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Formatted question description: https://leetcode.ca/all/1342.html
1342. Number of Steps to Reduce a Number to Zero (Easy)
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
Related Topics:
Bit Manipulation
Solution 1.
// OJ: https://leetcode.com/problems/number-of-steps-to-reduce-a-number-to-zero/
// Time: O(logN)
// Space: O(1)
class Solution {
public:
int numberOfSteps (int num) {
int cnt = 0;
while (num) {
cnt++;
if (num % 2 == 0) num /= 2;
else num--;
}
return cnt;
}
};
Java
-
class Solution { public int numberOfSteps (int num) { int count = 0; while (num > 0) { if (num % 2 == 0) num /= 2; else num--; count++; } return count; } } -
// OJ: https://leetcode.com/problems/number-of-steps-to-reduce-a-number-to-zero/ // Time: O(logN) // Space: O(1) class Solution { public: int numberOfSteps (int num) { int cnt = 0; while (num) { cnt++; if (num & 1) --num; else num >>= 1; } return cnt; } }; -
class Solution: def numberOfSteps(self, num: int) -> int: ans = 0 while num: if num & 1: num -= 1 else: num >>= 1 ans += 1 return ans