Total no of 1’s in numbers

Given an integer n, count the total number of digit 1 appearing in all positive integers less than or equal to n.

Examples:

Input : n = 13
Output : 6
Explanation:
Here, no <= 13 containing 1 are 1, 10, 11,
12, 13. So, total 1s are 6.

Input : n = 5
Output : 1
Here, no <= 13 containing 1 are 1.
So, total 1s are 1.

Approach 1:

  1. Iterate over i from 1 to n.
  2. Convert i to string and count ’1’ in each integer string.
  3. Add count of ’1’ in each string to the sum.

Below is the code for the above discussed approach.

C++

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// c++ code to count the frequency of 1
// in numbers less than or equal to
// the given number.
#include <bits/stdc++.h>
using namespace std;
int countDigitOne(int n)
{
    int countr = 0;
    for (int i = 1; i <= n; i++) {
        string str = to_string(i);
        countr += count(str.begin(), str.end(), '1');
    }
    return countr;
}
  
// driver function
int main()
{
    int n = 13;
    cout << countDigitOne(n) << endl;
    n = 131;
    cout << countDigitOne(n) << endl;
    n = 159;
    cout << countDigitOne(n) << endl;
    return 0;
}

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Java

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// Java code to count the frequency of 1
// in numbers less than or equal to
// the given number.
class GFG
{
static int countDigitOne(int n)
{
    int countr = 0;
    for (int i = 1; i <= n; i++)
    {
        String str = String.valueOf(i);
        countr += str.split("1", -1) . length - 1;
    }
    return countr;
}
  
// Driver Code
public static void main(String[] args)
{
    int n = 13;
    System.out.println(countDigitOne(n));
    n = 131;
    System.out.println(countDigitOne(n));
    n = 159;
    System.out.println(countDigitOne(n));
}
}
  
// This code is contributed by mits

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Python3

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# Python3 code to count the frequency 
# of 1 in numbers less than or equal 
# to the given number.
  
def countDigitOne(n):
    countr = 0;
    for i in range(1, n + 1):
        str1 = str(i);
        countr += str1.count("1");
  
    return countr;
  
# Driver Code
n = 13;
print(countDigitOne(n));
  
n = 131;
print(countDigitOne(n));
  
n = 159;
print(countDigitOne(n));
  
# This code is contributed by mits

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C#

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// C# code to count the frequency of 1
// in numbers less than or equal to
// the given number.
using System;
class GFG
{
static int countDigitOne(int n)
{
    int countr = 0;
    for (int i = 1; i <= n; i++)
    {
        string str = i.ToString();
        countr += str.Split("1") . Length - 1;
    }
    return countr;
}
  
// Driver Code
public static void Main()
{
    int n = 13;
    Console.WriteLine(countDigitOne(n));
    n = 131;
    Console.WriteLine(countDigitOne(n));
    n = 159;
    Console.WriteLine(countDigitOne(n));
}
}
  
// This code is contributed by mits

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PHP

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<?php
// PHP code to count the frequency of 1
// in numbers less than or equal to
// the given number.
  
function countDigitOne($n)
{
    $countr = 0;
    for ($i = 1; $i <= $n; $i++)
    {
        $str = strval($i);
        $countr += substr_count($str, '1');
    }
    return $countr;
}
  
// Driver Code
$n = 13;
echo countDigitOne($n) . "\n";
  
$n = 131;
echo countDigitOne($n) . "\n";
  
$n = 159;
echo countDigitOne($n) . "\n";
  
// This code is contributed by mits
?>

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Output:

6
66
96

Time complexity: O(nlog(n))

Approach 2:

  1. Initialize countr as 0.
  2. Iterate over i from 1 to n incrementing by 10 each time.
  3. Add (n / (i * 10 ) ) * i to countr after each (i*10) interval.
  4. Add min( max((n mod (i*10) – i + 1, 0), i) to countr.

Below is the code for the above discussed approach.

C++

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// c++ code to count the frequency of 1
// in numbers less than or equal to
// the given number.
#include <bits/stdc++.h>
using namespace std;
  
// function to count the frequency of 1.
int countDigitOne(int n)
{
    int countr = 0;
    for (int i = 1; i <= n; i *= 10) {
        int divider = i * 10;
        countr += (n / divider) * i + 
               min(max(n % divider - i + 1, 0), i);
    }
    return countr;
}
  
// driver function
int main()
{
    int n = 13;
    cout << countDigitOne(n) << endl;
    n = 113;
    cout << countDigitOne(n) << endl;
    n = 205;
    cout << countDigitOne(n) << endl;
}

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Java

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/// Java code to count the 
// frequency of 1 in numbers
// less than or equal to
// the given number.
import java.io.*;
  
class GFG
{
      
// function to count 
// the frequency of 1.
static int countDigitOne(int n)
{
    int countr = 0;
    for (int i = 1;
             i <= n; i *= 10
    {
        int divider = i * 10;
        countr += (n / divider) * i + 
                Math.min(Math.max(n % 
                         divider - i + 
                            1, 0), i);
    }
    return countr;
}
  
// Driver Code
public static void main (String[] args) 
{
    int n = 13;
    System.out.println(countDigitOne(n));
      
    n = 113;
    System.out.println(countDigitOne(n));
      
    n = 205;
    System.out.println(countDigitOne(n));
}
}
  
// This code is contributed by akt_mit

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Python3

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# Python3 code to count the 
# frequency of 1 in numbers 
# less than or equal to
# the given number.
  
# function to count the 
# frequency of 1.
def countDigitOne(n):
    countr = 0;
    i = 1;
    while(i <= n):
        divider = i * 10;
        countr += (int(n / divider) * i +
                 min(max(n % divider -i + 
                              1, 0), i));
        i *= 10;
      
    return countr;
  
# Driver Code
n = 13;
print(countDigitOne(n));
n = 113;
print(countDigitOne(n));
n = 205;
print(countDigitOne(n));
  
# This code is contributed by mits

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C#

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// C# code to count the 
// frequency of 1 in numbers
// less than or equal to
// the given number.
using System;
  
class GFG
{
      
// function to count 
// the frequency of 1.
static int countDigitOne(int n)
{
    int countr = 0;
    for (int i = 1; 
             i <= n; i *= 10) 
    {
        int divider = i * 10;
        countr += (n / divider) * i + 
                   Math.Min(Math.Max(n % divider - 
                                     i + 1, 0), i);
    }
    return countr;
}
  
// Driver Code
public static void Main() 
{
    int n = 13;
    Console.WriteLine(countDigitOne(n));
      
    n = 113;
    Console.WriteLine(countDigitOne(n));
      
    n = 205;
    Console.WriteLine(countDigitOne(n));
}
}
  
// This code is contributed by mits

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PHP

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<?php
// PHP code to count the 
// frequency of 1 in numbers 
// less than or equal to
// the given number.
  
// function to count the 
// frequency of 1.
function countDigitOne($n)
{
    $countr = 0;
    for ($i = 1; $i <= $n; $i *= 10)
    {
        $divider = $i * 10;
        $countr += (int)($n / $divider) * $i
                       min(max($n % $divider
                              $i + 1, 0), $i);
    }
      
    return $countr;
}
  
// Driver Code
$n = 13;
echo countDigitOne($n), "\n";
$n = 113;
echo countDigitOne($n), "\n";
$n = 205;
echo countDigitOne($n), "\n";
  
// This code is contributed by ajit
?>

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Output:

6
40
141

Time complexity: O(log(n))



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