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

24 Aug, 2021
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++
Java
Python3
C#
PHP
Javascript
// 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 numberint 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 nint 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 Codeint main(){    cout << numberOf2sinRange(22);    cout << endl;    cout << numberOf2sinRange(100);    return 0;}
// Java program to count 2s from 0 to n class GFG{     // Counts the number of '2'// digits in a single numberstatic 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 nstatic 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 codepublic 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.
# Python3 program to count# 2s from 0 to n # Counts the number of# '2' digits in a# single numberdef 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 ndef 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.
// C# program to count 2s from 0 to nusing System; class GFG{     // Counts the number of '2' digits// in a single numberstatic 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 nstatic 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 codepublic static void Main(){    Console.Write(numberOf2sinRange(22));    Console.WriteLine();    Console.Write(numberOf2sinRange(100));}} // This code is contributed by nitin mittal
<?php// php program to count 2s from 0 to n // Counts the number of '2'// digits in a single numberfunction number0f2s($n){    $count = 0;    while ($n > 0)    {        if ($n % 10 == 2)            $count++;         $n = $n / 10;    }    return $count;} // Counts the number of '2'// digits between 0 and nfunction numberOf2sinRange($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.?>
<script>     // JavaScript program to count 2s from 0 to n         // Counts the number of '2'    // digits in a single number    function number0f2s(n)    {        let count = 0;        while (n > 0)        {            if (n % 10 == 2)                count++;             n = parseInt(n / 10, 10);        }        return count;    }     // Counts the number of '2'    // digits between 0 and n    function numberOf2sinRange(n)    {        // Initialize result        let count = 0 ;         // Count 2's in every number        // from 2 to n        for (let i = 2; i <= n; i++)            count += number0f2s(i);         return count;    }         document.write(numberOf2sinRange(22) + "</br>");    document.write(numberOf2sinRange(100));     </script>
6
20
 
Below is an alternate implementation of this approach.
C++
Java
Python3
C#
Javascript
// Write C++ code here#include <bits/stdc++.h>using namespace std;int numberOf2sinRange(int n){    string s = "";    for(int i = 0; i < n + 1; i++)        s += to_string(i);           int count = 0;    for(int i = 0; i < s.length(); i++)    {        if(s[i] == '2')        {            count++;        }    }    return count;} // Driver codeint main(){    int n = 30;    cout << numberOf2sinRange(n);    return 0;} // This code is contributed by divyeshrabadiya07.
// Java code for above implementationclass GFG{  static int numberOf2sinRange(int n)  {    String s = "";    for(int i = 0; i < n + 1; i++)      s += String.valueOf(i);          int count = 0;    for(int i = 0; i < s.length(); i++)    {      if(s.charAt(i) == '2')      {        count++;      }    }    return count;  }   // Driver code  public static void main(String[] args) {    int n = 30;    System.out.println(numberOf2sinRange(n));  }} // This code is contributed by divyesh072019
# Write Python3 code heredef numberOf2sinRange(n):    s = ""    for i in range(0,n+1):        s+=str(i)    return(list(s).count('2')) # Driver coden = 30print(numberOf2sinRange(n))
// C# code for above implementationusing System;class GFG{  static int numberOf2sinRange(int n)  {    string s = "";    for(int i = 0; i < n + 1; i++)      s += i + "";      int count = 0;    for(int i = 0; i < s.Length; i++)    {      if(s[i] == '2')      {        count++;      }    }    return count;  }   // Driver code  public static void Main(string[] args)  {    int n = 30;    Console.Write(numberOf2sinRange(n));  }} // This code is contributed by rutvik_56.
<script> // Javascript code for above implementationfunction numberOf2sinRange(n){    var s = "";    for(var i = 0; i < n + 1; i++)        s += i.toString();             var count = 0;    for(var i = 0; i < s.length; i++)    {        if (s.charAt(i) == '2')        {            count++;        }    }    return count;} // Driver codevar n = 30; document.write(numberOf2sinRange(n)); // This code is contributed by Princi Singh </script>
13
 
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++
Java
Python3
C#
PHP
Javascript
// 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 digitint 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 nint 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 Codeint main(){    cout << numberOf2sinRange(22) << endl;    cout << numberOf2sinRange(100);    return 0;}
// Java program to count 2s from 0 to nclass 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
# Python3 program to count 2s from 0 to n # Counts the number of 2s in a# number at d-th digitdef 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 ndef 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 Codeprint(numberOf2sinRange(22));print(numberOf2sinRange(100)); # This code is contributed by mits
// C# program to count 2s from 0 to nusing System; class GFG{ // Counts the number of 2s// in a number at d-th digitstatic 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 nstatic 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 Codepublic static void Main(){    Console.WriteLine(numberOf2sinRange(22));    Console.WriteLine(numberOf2sinRange(100));}} // This code is contributed by mits
<?php// PHP program to count 2s from 0 to n // Counts the number of 2s in a number// at d-th digitfunction 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 $roundup / 10;} // Counts the number of '2' digits// between 0 and nfunction numberOf2sinRange($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 Codeprint(numberOf2sinRange(22) . "\n");print(numberOf2sinRange(100) . "\n"); // This code is contributed by mits?>
<script>// javascript program to count 2s from 0 to n      // Counts the number of 2s    // in a number at d-th digit    function count2sinRangeAtDigit(number , d)    {        var powerOf10 = parseInt( Math.pow(10, d));        var nextPowerOf10 = powerOf10 * 10;        var right = number % powerOf10;         var roundDown = number - number % nextPowerOf10;        var roundup = roundDown + nextPowerOf10;         var digit = parseInt(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        var convert;        convert = number.toString();        var s = convert;        var len = s.length;         // Traverse every digit and        // count for every digit        var count = 0;        for (digit = 0; digit < len; digit++)        {            count += count2sinRangeAtDigit(number, digit);        }         return count;    }     // Driver Code    document.write(numberOf2sinRange(22));    document.write("<br>"+numberOf2sinRange(100)); // This code is contributed by 29AjayKumar</script>
6
20
 
Time Complexity:O(logN)
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