Given a number d, representing the number of digits of a positive integer. Find the total count of positive integer (consisting of d digits exactly) which have at-least one zero in them.
Examples:
Input : d = 1
Output : 0
There's no natural number of 1 digit that
contains a zero.
Input : d = 2
Output : 9
The numbers are, 10, 20, 30, 40, 50, 60,
70, 80 and 90.
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One Simple Solution is to traverse through all d digit positive numbers. For every number, traverse through its digits and if there is any 0 digit, increment count (similar to this).
Following are some observations:
- There are exactly d digits.
- The number at most significant place can’t be a zero (no leading zeroes allowed).
- All the other places except the most significant one can contain zero .
So considering the above points, let’s find the total count of numbers having d digits:
We can place any of {1, 2, ... 9} in D1 Hence D1 can be filled in 9 ways. Apart from D1 all the other places can be 10 ways. (we can place 0 as well) Hence the total numbers having d digits can be given as: Total = 9*10d-1 Now, let's find the numbers having d digits, that don't contain zero at any place. In this case, all the places can be filled in 9 ways. Hence count of such numbers is given by: Non_Zero = 9d Now the count of numbers having at least one zero can be obtained by subtracting Non_Zero from Total. Hence Answer would be given by: 9*(10d-1 -9d-1)
Below is the program for the same.
C++
//C++ program to find the count of positive integer of a// given number of digits that contain atleast one zero#include<bits/stdc++.h>using namespace std;// Returns count of 'd' digit integers have 0 as a digitint findCount(int d){ return 9*(pow(10,d-1) - pow(9,d-1));}// Driver Codeint main(){ int d = 1; cout << findCount(d) << endl; d = 2; cout << findCount(d) << endl; d = 4; cout << findCount(d) << endl; return 0;} |
Java
// Java program to find the count// of positive integer of a// given number of digits // that contain atleast one zeroimport java.io.*;class GFG { // Returns count of 'd' digit // integers have 0 as a digit static int findCount(int d) { return 9 * ((int)(Math.pow(10, d - 1)) - (int)(Math.pow(9, d - 1))); } // Driver Code public static void main(String args[]) { int d = 1; System.out.println(findCount(d)); d = 2; System.out.println(findCount(d)); d = 4; System.out.println(findCount(d)); }}// This code is contributed by Nikita Tiwari. |
Python3
# Python 3 program to find the# count of positive integer of a# given number of digits that# contain atleast one zeroimport math# Returns count of 'd' digit# integers have 0 as a digitdef findCount(d) : return 9*((int)(math.pow(10,d-1)) - (int)(math.pow(9,d-1)));# Driver Coded = 1print(findCount(d)) d = 2print(findCount(d))d = 4print(findCount(d))# This code is contributed by Nikita Tiwari. |
C#
// C# program to find the count// of positive integer of a// given number of digits // that contain atleast one zero.using System;class GFG { // Returns count of 'd' digit // integers have 0 as a digit static int findCount(int d) { return 9 * ((int)(Math.Pow(10, d - 1)) - (int)(Math.Pow(9, d - 1))); } // Driver Code public static void Main() { int d = 1; Console.WriteLine(findCount(d)); d = 2; Console.WriteLine(findCount(d)); d = 4; Console.WriteLine(findCount(d)); }}// This code is contributed by nitin mittal. |
Javascript
<script>// JavaScript program to find the count// of positive integer of a// given number of digits // that contain atleast one zero // Returns count of 'd' digit // integers have 0 as a digit function findCount(d) { return 9 * ((Math.pow(10, d - 1)) - (Math.pow(9, d - 1))); }// Driver Code let d = 1; document.write(findCount(d) + "<br/>"); d = 2; document.write(findCount(d) + "<br/>"); d = 4; document.write(findCount(d) + "<br/>"); // This code is contributed by target_2.</script> |
PHP
<?php// PHP program to find the count// of positive integer of a given // number of digits that contain// atleast one zero// Returns count of 'd' digit // integers have 0 as a digitfunction findCount($d){ return 9 * (pow(10, $d - 1) - pow(9, $d - 1));}// Driver Code{ $d = 1; echo findCount($d),"\n"; $d = 2; echo findCount($d),"\n"; $d = 4; echo findCount($d), "\n"; return 0;}// This code is contributed by nitin mittal?> |
Output :
0
9
2439
Time Complexity : O(logd) for given d
Auxiliary Space: O(1)
This article is contributed by Ashutosh Kumar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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