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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 // c++ code to count the frequency of 1 // in numbers less than or equal to // the given number. #include 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; }

Java

 // 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

Python3

 # 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

C#

 // 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

PHP



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++

 // c++ code to count the frequency of 1 // in numbers less than or equal to // the given number. #include 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; }

Java

 /// 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

Python3

 # 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

C#

 // 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

PHP



Output:

6
40
141

Time complexity: O(log(n))

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