Given a skewed tree (Every node has at most one child) with N nodes and K colors. You have to assign a color from 1 to K to each node such that parent and child has different colors. Find the maximum number of ways of coloring the nodes.
Examples:
Input : N = 2, K = 2.
Output :
Let A1 and A2 be the two nodes.
Let A1 is parent of A2.
Colors are Red and Blue.
Case 1: A1 is colored Red
and A2 is colored Blue.
Case 2: A1 is colored Blue
and A2 is colored Red.
No. of ways : 2
Input : N = 3, K = 3.
Output :
A1, A2, A3 are the nodes.
A1 is parent of A2
and A2 is parent of A3.
Let colors be R, B, G.
A1 can choose any three colors
and A2 can choose
any other two colors
and A3 can choose
any other two colors
than its parents.
No. of ways: 12
Note that only the root and children (children, grand children, grand grand children …. and all) should have different colors. The root of the tree can choose any of the K colors so K ways. Every other node can choose other K-1 colors other than its parent. So every node has K-1 choices.
Here, we select the tree as every node as only one child. We can choose any of the K colors for the root of the tree so K ways. And we are left with K-1 colors for its child. So for every child we can assign a color other than its parent. Thus, for each of the N-1 nodes we are left with K-1 colors. Thus the answer is K*(K-1)^(N-1).
We can find the answer by using normal power function which takes O(N) time complexity. But for better time complexity we use Faster Exponentiation function which takes O(log N) time complexity.
Implementation:
C++
// C++ program to count number of ways to color// a N node skewed tree with k colors such that// parent and children have different colors.#include <bits/stdc++.h>using namespace std;// fast_way is recursive// method to calculate powerint fastPow(int N, int K){ if (K == 0) return 1; int temp = fastPow(N, K / 2); if (K % 2 == 0) return temp * temp; else return N * temp * temp;}int countWays(int N, int K){ return K * fastPow(K - 1, N - 1);}// driver programint main(){ int N = 3, K = 3; cout << countWays(N, K); return 0;} |
Java
// Java program to count number of ways to color// a N node skewed tree with k colors such that// parent and children have different colors.import java.io.*;class GFG { // fast_way is recursive // method to calculate power static int fastPow(int N, int K) { if (K == 0) return 1; int temp = fastPow(N, K / 2); if (K % 2 == 0) return temp * temp; else return N * temp * temp; } static int countWays(int N, int K) { return K * fastPow(K - 1, N - 1); } // Driver program public static void main(String[] args) { int N = 3, K = 3; System.out.println(countWays(N, K)); }}// This code is contributed by vt_m. |
Python3
# Python3 program to count # number of ways to color # a N node skewed tree with # k colors such that parent # and children have different # colors.# fast_way is recursive# method to calculate powerdef fastPow(N, K): if (K == 0): return 1; temp = fastPow(N, int(K / 2)); if (K % 2 == 0): return temp * temp; else: return N * temp * temp;def countWays(N, K): return K * fastPow(K - 1, N - 1);# Driver CodeN = 3; K = 3;print(countWays(N, K));# This code is contributed by mits |
C#
// C# program to count number of ways// to color a N node skewed tree with// k colors such that parent and// children have different colorsusing System;class GFG { // fast_way is recursive // method to calculate power static int fastPow(int N, int K) { if (K == 0) return 1; int temp = fastPow(N, K / 2); if (K % 2 == 0) return temp * temp; else return N * temp * temp; } static int countWays(int N, int K) { return K * fastPow(K - 1, N - 1); } // Driver code public static void Main() { int N = 3, K = 3; Console.WriteLine(countWays(N, K)); }}// This code is contributed by vt_m. |
PHP
<?php// PHP program to count number // of ways to color a N node // skewed tree with k colors // such that parent and children// have different colors.// fast_way is recursive// method to calculate powerfunction fastPow($N, $K){ if ($K == 0) return 1; $temp = fastPow($N, $K / 2); if ($K % 2 == 0) return $temp * $temp; else return $N * $temp * $temp;}function countWays($N, $K){ return $K * fastPow($K - 1, $N - 1);} // Driver Code $N = 3; $K = 3; echo countWays($N, $K);// This code is contributed by ajit?> |
Javascript
<script>// Javascript program to count number of ways to color// a N node skewed tree with k colors such that// parent and children have different colors. // fast_way is recursive // method to calculate power function fastPow(N, K) { if (K == 0) return 1; let temp = fastPow(N, Math.floor(K / 2)); if (K % 2 == 0) return temp * temp; else return N * temp * temp; } function countWays(N, K) { return K * fastPow(K - 1, N - 1); } // Driver code let N = 3, K = 3; document.write(countWays(N, K)); // This code is contributed by sanjoy_62.</script> |
Javascript
<script>// Javascript program to count number of ways to color// a N node skewed tree with k colors such that// parent and children have different colors. // fast_way is recursive // method to calculate power function fastPow(N, K) { if (K == 0 || N == 0) return 1; return K *(K - 1)*(N - 1); } // Driver code let N = 3, K = 3; document.write(fastPow(N, K)); // This code is contributed by sanjoy_62.</script> |
12
Time Complexity: O(log N)
Auxiliary Space: O(1)
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