# 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 required``int` `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 code``int` `main()``{``    ``int` `p = 2, q = 3, n = 20;``    ``cout << minOperations(p, q, n) << ``"\n"``;``    ``return` `0;``}`

## Java

 `// Java program for the above approach``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``;``        ``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 required``def` `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 code``if` `__name__ ``=``=` `"__main__"``:` `    ``p ``=` `2``    ``q ``=` `3``    ``n ``=` `20``    ``print``(minOperations(p, q, n))` `    ``# This code is contributed by rakeshsahni`

## Javascript

 ``

## C#

 `// C# program for the above approach``using` `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)

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