Minimize operations till a or b exceeds N by replacing a or b with their sum

• Last Updated : 01 Nov, 2021

Given three integers a, b, and N. The task is to find minimum addition operations between a and b, such that after applying the operations, either of a or b becomes greater than N. An addition operation is defined as replacing either of a or b with their sum and keeping another one intact.

Examples:

Input: a = 2, b = 3, N = 20
Output: 4
Explanation

• Adding 2 and 3, 2 + 3 = 5 and replacing 2 with 5,  now a = 5, b = 3
• Again add a and b 5 + 3 = 8 replace b with 8, now a = 5, b = 8
• Again add a and b 5 + 8 = 13 replace a with 13. now a = 13, b = 8
• Again add a and b 13 + 8 = 21 replace b with 21, now a = 13, b = 21  Here, (b>=n) therefore minimum operations required are 4

Input: a = 2, b = 3, N = 5
Output: 1
Explanation: After replacing 2 with 2+3, a becomes 5 and b becomes 3, therefore minimum operations required are 1

Approach: The idea is to add a and b and store their sum in the minimum of a and b, every time until any of the numbers is greater than N. Reason behind it is making the minimum element largest every time, makes their sum high and thereby reducing the number of operations required.

Below is the implementation of the above approach:

C++

 // C++ program for the above approach #include using namespace std; // Function to print the minimum number// of operations requiredint minOperations(int a, int b, int n){     // Store the count of operations    int count = 0;     while (1) {         // If any value is greater than N        // return count        if (n <= a or n <= b) {            return count;            break;        }        else {            int sum = a + b;            if (a < b)                a = sum;            else                b = sum;        }        count++;    }    return count;} // Driver codeint main(){    int p = 2, q = 3, n = 20;    cout << minOperations(p, q, n) << "\n";    return 0;}

Java

 // Java program for the above approachclass GFG {     // Function to print the minimum number    // of operations required    public static int minOperations(int a, int b, int n) {         // Store the count of operations        int count = 0;         while (true) {             // If any value is greater than N            // return count            if (n <= a || n <= b) {                return count;            } else {                int sum = a + b;                if (a < b)                    a = sum;                else                    b = sum;            }            count++;        }    }     // Driver code    public static void main(String args[]) {        int p = 2, q = 3, n = 20;        System.out.println(minOperations(p, q, n));    }} // This code is contributed by saurabh_jaiswal.

Python3

 # python program for the above approach # Function to print the minimum number# of operations requireddef minOperations(a, b, n):     # Store the count of operations    count = 0     while (1):         # If any value is greater than N        # return count        if (n <= a or n <= b):            return count            break         else:            sum = a + b            if (a < b):                a = sum            else:                b = sum         count += 1     return count # Driver codeif __name__ == "__main__":     p = 2    q = 3    n = 20    print(minOperations(p, q, n))     # This code is contributed by rakeshsahni



C#

 // C# program for the above approachusing System; public class GFG {     // Function to print the minimum number    // of operations required    public static int minOperations(int a, int b, int n) {         // Store the count of operations        int count = 0;         while (true) {             // If any value is greater than N            // return count            if (n <= a || n <= b) {                return count;            } else {                int sum = a + b;                if (a < b)                    a = sum;                else                    b = sum;            }            count++;        }    }     // Driver code    public static void Main(string []args) {        int p = 2, q = 3, n = 20;        Console.WriteLine(minOperations(p, q, n));    }} // This code is contributed by AnkThon

Output
4

Time Complexity: O(min(log(max(a, N), log(max(b, N)))
Auxiliary Space: O(1)

My Personal Notes arrow_drop_up