Find a number K having sum of numbers obtained by repeated removal of last digit of K is N

Difficulty Level : Hard
Last Updated : 26 May, 2021

Given an integer N, the task is to find an integer K such that the sum of the numbers formed by repeated removal of the last digit of K is equal to N.

Examples:

Input: N = 136
Output: 123
Explanation:

The numbers formed by repeatedly removing the last digit of 123 are {123, 12, 1}.
Therefore, the sum of these numbers = 123 + 12 + 1 = 136( = N).

Input: N = 107
Output: 98
Explanation:
The numbers formed by repeatedly removing the last digit of 98 are {98, 9}.
Therefore, the sum of these numbers = 98 + 7 = 107( = N).

Approach: The approach is based on the following observations:

Consider K = 123.
The possible numbers formed from 123 are 1, 12, and 123.
Now, 123 can be expressed as 100 + 20 + 3. If all the other numbers are expressed similarly, then the idea is to know the position and frequency of each digit in all the numbers combined, to get the total sum as N.

Digit Frequency of each digit Sum
units tens hundreds
1 1 1 1 1*1 + 1*10 + 1*100 = 111
2 1 1 2*1 + 2*10 = 22
3 1 3*1 = 3

Now, for the given number N of length L. Divide the number with L number of 1s to get the highest place digit.
Calculate the remainder which will be our newly formed N.
Again divide the newly formed N with (L – 1) number of 1s to get 2nd highest place digit and continue till the L becomes 0.

Follow the steps below to solve the problem:

Let L be the count of digits in the given number N.
Initialize string str as L numbers of 1s in it.
Initialize a variable ans as zero that will store the resultant number K.
Iterate until the string str is not empty and follow the steps below:
- Convert the string str to number using the function stoi() and store it in M.
- Divide N by M and update ans as:

ans = ans*10 + (N/M)

Update N to N % M.
Remove the last character from the string str.
After the above steps, print the value stored in the ans which is the required value of K.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the value of K
int findK(int N, int l)
{
    // Stores l number of 1s
    string ones = "";
 
    while (l--) {
 
        // Storing 1's
        ones = ones + '1';
    }
 
    // Stores the value of K
    int ans = 0;
 
    // Iterate until ones is empty
    while (ones != "") {
 
        // Convert ones to number
        int m = stoi(ones);
 
        // Update the ans
        ans = (ans * 10) + (N / m);
 
        // Update N to N%m
        N = N % m;
 
        // Removing last digit from ones
        ones.pop_back();
    }
 
    // Return the value of K
    return ans;
}
 
// Driver Code
int main()
{
    // Given number N
    int N = 136;
 
    // Number of digits in N
    int L = to_string(N).length();
 
    // Funtion Call
    cout << findK(N, L);
 
    return 0;
}

Java

// Java program for
// the above approach
import java.util.*;
class GFG{
 
// Function to find the
// value of K
static int findK(int N,
                 int l)
{
  // Stores l number of 1s
  String ones = "";
 
  while (l-- > 0)
  {
    // Storing 1's
    ones += '1';
  }
 
  // Stores the value of K
  int ans = 0;
 
  // Iterate until ones is empty
  while (!ones.equals(""))
  {
    // Convert ones to number
    int m = Integer.valueOf(ones);
 
    // Update the ans
    ans = (ans * 10) + (N / m);
 
    // Update N to N%m
    N = N % m;
 
    // Removing last digit from ones
    ones = ones.substring(0,
           ones.length() - 1);
  }
 
  // Return the value of K
  return ans;
}
 
// Driver Code
public static void main(String[] args)
{
  // Given number N
  int N = 136;
 
  // Number of digits in N
  int L = String.valueOf(N).length();
 
  // Funtion Call
  System.out.print(findK(N, L));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for
# the above approach
 
# Function to find
# the value of K
def findK(N, l):
 
    # Stores l number of 1s
    ones = ""
 
    while (l):
 
        #  Storing 1's
        ones = ones + '1'
        l -= 1
    
    # Stores the value of K
    ans = 0
     
    # Iterate until ones
    # is empty
    while (ones != ""):
 
        # Convert ones to number
        m = int(ones)
 
        # Update the ans
        ans = (ans * 10) + (N // m)
 
        # Update N to N%m
        N = N % m
 
        # Removing last digit from ones
        ones = ones.replace(ones[-1], "", 1)
     
    # Return the value of K
    return ans
 
# Driver Code
if __name__ == "__main__":
   
    # Given number N
    N = 136
 
    # Number of digits in N
    L = len(str(N))
 
    # Funtion Call
    print (findK(N, L))
 
# This code is contributed by Chitranayal

C#

// C# program for
// the above approach
using System;
class GFG{
 
// Function to find the
// value of K
static int findK(int N,
                 int l)
{
  // Stores l number of 1s
  String ones = "";
 
  while (l-- > 0)
  {
    // Storing 1's
    ones += '1';
  }
 
  // Stores the value of K
  int ans = 0;
 
  // Iterate until ones is empty
  while (!ones.Equals(""))
  {
    // Convert ones to number
    int m = Int32.Parse(ones);
 
    // Update the ans
    ans = (ans * 10) + (N / m);
 
    // Update N to N%m
    N = N % m;
 
    // Removing last digit from ones
    ones = ones.Substring(0,
           ones.Length - 1);
  }
 
  // Return the value of K
  return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given number N
  int N = 136;
 
  // Number of digits in N
  int L = String.Join("", N).Length;
 
  // Funtion Call
  Console.Write(findK(N, L));
}
}
 
// This code is contributed by Princi Singh

Javascript

<script>
 
      // JavaScript program for
      // the above approach
      // Function to find the
      // value of K
      function findK(N, l) {
        // Stores l number of 1s
        var ones = "";
 
        while (l) {
          // Storing 1's
          ones += "1";
          l -= 1;
        }
 
        // Stores the value of K
        var ans = 0;
 
        // Iterate until ones is empty
        while (ones !== "") {
          // Convert ones to number
          var m = parseInt(ones);
 
          // Update the ans
          ans = parseInt(ans * 10 + N / m);
 
          // Update N to N%m
          N = N % m;
 
          // Removing last digit from ones
          ones = ones.substring(0, ones.length - 1);
        }
 
        // Return the value of K
        return ans;
      }
 
      // Driver Code
      // Given number N
      var N = 136;
 
      // Number of digits in N
      var L = N.toString().length;
 
      // Funtion Call
      document.write(findK(N, L));
       
</script>

Output:

Time Complexity: O(log₁₀N)
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live

My Personal Notes