2169. Count Operations to Obtain Zero 

Description

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

Solutions

Solution 1: Simulation

We can directly simulate this process by repeatedly performing the following operations:

If $\textit{num1} \ge \textit{num2}$, then $\textit{num1} = \textit{num1} - \textit{num2}$;
Otherwise, $\textit{num2} = \textit{num2} - \textit{num1}$.
Each time an operation is performed, increment the operation count by one.

When either $\textit{num1}$ or $\textit{num2}$ becomes $0$, stop the loop and return the operation count.

The time complexity is $O(m)$, where $m$ is the maximum of $\textit{num1}$ and $\textit{num2}$. The space complexity is $O(1)$.

Python3

class Solution:
    def countOperations(self, num1: int, num2: int) -> int:
        ans = 0
        while num1 and num2:
            if num1 >= num2:
                num1 -= num2
            else:
                num2 -= num1
            ans += 1
        return ans

Java

class Solution {
    public int countOperations(int num1, int num2) {
        int ans = 0;
        for (; num1 != 0 && num2 != 0; ++ans) {
            if (num1 >= num2) {
                num1 -= num2;
            } else {
                num2 -= num1;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int countOperations(int num1, int num2) {
        int ans = 0;
        for (; num1 && num2; ++ans) {
            if (num1 >= num2) {
                num1 -= num2;
            } else {
                num2 -= num1;
            }
        }
        return ans;
    }
};

Go

func countOperations(num1 int, num2 int) (ans int) {
	for ; num1 != 0 && num2 != 0; ans++ {
		if num1 >= num2 {
			num1 -= num2
		} else {
			num2 -= num1
		}
	}
	return
}

TypeScript

function countOperations(num1: number, num2: number): number {
    let ans = 0;
    for (; num1 && num2; ++ans) {
        if (num1 >= num2) {
            num1 -= num2;
        } else {
            num2 -= num1;
        }
    }
    return ans;
}

Rust

impl Solution {
    pub fn count_operations(mut num1: i32, mut num2: i32) -> i32 {
        let mut ans = 0;
        while num1 != 0 && num2 != 0 {
            ans += 1;
            if num1 >= num2 {
                num1 -= num2;
            } else {
                num2 -= num1;
            }
        }
        ans
    }
}

JavaScript

/**
 * @param {number} num1
 * @param {number} num2
 * @return {number}
 */
var countOperations = function (num1, num2) {
    let ans = 0;
    for (; num1 && num2; ++ans) {
        if (num1 >= num2) {
            num1 -= num2;
        } else {
            num2 -= num1;
        }
    }
    return ans;
};

Solution 2

Python3

class Solution:
    def countOperations(self, num1: int, num2: int) -> int:
        ans = 0
        while num1 and num2:
            if num1 >= num2:
                ans += num1 // num2
                num1 %= num2
            else:
                ans += num2 // num1
                num2 %= num1
        return ans

Solution 2: Mathematics

Following the simulation process in Solution 1, we notice that if $\textit{num1}$ is much larger than $\textit{num2}$, each operation will only reduce the value of $\textit{num1}$ slightly, leading to an excessive number of operations. We can optimize this process by directly adding the quotient of $\textit{num1}$ divided by $\textit{num2}$ to the answer in each operation, then taking the remainder of $\textit{num1}$ divided by $\textit{num2}$. This reduces the number of operations.

The time complexity is $O(\log m)$, where $m$ is the maximum of $\textit{num1}$ and $\textit{num2}$. The space complexity is $O(1)$.

Python3

class Solution:
    def countOperations(self, num1: int, num2: int) -> int:
        ans = 0
        while num1 and num2:
            if num1 >= num2:
                ans += num1 // num2
                num1 %= num2
            else:
                ans += num2 // num1
                num2 %= num1
        return ans

Java

class Solution {
    public int countOperations(int num1, int num2) {
        int ans = 0;
        while (num1 != 0 && num2 != 0) {
            if (num1 >= num2) {
                ans += num1 / num2;
                num1 %= num2;
            } else {
                ans += num2 / num1;
                num2 %= num1;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int countOperations(int num1, int num2) {
        int ans = 0;
        while (num1 && num2) {
            if (num1 >= num2) {
                ans += num1 / num2;
                num1 %= num2;
            } else {
                ans += num2 / num1;
                num2 %= num1;
            }
        }
        return ans;
    }
};

Go

func countOperations(num1 int, num2 int) (ans int) {
	for num1 != 0 && num2 != 0 {
		if num1 >= num2 {
			ans += num1 / num2
			num1 %= num2
		} else {
			ans += num2 / num1
			num2 %= num1
		}
	}
	return
}

TypeScript

function countOperations(num1: number, num2: number): number {
    let ans = 0;
    while (num1 && num2) {
        if (num1 >= num2) {
            ans += (num1 / num2) | 0;
            num1 %= num2;
        } else {
            ans += (num2 / num1) | 0;
            num2 %= num1;
        }
    }
    return ans;
}

Rust

impl Solution {
    pub fn count_operations(mut num1: i32, mut num2: i32) -> i32 {
        let mut ans = 0;
        while num1 != 0 && num2 != 0 {
            if num1 >= num2 {
                ans += num1 / num2;
                num1 %= num2;
            } else {
                ans += num2 / num1;
                num2 %= num1;
            }
        }
        ans
    }
}

JavaScript

/**
 * @param {number} num1
 * @param {number} num2
 * @return {number}
 */
var countOperations = function (num1, num2) {
    let ans = 0;
    while (num1 && num2) {
        if (num1 >= num2) {
            ans += (num1 / num2) | 0;
            num1 %= num2;
        } else {
            ans += (num2 / num1) | 0;
            num2 %= num1;
        }
    }
    return ans;
};