Number of occurrences of 2 as a digit in numbers from 0 to n

Count the number of 2s as digit in all numbers from 0 to n.

Examples :

Input : 22
Output : 6
Explanation: Total 2s that appear as digit
             from 0 to 22 are (2, 12, 20,
             21, 22);

Input : 100
Output : 20
Explanation: total 2's comes between 0 to 100 
are (2, 12, 20, 21, 22..29, 32, 42, 52, 62, 72,
82, 92);

A Simple Brute force solution is to iterate through all numbers from 0 to n. For every number being visited, count the number of 2’s in it. Finally return total count.

Below is implementation of the idea.

C++

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// C++ program to count 2s from 0 to n
#include <bits/stdc++.h>
using namespace std;
  
// Counts the number of '2'
// digits in a single number
int number0f2s(int n)
{
    int count = 0;
    while (n > 0)
    {
        if (n % 10 == 2)
            count++;
  
        n = n / 10;
    }
    return count;
}
  
// Counts the number of '2' 
// digits between 0 and n 
int numberOf2sinRange(int n)
{
    // Initialize result
    int count = 0 ; 
  
    // Count 2's in every number
    // from 2 to n
    for (int i = 2; i <= n; i++)
        count += number0f2s(i);
  
    return count;
}
  
// Driver Code
int main()
{
    cout << numberOf2sinRange(22);
    cout << endl;
    cout << numberOf2sinRange(100);
    return 0;
}

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Java

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// Java program to count 2s from 0 to n
  
class GFG 
{
      
// Counts the number of '2' 
// digits in a single number 
static int number0f2s(int n) 
{
    int count = 0;
    while (n > 0
    {
        if (n % 10 == 2)
        count++;
  
        n = n / 10;
    }
    return count;
}
  
// Counts the number of '2' 
// digits between 0 and n 
static int numberOf2sinRange(int n) 
{
      
    // Initialize result
    int count = 0
  
    // Count 2's in every number 
    // from 2 to n
    for (int i = 2; i <= n; i++)
    count += number0f2s(i);
  
    return count;
}
  
// Driver code
public static void main(String[] args) 
{
    System.out.print(numberOf2sinRange(22));
    System.out.println();
    System.out.print(numberOf2sinRange(100));
}
}
  
// This code is contributed by Anant Agarwal.

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Python3

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# Python3 program to count
# 2s from 0 to n
  
# Counts the number of
# '2' digits in a
# single number 
def number0f2s(n):
  
    count = 0
    while (n > 0):
      
        if (n % 10 == 2):
            count = count + 1
  
        n = n // 10
      
    return count
  
  
# Counts the number of '2' 
# digits between 0 and n 
def numberOf2sinRange(n):
  
    # Initialize result
    count = 0 
  
    # Count 2's in every number 
    # from 2 to n
    for i in range(2, n + 1):
        count = count + number0f2s(i)
  
    return count
  
# Driver code
  
print(numberOf2sinRange(22))
print(numberOf2sinRange(100))
  
# This code is contributed
# by Anant Agarwal.

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C#

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// C# program to count 2s from 0 to n
using System; 
  
class GFG 
{
      
// Counts the number of '2' digits
// in a single number
static int number0f2s(int n) 
{
    int count = 0;
    while (n > 0) 
    {
        if (n % 10 == 2)
            count++;
  
    n = n / 10;
    }
    return count;
}
  
// Counts the number of '2' digits 
// between 0 and n
static int numberOf2sinRange(int n) 
{
      
    // Initialize result
    int count = 0; 
  
    // Count 2's in every number 
    // from 2 to n
    for (int i = 2; i <= n; i++)
    count += number0f2s(i);
  
    return count;
}
  
// Driver code
public static void Main() 
{
    Console.Write(numberOf2sinRange(22));
    Console.WriteLine();
    Console.Write(numberOf2sinRange(100));
}
}
  
// This code is contributed by nitin mittal

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PHP

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<?php
// php program to count 2s from 0 to n
  
// Counts the number of '2'
// digits in a single number
function number0f2s($n)
{
    $count = 0;
    while ($n > 0)
    {
        if ($n % 10 == 2)
            $count++;
  
        $n = $n / 10;
    }
    return $count;
}
  
// Counts the number of '2' 
// digits between 0 and n 
function numberOf2sinRange($n)
{
      
    // Initialize result
    $count = 0 ; 
  
    // Count 2's in every number
    // from 2 to n
    for ($i = 2; $i <= $n; $i++)
        $count += number0f2s($i);
  
    return $count;
}
  
// Driver Code
    echo (numberOf2sinRange(22));
    echo "\n";
    echo numberOf2sinRange(100);
  
// This code is contributed by 
// nitin mittal.
?>

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Output :

 6
 20

 

Improved Solution
The idea is to look at the problem digit by digit. Picture a sequence of numbers:

0  1  2  3  4  5  6  7  8  9
10 11 12 13 14 15 16 17 18 19
20 21 22 23 24 25 26 27 28 29
......
110 111 112 113 114 115 116 117 118 119 

We know that roughly one tenth of the time, the last digit will be a 2 since it happens once in any sequence of ten numbers. In fact, any digit is a 2 roughly one tenth of the time.

