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Find the number of positive integers less than or equal to N that have an odd number of digits

  • Difficulty Level : Basic
  • Last Updated : 10 May, 2021

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:
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++




// 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;
}

Java




// 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

Python3




# 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

C#




// 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

Javascript




<script>
// Java script implementation of the approach
 
 
// Function to return the number of
// positive integers less than or equal
// to N that have odd number of digits
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;
}
 
// Driver code
let n = 893;
 
    document.write(odd_digits(n));
 
 
// This code is contributed by sravan kumar Gottumukkala
</script>
Output: 
803

 


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