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++ 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 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 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# 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. |
<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>
|
50
Time Complexity: O(K)
Auxiliary Space: O(1)