Given a number s in their binary representation. Return the number of steps to reduce it to 1 under the following rules:
- If the current number is even, you have to divide it by 2.
- If the current number is odd, you have to add 1 to it.
It’s guaranteed that you can always reach to one for all testcases.
Example 1:
Input: s = "1101" Output: 6 Explanation: "1101" corressponds to number 13 in their decimal representation. Step 1) 13 is odd, add 1 and obtain 14. Step 2) 14 is even, divide by 2 and obtain 7. Step 3) 7 is odd, add 1 and obtain 8. Step 4) 8 is even, divide by 2 and obtain 4. Step 5) 4 is even, divide by 2 and obtain 2. Step 6) 2 is even, divide by 2 and obtain 1.
Example 2:
Input: s = "10" Output: 1 Explanation: "10" corressponds to number 2 in their decimal representation. Step 1) 2 is even, divide by 2 and obtain 1.
Example 3:
Input: s = "1" Output: 0
Constraints:
1 <= s.length <= 500sconsists of characters ‘0’ or ‘1’s[0] == '1'
Solution: Simulation
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int numSteps(string s) { int ans = 0; int carry = 0; // The highest bit must be 1, // process to the 2nd highest bit for (int i = s.length() - 1; i > 0; --i) { if (s[i] - '0' + carry == 1) { ans += 2; // odd: +1, even: / 2 carry = 1; // always has a carry } else { ans += 1; // even: / 2 // carray remains e.g. 1 + 1 = 10, or 0 + 0 = 00 } } // If there is a carry, then it's even, one more step. return ans + carry; } }; |
Python3
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# Author: Huahua class Solution: def numSteps(self, s: str) -> int: ans, carry = 0, 0 for i in range(1, len(s)): if ord(s[-i]) - ord('0') + carry == 1: ans += 2 carry = 1 else: ans += 1 return ans + carry |