Given are two integers (x and n). The task is to find an array such that it contains the frequency of index numbers occurring in (x^1, x^2, …., x^(n-1), x^(n) ).
Examples:
Input: x = 15, n = 3 Output: 0 1 2 2 0 3 0 1 0 0 Numbers x^1 to x^n are 15, 225, 3375. So frequency array is 0 1 2 2 0 3 0 1 0 0. Input: x = 1, n = 5 Output: 0 5 0 0 0 0 0 0 0 0 Numbers x^1 to x^n are 1, 1, 1, 1, 1. So frequency of digits is 0 5 0 0 0 0 0 0 0 0.
Approach:
- Maintain a frequency count array to store the count of digits 0-9.
- Traverse through each digit from x^1 to x^n, for each digit add 1 to the corresponding index in the frequency count array.
- Print the frequency array
Below is the implementation of the above approach:
C++
// CPP implementation of above approach #include<bits/stdc++.h> using namespace std; // Function that traverses digits in a number and // modifies frequency count array void countDigits( double val, long arr[]) { while (( long )val > 0) { long digit = ( long )val % 10; arr[( int )digit]++; val = ( long )val / 10; } return ; } void countFrequency( int x, int n) { // Array to keep count of digits long freq_count[10]={0}; // Traversing through x^1 to x^n for ( int i = 1; i <= n; i++) { // For power function, both its parameters are // to be in double double val = pow (( double )x, ( double )i); // calling countDigits function on x^i countDigits(val, freq_count); } // Printing count of digits 0-9 for ( int i = 0; i <= 9; i++) { cout << freq_count[i] << " " ; } } // Driver code int main() { int x = 15, n = 3; countFrequency(x, n); } // This code is contributed by ihritik |
Java
// Java implementation of above approach import java.io.*; import java.util.*; public class GFG { // Function that traverses digits in a number and // modifies frequency count array static void countDigits( double val, long [] arr) { while (( long )val > 0 ) { long digit = ( long )val % 10 ; arr[( int )digit]++; val = ( long )val / 10 ; } return ; } static void countFrequency( int x, int n) { // Array to keep count of digits long [] freq_count = new long [ 10 ]; // Traversing through x^1 to x^n for ( int i = 1 ; i <= n; i++) { // For power function, both its parameters are // to be in double double val = Math.pow(( double )x, ( double )i); // calling countDigits function on x^i countDigits(val, freq_count); } // Printing count of digits 0-9 for ( int i = 0 ; i <= 9 ; i++) { System.out.print(freq_count[i] + " " ); } } // Driver code public static void main(String args[]) { int x = 15 , n = 3 ; countFrequency(x, n); } } |
Python 3
# Python 3 implementation # of above approach import math # Function that traverses digits # in a number and modifies # frequency count array def countDigits(val, arr): while (val > 0 ) : digit = val % 10 arr[ int (digit)] + = 1 val = val / / 10 return ; def countFrequency(x, n): # Array to keep count of digits freq_count = [ 0 ] * 10 # Traversing through x^1 to x^n for i in range ( 1 , n + 1 ) : # For power function, # both its parameters # are to be in double val = math. pow (x, i) # calling countDigits # function on x^i countDigits(val, freq_count) # Printing count of digits 0-9 for i in range ( 10 ) : print (freq_count[i], end = " " ); # Driver code if __name__ = = "__main__" : x = 15 n = 3 countFrequency(x, n) # This code is contributed # by ChitraNayal |
C#
// C# implementation of above approach using System; class GFG { // Function that traverses digits // in a number and modifies // frequency count array static void countDigits( double val, long [] arr) { while (( long )val > 0) { long digit = ( long )val % 10; arr[( int )digit]++; val = ( long )val / 10; } return ; } static void countFrequency( int x, int n) { // Array to keep count of digits long [] freq_count = new long [10]; // Traversing through x^1 to x^n for ( int i = 1; i <= n; i++) { // For power function, both its // parameters are to be in double double val = Math.Pow(( double )x, ( double )i); // calling countDigits // function on x^i countDigits(val, freq_count); } // Printing count of digits 0-9 for ( int i = 0; i <= 9; i++) { Console.Write(freq_count[i] + " " ); } } // Driver code public static void Main() { int x = 15, n = 3; countFrequency(x, n); } } // This code is contributed // by Shashank |
PHP
<?php // PHP implementation of above approach // Function that traverses digits // in a number and modifies // frequency count array function countDigits( $val , & $arr ) { while ( $val > 0) { $digit = $val % 10; $arr [(int)( $digit )] += 1; $val = (int)( $val / 10); } return ; } function countFrequency( $x , $n ) { // Array to keep count of digits $freq_count = array_fill (0, 10, 0); // Traversing through x^1 to x^n for ( $i = 1; $i < $n + 1; $i ++) { // For power function, // both its parameters // are to be in double $val = pow( $x , $i ); // calling countDigits // function on x^i countDigits( $val , $freq_count ); } // Printing count of digits 0-9 for ( $i = 0; $i < 10; $i ++) { echo $freq_count [ $i ] . " " ; } } // Driver code $x = 15; $n = 3; countFrequency( $x , $n ) // This code is contributed by mits ?> |
Javascript
<script> // Javascript implementation of above approach // Function that traverses digits in a number and // modifies frequency count array function countDigits(val,arr) { while (val > 0) { let digit = val % 10; arr[digit]++; val = Math.floor(val / 10); } return ; } function countFrequency(x,n) { // Array to keep count of digits let freq_count = new Array(10); for (let i=0;i<10;i++) { freq_count[i]=0; } // Traversing through x^1 to x^n for (let i = 1; i <= n; i++) { // For power function, both its parameters are // to be in double let val = Math.pow(x, i); // calling countDigits function on x^i countDigits(val, freq_count); } // Printing count of digits 0-9 for (let i = 0; i <= 9; i++) { document.write(freq_count[i] + " " ); } } // Driver code let x = 15, n = 3; countFrequency(x, n); // This code is contributed by avanitrachhadiya2155 </script> |
Output
0 1 2 2 0 3 0 1 0 0
Complexity Analysis:
- Time complexity: O(nlogn) since using a pow function “logn time complexity” inside a for loop
- Auxiliary Space: O(10)
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