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Find number formed in K steps by reducing N by 1 if last digit is 0 else divide by 10

  • Last Updated : 07 Jan, 2022

Given two integers N and K. Perform the following type of operations on N:

  • if the last digit of N is non-zero, decrease the number by one.
  • if the last digit of N is zero, divide the number by 10 (i.e. remove the last digit).

The task is to print the result after K such operations.

Examples:

Input: N = 512, K = 4
Output: 50
Explanation: Following are the operations performed K times to get the desired result.
Operation 1: Last digit of N i.e. 2 != 0. N is reduced by 1. ( N = 512  – 1 i.e.  511).
Operation 2: Last digit of N i.e. 1 != 0. N is reduced by 1. (N = 511 – 1 i.e.  510).
Operation 3: Last digit of N is 0. N is divided by 10. ( N = 510/10 i.e. 51).
Operation 4: Last digit of N i.e. 2 != 0. N is reduced by 1. (N = 51 – 1 i.e. 50).
Therefore, after 4 operations N = 50.

Input: N = 100, K = 2
Output: 1
Explanation: N is divided by 10 two times.

 

Approach: This problem is implementation-based and similar to the Last digit of a number. Follow the steps below to solve the given problem.

  • Repeatedly check the last digit of integer N.
  • If last digit is 0, divide N by 10.
  • If last digit is NOT 0, subtract 1 from N.
  • Repeat the above steps K times.

Below is the implementation for the above approach.

C++




// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to perform operations K times
int decreaseNum(int N, int K)
{
    while (K--) {
        // Last digit is 0
        if (N % 10 == 0)
            N /= 10;
 
        // Last digit is not 0
        else
            N--;
    }
    return N;
}
 
// Driver Code
int main()
{
    // Declaration and initialisation
    int N, K;
    N = 512;
    K = 4;
 
    // Function call
    cout << decreaseNum(N, K);
 
    return 0;
}

Java




// Java program of the above approach
import java.util.*;
class GFG {
 
  // Function to perform operations K times
  public static int decreaseNum(int N, int K)
  {
    while (true) {
      K -= 1;
       
      // Last digit is 0
      if (N % 10 == 0)
        N /= 10;
 
      // Last digit is not 0
      else
        N--;
 
      if (K == 0)
        break;
    }
    return N;
  }
 
  // Driver Code
  public static void main(String args[])
  {
     
    // Declaration and initialisation
    int N, K;
    N = 512;
    K = 4;
 
    // Function call
    System.out.println(decreaseNum(N, K));
 
  }
}
 
    // This code is contributed by rakeshsahni

Python3




# python3 for above approach
 
# def Function to perform operations K times
def decreaseNum(N, K):
 
    while True:
        K -= 1
        # Last digit is 0
        if (N % 10 == 0):
            N //= 10
 
            # Last digit is not 0
        else:
            N -= 1
 
        if K == 0:
            break
 
    return N
 
# Driver Code
if __name__ == "__main__":
 
        # Declaration and initialisation
    N = 512
    K = 4
 
    # Function call
    print(decreaseNum(N, K))
 
    # This code is contributed by rakeshsahni

C#




// C# program for the above approach
using System;
using System.Collections;
 
class GFG
{
 
  // Function to perform operations K times
  public static int decreaseNum(int N, int K)
  {
    while (true) {
      K -= 1;
 
      // Last digit is 0
      if (N % 10 == 0)
        N /= 10;
 
      // Last digit is not 0
      else
        N--;
 
      if (K == 0)
        break;
    }
    return N;
  }
 
  // Driver Code
  public static void Main()
  {
    // Declaration and initialisation
    int N = 512;
    int K = 4;
 
    // Function call
    Console.Write(decreaseNum(N, K));
 
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




  <script>
      // JavaScript code for the above approach
 
      // Function to perform operations K times
      function decreaseNum(N, K)
      {
          while (K--)
          {
           
              // Last digit is 0
              if (N % 10 == 0)
                  N /= 10;
 
              // Last digit is not 0
              else
                  N--;
          }
          return N;
      }
 
      // Driver Code
 
      // Declaration and initialisation
      let N, K;
      N = 512;
      K = 4;
 
      // Function call
      document.write(decreaseNum(N, K));
 
// This code is contributed by Potta Lokesh
  </script>

 
 

Output
50

 

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

 


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