Min operations to convert A to B by incrementing A in the given range

Given two integers A and B and a range defined by two integers C and D, the task is to convert A to B using the minimum number of operations possible. The only allowed operations are to add or subtract any integer greater than or equal to E to A. However, after each operation, the resulting value of A must lie in the range C to D (inclusive). If it is not possible to obtain B from A using these operations, return -1.

Examples:

Input: A = 4, B = 5, C = 0, D = 15, E = 5
Output: 2
Explanation: In first operation add 6 to A. A becomes 10. In second operation, subtract 5 from A. Now A becomes 5, which is equal to B. Note that after both operations, A lies in the range [C, D]. Also, the numbers added or subtracted (6 and 5) are greater than or equal to E.

Input: A = 3, B = 4, C = 3, D = 5, E = 6
Output: -1

Approach: To solve the problem follow the below idea:

• If A = B, in this case, 0 operations are required to convert A to B.
• If the absolute difference between A and B is greater than or equal to E, 1 operation is sufficient to make A equal to B. The integer value E = |B-A| can be added or subtracted to/from A to make it equal to B.
• There exists an integer X between C and D, such that |A-X| ≥ E and |B-X| ≥ E, in this case, A can be made B in 2 operations: A -> C -> B or A -> D -> B.
• If we cannot make A equal to B through one of the boundaries as in 3rd case, we will require 3 operations to do so, i.e. we will go through both the boundaries C and D: 
  A -> C -> D -> B or A -> D -> C -> B.
• If any of the above conditions are not satisfied, then it is not possible to convert A to B. Hence, we will return -1.

Below are the steps for the above approach:

• If A is equal to B, return 0.
• If the absolute difference between A and B is greater than or equal to E, return 1.
• If the difference between D and the greater of A and B is greater than or equal to E, or if the difference between the lesser of A and B and C is greater than or equal to E, return 2.
• Check if (D-B ≥ E && A-C ≥ E) or (D-A ≥ E && B-C ≥ E) is true, and return 3.
• Else returns -1.

Below is the code for the above approach:

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
#define int long long

// Function to find minimum operations
// to covert A to B
int minOperations(int A, int B, int C, int D, int E)
{

    // Case-1
    if (A == B) {
        return 0;
    }

    // Case-2
    else if (abs(A - B) >= E) {
        return 1;
    }

    // Case-3
    else if (D - max(A, B) >= E || min(A, B) - C >= E) {
        return 2;
    }

    // Case-4
    else if ((D - B >= E && A - C >= E)
             || (D - A >= E && B - C >= E)) {
        return 3;
    }

    // Case-5
    else {
        return -1;
    }
}

// Driver code
int32_t main()
{
    int A = 4, B = 5, C = 0, D = 15, E = 5;
    int answer = minOperations(A, B, C, D, E);

    // Function Call
    cout << answer << endl;
    return 0;
}

import java.util.*;

class Main {
    // Function to find minimum operations
    // to covert A to B
    public static long minOperations(long A, long B, long C,
                                     long D, long E)
    {
        // Case-1
        if (A == B) {
            return 0;
        }

        // Case-2
        else if (Math.abs(A - B) >= E) {
            return 1;
        }

        // Case-3
        else if (D - Math.max(A, B) >= E
                 || Math.min(A, B) - C >= E) {
            return 2;
        }

        // Case-4
        else if ((D - B >= E && A - C >= E)
                 || (D - A >= E && B - C >= E)) {
            return 3;
        }

        // Case-5
        else {
            return -1;
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        long A = 4, B = 5, C = 0, D = 15, E = 5;
        long answer = minOperations(A, B, C, D, E);

        // Function Call
        System.out.println(answer);
    }
}

# Python code for the above approach
import sys


def minOperations(A, B, C, D, E):
    # Case-1
    if A == B:
        return 0

    # Case-2
    elif abs(A - B) >= E:
        return 1

    # Case-3
    elif D - max(A, B) >= E or min(A, B) - C >= E:
        return 2

    # Case-4
    elif (D - B >= E and A - C >= E) or (D - A >= E and B - C >= E):
        return 3

    # Case-5
    else:
        return -1


# Driver code
if __name__ == '__main__':
    A = 4
    B = 5
    C = 0
    D = 15
    E = 5

    # Function call
    answer = minOperations(A, B, C, D, E)

    print(answer)

// C# code for the above approach

using System;

class GFG {
      // Function to find minimum operations
    // to covert A to B
    static int MinOperations(int A, int B, int C, int D, int E) {
        // Case-1
        if (A == B) {
            return 0;
        }

        // Case-2
        else if (Math.Abs(A - B) >= E) {
            return 1;
        }

        // Case-3
        else if (D - Math.Max(A, B) >= E || Math.Min(A, B) - C >= E) {
            return 2;
        }

        // Case-4
        else if ((D - B >= E && A - C >= E) || (D - A >= E && B - C >= E)) {
            return 3;
        }

        // Case-5
        else {
            return -1;
        }
    }

      // Driver's code
    static void Main(string[] args) {
          // Input
        int A = 4, B = 5, C = 0, D = 15, E = 5;
        int answer = MinOperations(A, B, C, D, E);

        // Function Call
        Console.WriteLine(answer);
    }
}

// Javascript code for the above approach

function minOperations(A, B, C, D, E) {

    // Case-1: if A and B are already equal, then no operations are required
    if (A == B) {
        return 0;
    }

    // Case-2: if the absolute difference between A and B 
    // is greater than or equal to E, then only one operation is required
    else if (Math.abs(A - B) >= E) {
        return 1;
    }

    // Case-3: if either A or B can be made equal to C or D by 
    // just one operation, then only two operations are required
    else if (D - Math.max(A, B) >= E || Math.min(A, B) - C >= E) {
        return 2;
    }

    // Case-4: if it is possible to make A and B both equal to C or D or 
    // any value in between C and D by two operations, then only three 
    // operations are required
    else if ((D - B >= E && A - C >= E) || (D - A >= E && B - C >= E)) {
        return 3;
    }

    // Case-5: if none of the above cases are applicable, 
    // then it is impossible to make A and B equal by performing 
    // any number of operations
    else {
        return -1;
    }
}

// Driver code
let A = 4, B = 5, C = 0, D = 15, E = 5;
let answer = minOperations(A, B, C, D, E);

// Function Call
console.log(answer);
//This Code is Contributed By Vikas Bishnoi (vikas20noi)

2

Time Complexity: O(1)
Auxiliary space: O(1)

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Article Tags :

• Greedy
• Mathematical
• DSA

Practice Tags :

• Greedy
• Mathematical