Find the number of positive integers less than or equal to N that have an odd number of digits
Given an integer N where 1 ? N ? 105, the task is to find the number of positive integers less than or equal to N that have an odd number of digits without leading zeros.
Examples:
Input: N = 11
Output: 9
1, 2, 3, …, 8 and 9 are the numbers ? 11
with odd number of digits.
Input: N = 893
Output: 803
Naive approach: Traverse from 1 to N and for each number check if it contains odd digits or not.
Efficient approach: For the values:
- When N < 10 then the count of valid numbers will be N.
- When N / 10 < 10 then 9.
- When N / 100 < 10 then 9 + N – 99.
- When N / 1000 < 10 then 9 + 900.
- When N / 10000 < 10 then 909 + N – 9999.
- Otherwise 90909.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int odd_digits( int n)
{
if (n < 10)
return n;
else if (n / 10 < 10)
return 9;
else if (n / 100 < 10)
return 9 + n - 99;
else if (n / 1000 < 10)
return 9 + 900;
else if (n / 10000 < 10)
return 909 + n - 9999;
else
return 90909;
}
int main()
{
int n = 893;
cout << odd_digits(n);
return 0;
}
|
Java
import java.io.*;
public class GFG
{
static int odd_digits( int n)
{
if (n < 10 )
return n;
else if (n / 10 < 10 )
return 9 ;
else if (n / 100 < 10 )
return 9 + n - 99 ;
else if (n / 1000 < 10 )
return 9 + 900 ;
else if (n / 10000 < 10 )
return 909 + n - 9999 ;
else
return 90909 ;
}
public static void main(String []args)
{
int n = 893 ;
System.out.println(odd_digits(n));
}
}
|
Python3
def odd_digits(n) :
if (n < 10 ) :
return n;
elif (n / 10 < 10 ) :
return 9 ;
elif (n / 100 < 10 ) :
return 9 + n - 99 ;
elif (n / 1000 < 10 ) :
return 9 + 900 ;
elif (n / 10000 < 10 ) :
return 909 + n - 9999 ;
else :
return 90909 ;
if __name__ = = "__main__" :
n = 893 ;
print (odd_digits(n));
|
C#
using System;
class GFG
{
static int odd_digits( int n)
{
if (n < 10)
return n;
else if (n / 10 < 10)
return 9;
else if (n / 100 < 10)
return 9 + n - 99;
else if (n / 1000 < 10)
return 9 + 900;
else if (n / 10000 < 10)
return 909 + n - 9999;
else
return 90909;
}
public static void Main(String []args)
{
int n = 893;
Console.WriteLine(odd_digits(n));
}
}
|
Javascript
<script>
function odd_digits( n)
{
if (n < 10)
return n;
else if (n / 10 < 10)
return 9;
else if (n / 100 < 10)
return 9 + n - 99;
else if (n / 1000 < 10)
return 9 + 900;
else if (n / 10000 < 10)
return 909 + n - 9999;
else
return 90909;
}
let n = 893;
document.write(odd_digits(n));
</script>
|
Time complexity: O(1)
Auxiliary space: O(1)
Last Updated :
20 Dec, 2022
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