Numbers in a Range with given Digital Root

Given an integer K and a range of consecutive numbers [L, R]. The task is to count the numbers from the given range which have digital root as K (1 ≤ K ≤ 9). Digital root is sum of digits of a number until it becomes a single digit number. For example, 256 -> 2 + 5 + 6 = 13 -> 1 + 3 = 4.

Examples:

Input: L = 10, R = 22, K = 3
Output: 2
12 and 21 are the only numbers from the range whose digit sum is 3.



Input: L = 100, R = 200, K = 5
Output: 11

Approach:

  • First thing is to note that for any number Sum of Digits is equal to Number % 9. If remainder is 0, then sum of digits is 9.
  • So if K = 9, then replace K with 0.
  • Task, now is to find count of numbers in range L to R with modulo 9 equal to K.
  • Divide the entire range into the maximum possible groups of 9 starting with L (TotalRange / 9), since in each range there will be exactly one number with modulo 9 equal to K.
  • Loop over rest number of elements from R to R – count of rest elements, and check if any number satisfies the condition.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
  
// Function to return the count
// of required numbers
int countNumbers(int L, int R, int K)
{
    if (K == 9)
        K = 0;
  
    // Count of numbers present
    // in given range
    int totalnumbers = R - L + 1;
  
    // Number of groups of 9 elements
    // starting from L
    int factor9 = totalnumbers / 9;
  
    // Left over elements not covered
    // in factor 9
    int rem = totalnumbers % 9;
  
    // One Number in each group of 9
    int ans = factor9;
  
    // To check if any number in rem
    // satisfy the property
    for (int i = R; i > R - rem; i--) {
        int rem1 = i % 9;
        if (rem1 == K)
            ans++;
    }
  
    return ans;
}
  
// Driver code
int main()
{
    int L = 10;
    int R = 22;
    int K = 3;
    cout << countNumbers(L, R, K);
  
    return 0;
}

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Java

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// Java implementation of the approach
  
class GFG {
  
// Function to return the count
// of required numbers
    static int countNumbers(int L, int R, int K) {
        if (K == 9) {
            K = 0;
        }
  
        // Count of numbers present
        // in given range
        int totalnumbers = R - L + 1;
  
        // Number of groups of 9 elements
        // starting from L
        int factor9 = totalnumbers / 9;
  
        // Left over elements not covered
        // in factor 9
        int rem = totalnumbers % 9;
  
        // One Number in each group of 9
        int ans = factor9;
  
        // To check if any number in rem
        // satisfy the property
        for (int i = R; i > R - rem; i--) {
            int rem1 = i % 9;
            if (rem1 == K) {
                ans++;
            }
        }
  
        return ans;
    }
  
// Driver code
    public static void main(String[] args) {
        int L = 10;
        int R = 22;
        int K = 3;
        System.out.println(countNumbers(L, R, K));
    }
}
/* This code contributed by PrinciRaj1992 */

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Python3

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# Python3 implementation of the approach
  
# Function to return the count
# of required numbers
def countNumbers(L, R, K):
  
    if (K == 9):
        K = 0
  
    # Count of numbers present
    # in given range
    totalnumbers = R - L + 1
  
    # Number of groups of 9 elements
    # starting from L
    factor9 = totalnumbers // 9
  
    # Left over elements not covered
    # in factor 9
    rem = totalnumbers % 9
  
    # One Number in each group of 9
    ans = factor9
  
    # To check if any number in rem
    # satisfy the property
    for i in range(R, R - rem, -1):
        rem1 = i % 9
        if (rem1 == K):
            ans += 1
      
    return ans
  
# Driver code
L = 10
R = 22
K = 3
print(countNumbers(L, R, K))
  
# This code is contributed 
# by mohit kumar

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C#

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// C# implementation of the approach
using System ;
  
class GFG 
{
  
    // Function to return the count
    // of required numbers
    static int countNumbers(int L, int R, int K)
    {
        if (K == 9) 
        {
            K = 0;
        }
  
        // Count of numbers present
        // in given range
        int totalnumbers = R - L + 1;
  
        // Number of groups of 9 elements
        // starting from L
        int factor9 = totalnumbers / 9;
  
        // Left over elements not covered
        // in factor 9
        int rem = totalnumbers % 9;
  
        // One Number in each group of 9
        int ans = factor9;
  
        // To check if any number in rem
        // satisfy the property
        for (int i = R; i > R - rem; i--)
        {
            int rem1 = i % 9;
            if (rem1 == K)
            {
                ans++;
            }
        }
  
        return ans;
    }
  
    // Driver code
    public static void Main() 
    {
        int L = 10;
        int R = 22;
        int K = 3;
          
        Console.WriteLine(countNumbers(L, R, K));
    }
}
  
/* This code is contributed by Ryuga */

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PHP

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<?php
// PHP implementation of the approach
  
// Function to return the count
// of required numbers
function countNumbers($L, $R, $K)
{
    if ($K == 9)
        $K = 0;
  
    // Count of numbers present
    // in given range
    $totalnumbers = $R - $L + 1;
  
    // Number of groups of 9 elements
    // starting from L
    $factor9 = intval($totalnumbers / 9);
  
    // Left over elements not covered
    // in factor 9
    $rem = $totalnumbers % 9;
  
    // One Number in each group of 9
    $ans = $factor9;
  
    // To check if any number in rem
    // satisfy the property
    for ($i = $R; $i > $R - $rem; $i--) 
    {
        $rem1 = $i % 9;
        if ($rem1 == $K)
            $ans++;
    }
  
    return $ans;
}
  
// Driver code
$L = 10;
$R = 22;
$K = 3;
echo countNumbers($L, $R, $K);
  
// This code is contributed by Ita_c
?>

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Output:

2


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