<reponame>dvlpsh/leetcode-1<filename>python/1342_Number_of_Steps_to_Reduce_a_Number_to_Zero.py
'''
Given a non-negative integer num, return the number of steps to reduce it to zero. 
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 <= 10^6
'''

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

    # def numberOfSteps (self, num: int) -> int:
    #     ans = 0
    #     # check the number of 1
    #     while num:
    #         ans += (num & 1) + 1
    #         num >>= 1
    #     return ans - 1

    # def numberOfSteps (self, num: int) -> int:
    #     ones, zeros = self.bitCount(num)
    #     if ones == 0:
    #         return 0
    #     else:
    #         return 2 * ones + zeros - 1

'''
Given a non-negative integer num, return the number of steps to reduce it to zero. 
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 <= 10^6
'''

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

    # def numberOfSteps (self, num: int) -> int:
    #     ans = 0
    #     # check the number of 1
    #     while num:
    #         ans += (num & 1) + 1
    #         num >>= 1
    #     return ans - 1

    # def numberOfSteps (self, num: int) -> int:
    #     ones, zeros = self.bitCount(num)
    #     if ones == 0:
    #         return 0
    #     else:
    #         return 2 * ones + zeros - 1
Given a non-negative integer num, return the number of steps to reduce it to zero.  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 <= 10^6 # def numberOfSteps (self, num: int) -> int: #     ans = 0 #     # check the number of 1 #     while num: #         ans += (num & 1) + 1 #         num >>= 1 #     return ans - 1 # def numberOfSteps (self, num: int) -> int: #     ones, zeros = self.bitCount(num) #     if ones == 0: #         return 0 #     else: #         return 2 * ones + zeros - 1