Minimum decrements to make integer A divisible by integer B

0 0 3 minutes read

Given two positive integers A and B where A is greater than B. In one move one can decrease A by 1 which implies that after one move A is equal to A – 1. The task is to find the minimum number of moves required to make A divisible by B in constant time.

Examples:

Input: A = 10, B = 3
Output: 1
Explanation: Only one move is required A = A – 1 = 9, which is divisible by 3.

Input : A = 10, B = 10
Output : 0
Explanation: Since A is equal to B therefore zero move required.

Approach:
To solve the problem mentioned above we take the modulus of the numbers that are A % B and the result is stored in a variable that is the required answer.
Below is the implementation of the above approach:

C++

// C++ implementation to count
// Total numbers moves to make
// integer A divisible by integer B
 
#include <bits/stdc++.h>
using namespace std;
 
// Function that print number
// of moves required
void movesRequired(int a, int b)
{
    // calculate modulo
    int total_moves = a % b;
 
    // print the required answer
    cout << total_moves << "\n";
}
 
// Driver Code
int main()
{
 
    // initialise A and B
    int A = 10, B = 3;
 
    movesRequired(A, B);
 
    return 0;
}

Java

// Java implementation to count
// total numbers moves to make
// integer A divisible by integer B
import java.util.*;
 
class GFG{
 
// Function that print number
// of moves required
static void movesRequired(int a, int b)
{
     
    // Calculate modulo
    int total_moves = a % b;
 
    // Print the required answer
    System.out.println(total_moves);
}
 
// Driver code
public static void main(String[] args)
{
     
    // Initialise A and B
    int A = 10, B = 3;
 
    movesRequired(A, B);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 implementation to count
# total numbers moves to make
# integer A divisible by integer B
 
# Function that print number
# of moves required
def movesRequired(a, b):
     
    # Calculate modulo
    total_moves = a % b
 
    # Print the required answer
    print(total_moves)
 
# Driver Code
if __name__ == '__main__':
     
    # Initialise A and B
    A = 10
    B = 3
 
    movesRequired(A, B)
 
# This code is contributed by Samarth

C#

// C# implementation to count
// total numbers moves to make
// integer A divisible by integer B
using System;
class GFG
{
 
// Function that print number
// of moves required
static void movesRequired(int a, int b)
{
     
    // Calculate modulo
    int total_moves = a % b;
 
    // Print the required answer
    Console.Write(total_moves);
}
 
// Driver code
public static void Main(String []args)
{
 
    // Initialise A and B
    int A = 10, B = 3;
 
    movesRequired(A, B);
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

<script>
    // Javascript implementation to count 
    // Total numbers moves to make 
    // integer A divisible by integer B 
     
    // Function that print number 
    // of moves required 
    function movesRequired(a, b) 
    { 
        // calculate modulo 
        let total_moves = a % b; 
 
        // print the required answer 
        document.write(total_moves); 
    } 
     
    // initialise A and B 
    let A = 10, B = 3; 
   
    movesRequired(A, B); 
 
// This code is contributed by divyesh072019.
</script>

Output:

Time Complexity: O(1), as constant operations are being performed.
Auxiliary Space: O(1), constant space is required for the above approach.

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!