We are given two integers n and d, we need to count all n digit numbers that do not have digit d.
Example :
Input : n = 2, d = 7 Output : 72 All two digit numbers that don't have 7 as digit are 10, 11, 12, 13, 14, 15, 16, 18, ..... Input : n = 3, d = 9 Output : 648
A simple solution is to traverse through all d digit numbers. For every number, check if it has x as digit or not.
Efficient approach In this method, we check first if excluding digit d is zero or non-zero. If it is zero then, we have 9 numbers (1 to 9) as first number otherwise we have 8 numbers(excluding x digit and 0). Then for all other digits, we have 9 choices i.e (0 to 9 excluding d digit). We simple call digitNumber function with n-1 digits as first number we already find it can be 8 or 9 and simple multiply it. For other numbers we check if digits are odd or even if it is odd we call digitNumber function with (n-1)/2 digits and multiply it by 9 otherwise we call digitNumber function with n/2 digits and store them in result and take result square.
Illustration :
Number from 1 to 7 excluding digit 9. digits multiple number 1 8 8 2 8*9 72 3 8*9*9 648 4 8*9*9*9 5832 5 8*9*9*9*9 52488 6 8*9*9*9*9*9 472392 7 8*9*9*9*9*9*9 4251528
As we can see, in each step we are half the number of digits. Suppose we have 7 digits in this we call function from main with 6(7-1) digits. when we half the digits we left with 3(6/2) digits. Because of this we have to multiply result due to 3 digits with itself to get result for 6 digits. Similarly for 3 digits we have odd digits, we have odd digits. We find result with 1 ((3-1)/2) digits and find result square and multiply it with 9, because we find result for d-1 digits.
// C++ Implementation of above method #include <bits/stdc++.h> #define mod 1000000007 using namespace std;
// Finding number of possible number with // n digits excluding a particular digit long long digitNumber( long long n) {
// Checking if number of digits is zero
if (n == 0)
return 1;
// Checking if number of digits is one
if (n == 1)
return 9;
// Checking if number of digits is odd
if (n % 2) {
// Calling digitNumber function
// with (digit-1)/2 digits
long long temp = digitNumber((n - 1) / 2) % mod;
return (9 * (temp * temp) % mod) % mod;
} else {
// Calling digitNumber function
// with n/2 digits
long long temp = digitNumber(n / 2) % mod;
return (temp * temp) % mod;
}
} int countExcluding( int n, int d)
{ // Calling digitNumber function
// Checking if excluding digit is
// zero or non-zero
if (d == 0)
return (9 * digitNumber(n - 1)) % mod;
else
return (8 * digitNumber(n - 1)) % mod;
} // Driver function to run above program int main() {
// Initializing variables
long long d = 9;
int n = 3;
cout << countExcluding(n, d) << endl;
return 0;
} |
// Java Implementation of above method import java.lang.*;
class GFG {
static final int mod = 1000000007 ;
// Finding number of possible number with // n digits excluding a particular digit static int digitNumber( long n) {
// Checking if number of digits is zero
if (n == 0 )
return 1 ;
// Checking if number of digits is one
if (n == 1 )
return 9 ;
// Checking if number of digits is odd
if (n % 2 != 0 ) {
// Calling digitNumber function
// with (digit-1)/2 digits
int temp = digitNumber((n - 1 ) / 2 ) % mod;
return ( 9 * (temp * temp) % mod) % mod;
}
else {
// Calling digitNumber function
// with n/2 digits
int temp = digitNumber(n / 2 ) % mod;
return (temp * temp) % mod;
}
} static int countExcluding( int n, int d) {
// Calling digitNumber function
// Checking if excluding digit is
// zero or non-zero
if (d == 0 )
return ( 9 * digitNumber(n - 1 )) % mod;
else
return ( 8 * digitNumber(n - 1 )) % mod;
} // Driver function to run above program public static void main(String[] args) {
// Initializing variables
int d = 9 ;
int n = 3 ;
System.out.println(countExcluding(n, d));
} } // This code is contributed by Anant Agarwal. |
# Python Implementation # of above method mod = 1000000007
# Finding number of # possible number with # n digits excluding # a particular digit def digitNumber(n):
# Checking if number
# of digits is zero
if (n = = 0 ):
