Given two integers N and M, the task is count the total words of M character length formed by the given N distinct characters such that the words have at least one character repeated more than once.
Examples:
Input: N = 3, M = 2
Output: 3
Suppose the characters are {‘a’, ‘b’, ‘c’}
All 2 length words that can be formed with these characters
are “aa”, “ab”, “ac”, “ba”, “bb”, “bc”, “ca”, “cb” and “cc”.
Out of these words only “aa”, “bb” and “cc” have
at least one character repeated more than once.Input: N = 10, M = 5
Output: 69760
Approach:
Total number of M character words possible from N characters, total = NM.
Total number of M character words possible from N characters where no character repeats itself, noRepeat = NPM.
So, total words where at least a single character appear more than once is total – noRepeat i.e. NM – NPM.
Below is the implementation of the above approach:
C++
// C++ implementation for the above approach#include <math.h> #include <iostream>using namespace std;// Function to return the// factorial of a numberint fact(int n){ if (n <= 1) return 1; return n * fact(n - 1);}// Function to return the value of nPrint nPr(int n, int r){ return fact(n) / fact(n - r);}// Function to return the total number of// M length words which have at least a// single character repeated more than onceint countWords(int N, int M){ return pow(N, M) - nPr(N, M);}// Driver codeint main(){ int N = 10, M = 5; cout << (countWords(N, M)); return 0;}// This code is contributed by jit_t |
Java
// Java implementation of the approachclass GFG { // Function to return the // factorial of a number static int fact(int n) { if (n <= 1) return 1; return n * fact(n - 1); } // Function to return the value of nPr static int nPr(int n, int r) { return fact(n) / fact(n - r); } // Function to return the total number of // M length words which have at least a // single character repeated more than once static int countWords(int N, int M) { return (int)Math.pow(N, M) - nPr(N, M); } // Driver code public static void main(String[] args) { int N = 10, M = 5; System.out.print(countWords(N, M)); }} |
Python3
# Python3 implementation for the above approach# Function to return the# factorial of a numberdef fact(n): if (n <= 1): return 1; return n * fact(n - 1);# Function to return the value of nPrdef nPr(n, r): return fact(n) // fact(n - r);# Function to return the total number of# M length words which have at least a# single character repeated more than oncedef countWords(N, M): return pow(N, M) - nPr(N, M);# Driver codeN = 10; M = 5;print(countWords(N, M));# This code is contributed by Code_Mech |
C#
// C# implementation of the approachusing System;class GFG{ // Function to return the // factorial of a number static int fact(int n) { if (n <= 1) return 1; return n * fact(n - 1); } // Function to return the value of nPr static int nPr(int n, int r) { return fact(n) / fact(n - r); } // Function to return the total number of // M length words which have at least a // single character repeated more than once static int countWords(int N, int M) { return (int)Math.Pow(N, M) - nPr(N, M); } // Driver code static public void Main () { int N = 10, M = 5; Console.Write(countWords(N, M)); }}// This code is contributed by ajit. |
Javascript
// javascript implementation of the approach // Function to return the // factorial of a number function fact(n) { if (n <= 1) return 1; return n * fact(n - 1); } // Function to return the value of nPr function nPr( n, r) { return fact(n) / fact(n - r); } // Function to return the total number of // M length words which have at least a // single character repeated more than once function countWords( N, M) { return Math.pow(N, M) - nPr(N, M); } // Driver code var N = 10 ; var M = 5; document.write(countWords(N, M)); // This code is contributed by bunnyram19. |
69760
Time Complexity: O(n)
Auxiliary Space: O(N), for recursive stack space.
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