Problem
Given the binary representation of an integer as a string s, 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 is guaranteed that you can always reach one for all test cases.
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 (Java)
class Solution {
public int numSteps(String s) {
int steps = 0;
int carry = 0;
for (int i = s.length() - 1; i >= 1; i--) {
if (carry == 0) {
if (s.charAt(i) == '1') {
steps += 2;
carry = 1;
} else {
steps++;
}
} else {
if (s.charAt(i) == '0') {
steps += 2;
} else {
steps++;
}
}
}
return steps + carry;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).