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1342. Number of Steps to Reduce a Number to Zero

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

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

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class Solution {

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

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

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

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

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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

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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)
        )

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class Solution {

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

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

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

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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)
        }
    }
}

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