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Formatted question description: https://leetcode.ca/all/1404.html

# 1404. Number of Steps to Reduce a Number in Binary Representation to One (Medium)

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 <= 500
• s consists of characters '0' or '1'
• s == '1'

Related Topics:
String, Bit Manipulation

## Solution 1.

• class Solution {
public int numSteps(String s) {
int steps = 0;
StringBuffer sb = new StringBuffer(s);
while (sb.length() > 1) {
char lastChar = sb.charAt(sb.length() - 1);
if (lastChar == '0')
sb.deleteCharAt(sb.length() - 1);
else
steps++;
}
return steps;
}

int length = sb.length();
int lastZeroIndex = -1;
for (int i = length - 1; i >= 0; i--) {
if (sb.charAt(i) == '0') {
lastZeroIndex = i;
break;
}
}
if (lastZeroIndex < 0) {
for (int i = length - 1; i >= 0; i--)
sb.setCharAt(i, '0');
sb.insert(0, '1');
} else {
sb.setCharAt(lastZeroIndex, '1');
for (int i = lastZeroIndex + 1; i < length; i++)
sb.setCharAt(i, '0');
}
return sb;
}
}

############

class Solution {
public int numSteps(String s) {
boolean carry = false;
int ans = 0;
for (int i = s.length() - 1; i > 0; --i) {
char c = s.charAt(i);
if (carry) {
if (c == '0') {
c = '1';
carry = false;
} else {
c = '0';
}
}
if (c == '1') {
++ans;
carry = true;
}
++ans;
}
if (carry) {
++ans;
}
return ans;
}
}

• // OJ: https://leetcode.com/problems/number-of-steps-to-reduce-a-number-in-binary-representation-to-one/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int numSteps(string s) {
int ans = 0;
while (s != "1") {
if (s.back() == '1') {
int i = s.size() - 1;
while (i >= 0 && s[i] == '1') {
s[i] = '0';
--i;
}
if (i == -1) s.insert(s.begin(), '1');
else s[i] = '1';
} else s.pop_back();
++ans;
}
return ans;
}
};

• class Solution:
def numSteps(self, s: str) -> int:
carry = False
ans = 0
for c in s[:0:-1]:
if carry:
if c == '0':
c = '1'
carry = False
else:
c = '0'
if c == '1':
ans += 1
carry = True
ans += 1
if carry:
ans += 1
return ans

############

class Solution:
def numSteps(self, s: str) -> int:
step = 0
s_int = int(s, 2)
while s_int != 1:
if s_int % 2 == 0:
s_int //= 2
else:
s_int += 1
step += 1
return step

• func numSteps(s string) int {
ans := 0
carry := false
for i := len(s) - 1; i > 0; i-- {
c := s[i]
if carry {
if c == '0' {
c = '1'
carry = false
} else {
c = '0'
}
}
if c == '1' {
ans++
carry = true
}
ans++
}
if carry {
ans++
}
return ans
}