Numbers in a Range with given Digital Root
Last Updated :
22 Jun, 2022
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++
// 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;
}
// 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 */
# 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
// 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 */
<?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
?>
<script>
// Javascript 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
var totalnumbers = R - L + 1;
// Number of groups of 9 elements
// starting from L
var factor9 = totalnumbers / 9;
// Left over elements not covered
// in factor 9
var rem = totalnumbers % 9;
// One Number in each group of 9
var ans = factor9;
// To check if any number in rem
// satisfy the property
for(var i = R; i > R - rem; i--)
{
var rem1 = i % 9;
if (rem1 == K)
{
ans++;
}
}
return ans;
}
// Driver Code
var L = 10;
var R = 22;
var K = 3;
document.write(Math.round(countNumbers(L, R, K)));
// This code is contributed by Ankita saini
</script>
Time Complexity: O(R)
Auxiliary Space: O(1)
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