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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the number of
// positive integers less than or equal
// to N that have odd number of digits
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;
}
  
// Driver code
int main()
{
    int n = 893;
  
    cout << odd_digits(n);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG 
{
  
// Function to return the number of
// positive integers less than or equal
// to N that have odd number of digits
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;
}
  
// Driver code
public static void main(String []args) 
{
    int n = 893;
  
    System.out.println(odd_digits(n));
}
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python3 implementation of the approach 
  
# Function to return the number of 
# positive integers less than or equal 
# to N that have odd number of digits 
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
  
# Driver code 
if __name__ == "__main__"
  
    n = 893
  
    print(odd_digits(n)); 
  
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
                      
class GFG 
{
  
// Function to return the number of
// positive integers less than or equal
// to N that have odd number of digits
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;
}
  
// Driver code
public static void Main(String []args) 
{
    int n = 893;
  
    Console.WriteLine(odd_digits(n));
}
}
  
// This code is contributed by 29AjayKumar

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Output:

803


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Improved By : AnkitRai01, 29AjayKumar