We say “roughly” because there are (very common) boundary conditions. For example, between 1 and 100, the 10’s digit is a 2 exactly 1/10th of the time. However, between 1 and 37, the 10’s digit is a 2 much more than 1/10th of the time.

We can work out what exactly the ratio is by looking at the three cases individually: digit 2.

Case digits < 2
Consider the value x = 61523 and digit at index d = 3 (here indexes are considered from right and rightmost index is 0). We observe that x[d] = 1. There are 2s at the 3rd digit in the ranges 2000 – 2999, 12000 – 12999, 22000 – 22999, 32000 32999, 42000 – 42999, and 52000 – 52999. So there are 6000 2’s total in the 3rd digit. This is the same amount as if we were just counting all the 2s in the 3rd digit between 1 and 60000.

In other words, we can round down to the nearest 10d+1, and then divide by 10, to compute the number of 2s in the d-th digit.

if x[d) < 2: count2sinRangeAtDigit(x, d) =
  Compute y = round down to nearest 10d+1 
  return y/10 

Case digit > 2
Now, let’s look at the case where d-th digit (from right) of x is greater than 2 (x[d] > 2). We can apply almost the exact same logic to see that there are the same number of 2s in the 3rd digit in the range 0 – 63525 as there as in the range 0 – 70000. So, rather than rounding down, we round up.

if x[d) > 2: count2sinRangeAtDigit(x, d) =
  Compute y = round down to nearest 10d+1 
  return y / 10 

Case digit = 2
The final case may be the trickiest, but it follows from the earlier logic. Consider x = 62523 and d = 3. We know that there are the same ranges of 2s from before (that is, the ranges 2000 – 2999, 12000 – 12999, … , 52000 – 52999). How many appear in the 3rd digit in the final, partial range from 62000 – 62523? Well, that should be pretty easy. It’s just 524 (62000, 62001, … , 62523).

if x[d] = 2: count2sinRangeAtDigit(x, d) = 
   Compute y = round down to nearest 10d+1 
   Compute z = right side of x (i.e., x%  10d)
   return y/10 + z + 1

Now, all we need is to iterate through each digit in the number. Implementing this code is reasonably straightforward.

Below is the implementation of the idea.

C++

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// C++ program to count 2s from 0 to n
#include <bits/stdc++.h>
using namespace std;
  
// Counts the number of 2s 
// in a number at d-th digit
int count2sinRangeAtDigit(int number, int d)
{
    int powerOf10 = (int)pow(10, d);
    int nextPowerOf10 = powerOf10 * 10;
    int right = number % powerOf10;
  
    int roundDown = number - number % nextPowerOf10;
    int roundup = roundDown + nextPowerOf10;
  
    int digit = (number / powerOf10) % 10;
  
    // if the digit in spot digit is
    if (digit < 2)
        return roundDown / 10;
  
    if (digit == 2)
        return roundDown / 10 + right + 1;
  
    return roundup / 10;
}
  
// Counts the number of '2' digits between 0 and n 
int numberOf2sinRange(int number)
{
    // Convert integer to String 
    // to find its length
    stringstream convert;
    convert << number;
    string s = convert.str();
    int len = s.length();
  
    // Traverse every digit and 
    // count for every digit
    int count = 0;
    for (int digit = 0; digit < len; digit++)
        count += count2sinRangeAtDigit(number, digit);
  
    return count;
}
  
// Driver Code
int main()
{
    cout << numberOf2sinRange(22) << endl;
    cout << numberOf2sinRange(100);
    return 0;
}

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Java

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// Java program to count 2s from 0 to n 
class GFG 
{
  
    // Counts the number of 2s 
    // in a number at d-th digit 
    static int count2sinRangeAtDigit(int number, int d) 
    {
        int powerOf10 = (int) Math.pow(10, d);
        int nextPowerOf10 = powerOf10 * 10;
        int right = number % powerOf10;
  
        int roundDown = number - number % nextPowerOf10;
        int roundup = roundDown + nextPowerOf10;
  
        int digit = (number / powerOf10) % 10;
  
        // if the digit in spot digit is 
        if (digit < 2
        {
            return roundDown / 10;
        }
  
        if (digit == 2)
        {
            return roundDown / 10 + right + 1;
        }
        return roundup / 10;
    }
  
