Given a number of digits ‘N’ and base ‘B’, the task is to count all the ‘N’ digit numbers without leading zeros that are in base ‘B’
Examples:
Input: N = 2, B = 2 Output: 2 All possible numbers without leading zeros are 10 and 11. Input: N = 5, B = 8 Output: 28672
Approach:
- If the base is ‘B’ then every digit of the number can take any value within the range [0, B-1].
- So, B
‘N’ digit numbers are possible with base ‘B’ (including the numbers with leading zeros). - And, if we fix the first digit as ‘0’ then the rest of the ‘N-1’ digits can form a total of B
numbers. - So, the total number of ‘N’ digit numbers with base ‘B’ possible without leading zeros are B– B.
‘N’ digit numbers are possible with base ‘B’ (including the numbers with leading zeros).Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// function to count// all permutationsvoid countPermutations(int N, int B){ // count of // all permutations int x = pow(B, N); // count of permutations // with leading zeros int y = pow(B, N - 1); // Return the permutations // without leading zeros cout << x - y << "\n";}// Driver codeint main(){ int N = 6; int B = 4; countPermutations(N, B); return 0;} |
Java
// Java implementation of the approachclass GFG{// function to count// all permutationsstatic void countPermutations(int N, int B){ // count of // all permutations int x = (int)Math.pow(B, N); // count of permutations // with leading zeros int y = (int)Math.pow(B, N - 1); // Return the permutations // without leading zeros System.out.println(x - y);}// Driver codepublic static void main(String[] args){ int N = 6; int B = 4; countPermutations(N, B);}}// This code is contributed by mits |
Python3
# Python3 implementation of the approach # function to count all permutations def countPermutations(N, B): # count of all permutations x = B ** N # count of permutations # with leading zeros y = B ** (N - 1) # Return the permutations # without leading zeros print(x - y) # Driver code if __name__ == "__main__": N, B = 6, 4 countPermutations(N, B) # This code is contributed by Rituraj Jain |
C#
// C# implementation of the approachusing System;class GFG{// function to count// all permutationsstatic void countPermutations(int N, int B){ // count of // all permutations int x = (int)Math.Pow(B, N); // count of permutations // with leading zeros int y = (int)Math.Pow(B, N - 1); // Return the permutations // without leading zeros Console.WriteLine(x - y);}// Driver codepublic static void Main(){ int N = 6; int B = 4; countPermutations(N, B);}}// This code is contributed// by Akanksha Rai(Abby_akku) |
PHP
<?php// PHP implementation of the approach // function to count all permutations function countPermutations($N, $B) { // count of all permutations $x = pow($B, $N); // count of permutations // with leading zeros $y = pow($B, $N - 1); // Return the permutations // without leading zeros echo ($x - $y), "\n"; } // Driver code$N = 6; $B = 4; countPermutations($N, $B); // This code is contributed // by Sach_Code` ?> |
Javascript
<script>// Javascript implementation of the approach// function to count// all permutationsfunction countPermutations(N, B){ // count of // all permutations var x = Math.pow(B, N); // count of permutations // with leading zeros var y = Math.pow(B, N - 1); // Return the permutations // without leading zeros document.write( x - y );}// Driver codevar N = 6;var B = 4;countPermutations(N, B);</script> |
Output:
3072
Time Complexity: O(logn), since pow function takes logn time to find the power of a number to base n.
Auxiliary Space: O(1), since no extra space has been taken.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