return 1
# Checking if number
# of digits is one
if (n = = 1 ):
return 9
# Checking if number
# of digits is odd
if (n % 2 ! = 0 ):
# Calling digitNumber function
# with (digit-1)/2 digits
temp = digitNumber((n - 1 ) / / 2 ) % mod
return ( 9 * (temp * temp) % mod) % mod
else :
# Calling digitNumber function
# with n/2 digits
temp = digitNumber(n / / 2 ) % mod
return (temp * temp) % mod
def countExcluding(n,d):
# Calling digitNumber function
# Checking if excluding digit is
# zero or non-zero
if (d = = 0 ):
return ( 9 * digitNumber(n - 1 )) % mod
else :
return ( 8 * digitNumber(n - 1 )) % mod
# Driver code d = 9
n = 3
print (countExcluding(n, d))
# This code is contributed # by Anant Agarwal. |
// C# Implementation of above method using System;
class GFG {
static int mod = 1000000007;
// Finding number of possible number with
// n digits excluding a particular digit
static int digitNumber( long n) {
// Checking if number of digits is zero
if (n == 0)
return 1;
// Checking if number of digits is one
if (n == 1)
return 9;
// Checking if number of digits is odd
if (n % 2 != 0) {
// Calling digitNumber function
// with (digit-1)/2 digits
int temp = digitNumber((n - 1) / 2) % mod;
return (9 * (temp * temp) % mod) % mod;
}
else {
// Calling digitNumber function
// with n/2 digits
int temp = digitNumber(n / 2) % mod;
return (temp * temp) % mod;
}
}
static int countExcluding( int n, int d) {
// Calling digitNumber function
// Checking if excluding digit is
// zero or non-zero
if (d == 0)
return (9 * digitNumber(n - 1)) % mod;
else
return (8 * digitNumber(n - 1)) % mod;
}
// Driver function to run above program
public static void Main() {
// Initializing variables
int d = 9;
int n = 3;
Console.WriteLine(countExcluding(n, d));
}
} // This code is contributed by vt_m. |
<?php // PHP Implementation // of above method $mod = 1000000007;
// Finding number of // possible number with // n digits excluding // a particular digit function digitNumber( $n )
{ global $mod ;
// Checking if number
// of digits is zero
if ( $n == 0)
return 1;
// Checking if number
// of digits is one
if ( $n == 1)
return 9;
// Checking if number
// of digits is odd
if ( $n % 2 != 0)
{
// Calling digitNumber
// function with
// (digit-1)/2 digits;
$temp = digitNumber(( $n -
1) / 2) % $mod ;
return (9 * ( $temp * $temp ) %
$mod ) % $mod ;
}
else
{
// Calling digitNumber
// function with n/2 digits
$temp = digitNumber( $n /
2) % $mod ;
return ( $temp *
$temp ) % $mod ;
}
} function countExcluding( $n , $d )
{ global $mod ;
// Calling digitNumber
// function Checking if
// excluding digit is
// zero or non-zero
if ( $d == 0)
return (9 * digitNumber( $n -
1)) % $mod ;
else
return (8 * digitNumber( $n -
1)) % $mod ;
} // Driver code $d = 9;
$n = 3;
print (countExcluding( $n , $d ));
// This code is contributed by // Manish Shaw(manishshaw1) ?> |
<script> // JavaScript Implementation of above method const mod = 1000000007; // Finding number of possible number with // n digits excluding a particular digit function digitNumber(n) {
// Checking if number of digits is zero if (n == 0)
return 1;
// Checking if number of digits is one if (n == 1)
return 9;
// Checking if number of digits is odd if (n % 2) {
// Calling digitNumber function
// with (digit-1)/2 digits
let temp = digitNumber((n - 1) / 2) % mod;
return (9 * (temp * temp) % mod) % mod;
} else {
// Calling digitNumber function
// with n/2 digits
let temp = digitNumber(n / 2) % mod;
return (temp * temp) % mod;
} } function countExcluding(n, d)
{ // Calling digitNumber function // Checking if excluding digit is // zero or non-zero if (d == 0)
return (9 * digitNumber(n - 1)) % mod;
else return (8 * digitNumber(n - 1)) % mod;
} // Driver function to run above program // Initializing variables let d = 9; let n = 3; document.write(countExcluding(n, d) + "<br>" );
// This code is contributed by Surbhi Tyagi. </script> |
648
Time Complexity: O(log n), where n represents the given integer.
Auxiliary Space: O(logn), due to recursive stack space.