Skip to content
Related Articles

Related Articles

Improve Article

Count the minimum steps to reach 0 from the given integer N

  • Last Updated : 05 May, 2021

Given two integers N and K where K represents the number of jumps that we are allowed to make directly from N reducing N to N – K, our task is to count minimum steps to reach 0 following the given operations: 

  • We can jump by a amount of K from N that is N = N – K
  • Decrement N by 1 that is N = N -1.

Examples: 

Input: N = 11, K = 4 
Output:
Explanation: 
For the given value N we can perform the operation in the given sequence: 11 -> 7 -> 3 -> 2 -> 1 -> 0

Input: N = 6, K = 3 
Output:
Explanation: 
For the given value N we can perform the operation in the given sequence: 6 -> 3 -> 0. 
 

Approach: 
To solve the problem mentioned above we know that it will take N / K steps to directly jump from value N to least divisible value with K and N % K steps to decrement it by 1 such as to reduce the count to 0. So the total number of steps required to reach 0 from N will be 



(N / K) + (N % K)

Below is the implementation of the above approach:  

C++




// C++ program to Count the minimum steps
// to reach 0 from the given integer N
 
#include <bits/stdc++.h>
using namespace std;
 
// Function returns min step
// to reach 0 from N
int getMinSteps(int n, int jump)
{
    // Direct possible
    // reduction of value N
    int quotient = n / jump;
 
    // Remaining steps needs
    // to be reduced by 1
    int remainder = n % jump;
 
    // Summation of both the values
    int steps = quotient + remainder;
 
    // Return the final answer
    return steps;
}
 
// Driver code
int main()
{
    int N = 6, K = 3;
 
    cout << getMinSteps(N, K);
 
    return 0;
}

Java




// Java program to count the minimum steps
// to reach 0 from the given integer N
class GFG{
 
// Function returns min step
// to reach 0 from N
static int getMinSteps(int n, int jump)
{
     
    // Direct possible
    // reduction of value N
    int quotient = n / jump;
 
    // Remaining steps needs
    // to be reduced by 1
    int remainder = n % jump;
 
    // Summation of both the values
    int steps = quotient + remainder;
 
    // Return the final answer
    return steps;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 6, K = 3;
 
    System.out.print(getMinSteps(N, K));
}
}
 
// This code is contributed by Rohit_ranjan

Python3




# Python3 program to Count the minimum steps
# to reach 0 from the given integer N
 
# Function returns min step
# to reach 0 from N
def getMinSteps(n, jump):
 
    # Direct possible
    # reduction of value N
    quotient = int(n / jump)
 
    # Remaining steps needs
    # to be reduced by 1
    remainder = n % jump
 
    # Summation of both the values
    steps = quotient + remainder
 
    # Return the final answer
    return steps
 
# Driver code
N = 6
K = 3
 
print (getMinSteps(N, K))
 
# This code is contributed by PratikBasu

C#




// C# program to count the minimum steps
// to reach 0 from the given integer N
using System;
 
class GFG{
 
// Function returns min step
// to reach 0 from N
static int getMinSteps(int n, int jump)
{
     
    // Direct possible
    // reduction of value N
    int quotient = n / jump;
 
    // Remaining steps needs
    // to be reduced by 1
    int remainder = n % jump;
 
    // Summation of both the values
    int steps = quotient + remainder;
 
    // Return the final answer
    return steps;
}
 
// Driver code
public static void Main(string[] args)
{
    int N = 6, K = 3;
 
    Console.Write(getMinSteps(N, K));
}
}
 
// This code is contributed by rutvik_56

Javascript




<script>
// JavaScript program to Count the minimum steps
// to reach 0 from the given integer N
 
// Function returns min step
// to reach 0 from N
function getMinSteps(n, jump)
{
 
    // Direct possible
    // reduction of value N
    let quotient = Math.floor(n / jump);
 
    // Remaining steps needs
    // to be reduced by 1
    let remainder = n % jump;
 
    // Summation of both the values
    let steps = quotient + remainder;
 
    // Return the final answer
    return steps;
}
 
// Driver code
    let N = 6, K = 3;
    document.write(getMinSteps(N, K));
 
// This code is contributed by Surbhi Tyagi.
</script>
Output: 
2

 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.




My Personal Notes arrow_drop_up
Recommended Articles
Page :