution for total number of ways,

n * (m-1)^(n-1)is the actual answer.

Now, this can be either calculated by simple iteration or by the method of efficient power calculation in **O(logn)** time.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the above approach ` `#include <bits/stdc++.h> ` `#define modd 1000000007 ` `using` `namespace` `std; ` ` ` `// Function for finding the power ` `unsigned ` `long` `power(unsigned ` `long` `x, ` ` ` `unsigned ` `long` `y, unsigned ` `long` `p) ` `{ ` ` ` `unsigned ` `long` `res = 1; ` `// Initialize result ` ` ` ` ` `x = x % p; ` `// Update x if it is more than or ` ` ` `// equal to p ` ` ` ` ` `while` `(y > 0) { ` ` ` ` ` `// If y is odd, multiply x with result ` ` ` `if` `(y & 1) ` ` ` `res = (res * x) % p; ` ` ` ` ` `// y must be even now ` ` ` `y = y >> 1; ` `// y = y/2 ` ` ` `x = (x * x) % p; ` ` ` `} ` ` ` `return` `res; ` `} ` ` ` `// Function to calculate the number of ways ` `int` `ways(` `int` `n, ` `int` `m) ` `{ ` ` ` `// Answer must be modulo of 10^9 + 7 ` ` ` `return` `power(m - 1, n - 1, modd) * m % modd; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `n = 5, m = 5; ` ` ` `cout << ways(n, m); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the above approach ` ` ` `class` `GFG ` `{ ` ` ` `static` `final` `int` `modd = ` `1000000007` `; ` ` ` ` ` `// Function for finding the power ` ` ` `static` `long` `power(` `long` `x, ` `long` `y, ` `long` `p) ` ` ` `{ ` ` ` `long` `res = ` `1` `; ` `// Initialize result ` ` ` ` ` `// Update x if it is more than or ` ` ` `// equal to p ` ` ` `x = x % p; ` ` ` ` ` `while` `(y > ` `0` `) ` ` ` `{ ` ` ` `// If y is odd, multiply x with result ` ` ` `if` `(y % ` `2` `== ` `1` `) ` ` ` `{ ` ` ` `res = (res * x) % p; ` ` ` `} ` ` ` ` ` `// y must be even now ` ` ` `y = y >> ` `1` `; ` `// y = y/2 ` ` ` `x = (x * x) % p; ` ` ` `} ` ` ` `return` `res; ` ` ` `} ` ` ` ` ` `// Function to calculate the number of ways ` ` ` `static` `int` `ways(` `int` `n, ` `int` `m) ` ` ` `{ ` ` ` `// Answer must be modulo of 10^9 + 7 ` ` ` `return` `(` `int` `) (power(m - ` `1` `, n - ` `1` `, modd) ` ` ` `* m % modd); ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `n = ` `5` `, m = ` `5` `; ` ` ` `System.out.println(ways(n, m)); ` ` ` ` ` `} ` `} ` ` ` `// This code is contributed by 29AjayKumar ` |

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## Python3

`# Python3 implementation of the ` `# above approach ` ` ` `modd ` `=` `1000000007` ` ` `# Function for finding the power ` `def` `power(x, y, p): ` ` ` ` ` `res ` `=` `1` `# Initialize result ` ` ` ` ` `x ` `=` `x ` `%` `p ` `# Update x if it is more ` ` ` `# than or equal to p ` ` ` ` ` `while` `(y > ` `0` `): ` ` ` ` ` `# If y is odd, multiply x with result ` ` ` `if` `(y & ` `1` `): ` ` ` `res ` `=` `(res ` `*` `x) ` `%` `p ` ` ` ` ` `# y must be even now ` ` ` `y ` `=` `y >> ` `1` `# y = y/2 ` ` ` `x ` `=` `(x ` `*` `x) ` `%` `p ` ` ` ` ` `return` `res ` ` ` `# Function to calculate the number of ways ` `def` `ways(n, m): ` ` ` ` ` `# Answer must be modulo of 10^9 + 7 ` ` ` `return` `power(m ` `-` `1` `, n ` `-` `1` `, modd) ` `*` `m ` `%` `modd ` ` ` `# Driver code ` `n, m ` `=` `5` `, ` `5` `print` `(ways(n, m)) ` ` ` `# This code is contributed ` `# by Mohit Kumar 29 ` |

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## C#

`// C# implementation of the above approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `static` `int` `modd = 1000000007; ` ` ` ` ` `// Function for finding the power ` ` ` `static` `long` `power(` `long` `x, ` `long` `y, ` `long` `p) ` ` ` `{ ` ` ` `long` `res = 1; ` `// Initialize result ` ` ` ` ` `// Update x if it is more than or ` ` ` `// equal to p ` ` ` `x = x % p; ` ` ` ` ` `while` `(y > 0) ` ` ` `{ ` ` ` `// If y is odd, multiply x with result ` ` ` `if` `(y % 2 == 1) ` ` ` `{ ` ` ` `res = (res * x) % p; ` ` ` `} ` ` ` ` ` `// y must be even now ` ` ` `y = y >> 1; ` `// y = y/2 ` ` ` `x = (x * x) % p; ` ` ` `} ` ` ` `return` `res; ` ` ` `} ` ` ` ` ` `// Function to calculate the number of ways ` ` ` `static` `int` `ways(` `int` `n, ` `int` `m) ` ` ` `{ ` ` ` `// Answer must be modulo of 10^9 + 7 ` ` ` `return` `(` `int` `) (power(m - 1, n - 1, modd) ` ` ` `* m % modd); ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `static` `public` `void` `Main () ` ` ` `{ ` ` ` `int` `n = 5, m = 5; ` ` ` `Console.WriteLine(ways(n, m)); ` ` ` `} ` `} ` ` ` `// This code is contributed by ajit ` |

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## PHP

0)

{

// If y is odd, multiply

// x with result

if ($y & 1)

$res = ($res * $x) % $p;

// y must be even now

// y = $y/2

$y = $y >> 1;

$x = ($x * $x) % $p;

}

return $res;

}

// Function to calculate the number of ways

function ways($n, $m)

{

$modd =1000000007;

// Answer must be modulo of 10^9 + 7

return (power($m – 1, $n – 1,

$modd) * $m ) % $modd;

}

// Driver code

$n = 5;

$m = 5;

echo ways($n, $m);

// This code is contributed

// by Arnab Kundu

?>

**Output:**

1280

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