    // Counts the number of '2' digits between 0 and n 
    static int numberOf2sinRange(int number) 
    {
        // Convert integer to String 
        // to find its length 
        String convert;
        convert = String.valueOf(number);
        String s = convert;
        int len = s.length();
  
        // Traverse every digit and 
        // count for every digit 
        int count = 0;
        for (int digit = 0; digit < len; digit++)
        {
            count += count2sinRangeAtDigit(number, digit);
        }
  
        return count;
    }
  
    // Driver Code 
    public static void main(String[] args)
    {
        System.out.println(numberOf2sinRange(22));
        System.out.println(numberOf2sinRange(100));
    }
  
// This code is contributed by PrinciRaj1992

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Python3

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# Python3 program to count 2s from 0 to n 
  
# Counts the number of 2s in a 
# number at d-th digit 
def count2sinRangeAtDigit(number, d): 
  
    powerOf10 = int(pow(10, d)); 
    nextPowerOf10 = powerOf10 * 10
    right = number % powerOf10; 
  
    roundDown = number - number % nextPowerOf10; 
    roundup = roundDown + nextPowerOf10; 
  
    digit = (number // powerOf10) % 10
  
    # if the digit in spot digit is 
    if (digit < 2): 
        return roundDown // 10
  
    if (digit == 2): 
        return roundDown // 10 + right + 1
  
    return roundup // 10
  
# Counts the number of '2' digits
# between 0 and n 
def numberOf2sinRange(number):
      
    # Convert integer to String 
    # to find its length 
    s = str(number); 
    len1 = len(s); 
  
    # Traverse every digit and 
    # count for every digit 
    count = 0
    for digit in range(len1): 
        count += count2sinRangeAtDigit(number, 
                                       digit); 
  
    return count; 
  
# Driver Code 
print(numberOf2sinRange(22)); 
print(numberOf2sinRange(100)); 
  
# This code is contributed by mits

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C#

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// C# program to count 2s from 0 to n 
using System;
  
class GFG 
{
  
// Counts the number of 2s 
// in a number at d-th digit 
static int count2sinRangeAtDigit(int number, 
                                 int d) 
{
    int powerOf10 = (int) Math.Pow(10, d);
    int nextPowerOf10 = powerOf10 * 10;
    int right = number % powerOf10;
  
    int roundDown = number - number % nextPowerOf10;
    int roundup = roundDown + nextPowerOf10;
  
    int digit = (number / powerOf10) % 10;
  
    // if the digit in spot digit is 
    if (digit < 2) 
    {
        return roundDown / 10;
    }
  
    if (digit == 2)
    {
        return roundDown / 10 + right + 1;
    }
    return roundup / 10;
}
  
// Counts the number of '2' digits 
// between 0 and n 
static int numberOf2sinRange(int number) 
{
    // Convert integer to String 
    // to find its length 
    string convert;
    convert = number.ToString();
    string s = convert;
    int len = s.Length;
  
    // Traverse every digit and 
    // count for every digit 
    int count = 0;
    for (int digit = 0; digit < len; digit++)
    {
        count += count2sinRangeAtDigit(number,
                                       digit);
    }
  
    return count;
}
  
// Driver Code 
public static void Main()
{
    Console.WriteLine(numberOf2sinRange(22));
    Console.WriteLine(numberOf2sinRange(100));
}
  
// This code is contributed by mits

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PHP

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<?php
// PHP program to count 2s from 0 to n 
  
// Counts the number of 2s in a number 
// at d-th digit 
function count2sinRangeAtDigit($number, $d
    $powerOf10 = (int)pow(10, $d); 
    $nextPowerOf10 = $powerOf10 * 10; 
    $right = $number % $powerOf10
  
    $roundDown = $number - $number
                 $nextPowerOf10
    $roundup = $roundDown + $nextPowerOf10
  
    $digit = ($number / $powerOf10) % 10; 
  
    // if the digit in spot digit is 
    if ($digit < 2) 
        return $roundDown / 10; 
  
    if ($digit == 2) 
        return $roundDown / 10 + $right + 1; 
  
    return $roundup / 10; 
  
// Counts the number of '2' digits
// between 0 and n 
function numberOf2sinRange($number
    // Convert integer to String 
    // to find its length 
    $s = strval($number); 
    $len = strlen($s); 
  
    // Traverse every digit and 
    // count for every digit 
    $count = 0; 
    for ($digit = 0; $digit < $len; $digit++) 
        $count += count2sinRangeAtDigit($number
                                        $digit); 
  
    return $count
  
// Driver Code 
print(numberOf2sinRange(22) . "\n"); 
print(numberOf2sinRange(100) . "\n"); 
  
// This code is contributed by mits
?>

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Output:

 6
 20

Time Complexity:O(n)

This article is contributed by Mr. Somesh Awasthi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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