Count Unary Numbers in a Range

• Difficulty Level : Medium
• Last Updated : 26 Aug, 2021

Given two numbers A and B, A<=B, the task is to find the number of unary numbers between A and B, both inclusive.
Unary Number: Consider the number 28. If we take the sum of square of its digits, 2*2 + 8*8, we get 68. Taking the sum of squares of digits again, we get 6*6 + 8*8=100. Doing this again, we get 1*1 + 0*0 + 0*0 = 1. Any such number, which ultimately leads to 1, is called a unary number.

Examples:

Input : A = 1, B = 10
Output : 3

Input : A = 100, B = 150
Output : 7

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

The idea is to recursively calculate sum of squares of digits of the number and every time recurring down replace the number with calculated sum.
The base cases of the recursion will be:

• If the sum if reduced to either 1 or 7, then answer is true.
• If the sum if reduced to a single digit integer other than 1 and 7, answer is false.

Below is the recursive implementation of this problem:

C++

 // CPP program to count unary numbers // in a range    #include using namespace std;    // Function to check if a number is unary bool isUnary(int n) {     /// Base case. Note that if we repeat     // above process for 7, we get 1.     if (n == 1 || n == 7)         return true;     else if (n / 10 == 0)         return false;        /// rec case     int x, sum = 0;     while (n != 0) {         x = n % 10;         sum = sum + x * x;         n = n / 10;     }        isUnary(sum); }    // Function to count unary numbers // in a range int countUnary(int a, int b) {     int count = 0;        for (int i = a; i <= b; i++) {         if (isUnary(i) == 1)             count++;     }        return count; }    // Driver Code int main() {     int a = 1000, b = 1099;        cout << countUnary(a, b);        return 0; }

Java

 //Java program to count unary numbers // in a range    import java.io.*;    class GFG {        // Function to check if a number is unary static boolean isUnary(int n) {     /// Base case. Note that if we repeat     // above process for 7, we get 1.     if (n == 1 || n == 7)         return true;     else if (n / 10 == 0)         return false;        /// rec case     int x, sum = 0;     while (n != 0) {         x = n % 10;         sum = sum + x * x;         n = n / 10;     }    return isUnary(sum); }    // Function to count unary numbers // in a range static int countUnary(int a, int b) {     int count = 0;        for (int i = a; i <= b; i++) {         if (isUnary(i) == true)             count++;     }        return count; }    // Driver Code            public static void main (String[] args) {                   int a = 1000, b = 1099;     System.out.println (countUnary(a, b));        } //This code is contributed by ajit     }

Python3

 # Python 3 program to count unary numbers # in a range    # Function to check if a number is unary def isUnary(n):            # Base case. Note that if we repeat     # above process for 7, we get 1.     if (n == 1 or n == 7):         return True     elif (int(n / 10) == 0):         return False        # rec case     sum = 0     while (n != 0):         x = n % 10         sum = sum + x * x         n = int(n / 10)        return isUnary(sum)    # Function to count unary numbers # in a range def countUnary(a, b):     count = 0        for i in range(a, b + 1, 1):         if (isUnary(i) == 1):             count += 1        return count    # Driver Code if __name__ == '__main__':     a = 1000     b = 1099        print(countUnary(a, b))        # This code is contributed by # Sanjit_Prasad

C#

 //C# program to count unary numbers // in a range using System;                            public class GFG {         // Function to check if a number is unary static bool isUnary(int n) {     /// Base case. Note that if we repeat     // above process for 7, we get 1.     if (n == 1 || n == 7)         return true;     else if (n / 10 == 0)         return false;         /// rec case     int x, sum = 0;     while (n != 0) {         x = n % 10;         sum = sum + x * x;         n = n / 10;     }     return isUnary(sum); }     // Function to count unary numbers // in a range static int countUnary(int a, int b) {     int count = 0;         for (int i = a; i <= b; i++) {         if (isUnary(i) == true)             count++;     }         return count; }     // Driver Code             public static void Main () {                     int a = 1000, b = 1099;     Console.WriteLine(countUnary(a, b));         } //This code is contributed by 29AjayKumar  }

Javascript



Output:

13

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