LeetCode in Kotlin
1404. Number of Steps to Reduce a Number in Binary Representation to One
Medium
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
1to 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
class Solution {
fun numSteps(s: String): Int {
var steps = 0
var carry = 0
for (i in s.length - 1 downTo 1) {
if (carry == 0) {
if (s[i] == '1') {
steps += 2
carry = 1
} else {
steps++
}
} else {
if (s[i] == '0') {
steps += 2
} else {
steps++
}
}
}
return steps + carry
}
}