README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1300-1399/1342.Number%20of%20Steps%20to%20Reduce%20a%20Number%20to%20Zero/README_EN.md	1163	Biweekly Contest 19 Q1	Bit Manipulation Math

1342. Number of Steps to Reduce a Number to Zero

中文文档

Description

Given an integer num, return the number of steps to reduce it to zero.

In one step, if the current number is even, you have to divide it by 2, otherwise, you have to subtract 1 from it.

 

Example 1:

Input: num = 14
Output: 6
Explanation: 
Step 1) 14 is even; divide by 2 and obtain 7. 
Step 2) 7 is odd; subtract 1 and obtain 6.
Step 3) 6 is even; divide by 2 and obtain 3. 
Step 4) 3 is odd; subtract 1 and obtain 2. 
Step 5) 2 is even; divide by 2 and obtain 1. 
Step 6) 1 is odd; subtract 1 and obtain 0.

Example 2:

Input: num = 8
Output: 4
Explanation: 
Step 1) 8 is even; divide by 2 and obtain 4. 
Step 2) 4 is even; divide by 2 and obtain 2. 
Step 3) 2 is even; divide by 2 and obtain 1. 
Step 4) 1 is odd; subtract 1 and obtain 0.

Example 3:

Input: num = 123
Output: 12

 

Constraints:

• 0 <= num <= 106

Solutions

Solution 1

Python3

class Solution:
    def numberOfSteps(self, num: int) -> int:
        ans = 0
        while num:
            if num & 1:
                num -= 1
            else:
                num >>= 1
            ans += 1
        return ans

Java

class Solution {

    public int numberOfSteps(int num) {
        int ans = 0;
        while (num != 0) {
            num = (num & 1) == 1 ? num - 1 : num >> 1;
            ++ans;
        }
        return ans;
    }
}

C++

class Solution {
public:
    int numberOfSteps(int num) {
        int ans = 0;
        while (num) {
            num = num & 1 ? num - 1 : num >> 1;
            ++ans;
        }
        return ans;
    }
};

Go

func numberOfSteps(num int) int {
	ans := 0
	for num != 0 {
		if (num & 1) == 1 {
			num--
		} else {
			num >>= 1
		}
		ans++
	}
	return ans
}

TypeScript

function numberOfSteps(num: number): number {
    let ans = 0;
    while (num) {
        num = num & 1 ? num - 1 : num >>> 1;
        ans++;
    }
    return ans;
}

Rust

impl Solution {
    pub fn number_of_steps(mut num: i32) -> i32 {
        let mut count = 0;
        while num != 0 {
            if num % 2 == 0 {
                num >>= 1;
            } else {
                num -= 1;
            }
            count += 1;
        }
        count
    }
}

Solution 2

Python3

class Solution:
    def numberOfSteps(self, num: int) -> int:
        if num == 0:
            return 0
        return 1 + (
            self.numberOfSteps(num // 2)
            if num % 2 == 0
            else self.numberOfSteps(num - 1)
        )

Java

class Solution {

    public int numberOfSteps(int num) {
        if (num == 0) {
            return 0;
        }
        return 1 + numberOfSteps((num & 1) == 0 ? num >> 1 : num - 1);
    }
}

C++

class Solution {
public:
    int numberOfSteps(int num) {
        if (num == 0) return 0;
        return 1 + (num & 1 ? numberOfSteps(num - 1) : numberOfSteps(num >> 1));
    }
};

Go

func numberOfSteps(num int) int {
	if num == 0 {
		return 0
	}
	if (num & 1) == 0 {
		return 1 + numberOfSteps(num>>1)
	}
	return 1 + numberOfSteps(num-1)
}

Rust

impl Solution {
    pub fn number_of_steps(mut num: i32) -> i32 {
        if num == 0 {
            0
        } else if num % 2 == 0 {
            1 + Solution::number_of_steps(num >> 1)
        } else {
            1 + Solution::number_of_steps(num - 1)
        }
    }
}