09_Count_Operations_to_Obtain_Zero.cpp

// 2169. Count Operations to Obtain Zero

// You are given two non-negative integers num1 and num2.

// In one operation, if num1 >= num2, you must subtract num2 from num1, otherwise subtract num1 from num2.

// For example, if num1 = 5 and num2 = 4, subtract num2 from num1, thus obtaining num1 = 1 and num2 = 4. However, if num1 = 4 and num2 = 5, after one operation, num1 = 4 and num2 = 1.
// Return the number of operations required to make either num1 = 0 or num2 = 0.

 

// Example 1:

// Input: num1 = 2, num2 = 3
// Output: 3
// Explanation: 
// - Operation 1: num1 = 2, num2 = 3. Since num1 < num2, we subtract num1 from num2 and get num1 = 2, num2 = 3 - 2 = 1.
// - Operation 2: num1 = 2, num2 = 1. Since num1 > num2, we subtract num2 from num1.
// - Operation 3: num1 = 1, num2 = 1. Since num1 == num2, we subtract num2 from num1.
// Now num1 = 0 and num2 = 1. Since num1 == 0, we do not need to perform any further operations.
// So the total number of operations required is 3.
// Example 2:

// Input: num1 = 10, num2 = 10
// Output: 1
// Explanation: 
// - Operation 1: num1 = 10, num2 = 10. Since num1 == num2, we subtract num2 from num1 and get num1 = 10 - 10 = 0.
// Now num1 = 0 and num2 = 10. Since num1 == 0, we are done.
// So the total number of operations required is 1.
 

// Constraints:

// 0 <= num1, num2 <= 105

class Solution {
public:
    int countOperations(int x, int y) {
        int cnt=0, r;
        for( ; y>0; x=y, y=r){
            cnt+=x/y; 
            r=x%y;  
        }
        return cnt;
    }
};class Solution {
public:
    int countOperations(int x, int y) {
        int cnt=0, r;
        for( ; y>0; x=y, y=r){
            cnt+=x/y; 
            r=x%y;  
        }
        return cnt;
    }
};