Problem

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.

Examples

Example 1:

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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:

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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.

Solution

Method 1 - With Right Shift Operator

  • We keep on dividing the number by 2, but we have to check if number is odd or even.
  • Process the binary number from right to left and check if:
    • Encountering a 0 (even number): divide by 2 increases result by 1 (result += 1)
    • Encountering a 1 (odd number): increases result by 2, (result += 2). The reason is 1 operation for subtracting, and now it becomes even, so we divide again. Hence 2 operations.
    • only exception is the leftmost 1, because now number is 1, and we treat it is as odd. So, we do result+=2, but actually it just increase result by 1. So, we just do a “-1” and number becomes 0 already.

Code

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public int numberOfSteps (int num) {
	if (num == 0) {
		return 0;
	}
	int ans = 0;
	while (num != 0) {
		ans += (num & 1) == 0 ? 1 : 2;
		num >>= 1;
	}
	return ans - 1;
}

Method 2 - Without Bit Manipulation

Code

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public int numberOfSteps(int num) {
	int ans = 0;

	while (num > 0) {
		if (num % 2 == 0) {
			num /= 2;
		} else {
			num--;
		}

		ans++;
	}

	return ans;
}