Question

1702. Maximum Binary String After Change

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Description

You are given a binary string binary consisting of only 0's or 1's. You can apply each of the following operations any number of times:

• Operation 1: If the number contains the substring "00", you can replace it with "10".
  • For example, "00010" -> "10010"
• Operation 2: If the number contains the substring "10", you can replace it with "01".
  • For example, "00010" -> "00001"

_Return the maximum binary string you can obtain after any number of operations. Binary string x is greater than binary string y if x's decimal representation is greater than y's decimal representation._

 

Example 1:

Input: binary = "000110"
Output: "111011"
Explanation: A valid transformation sequence can be:
"000110" -> "000101" 
"000101" -> "100101" 
"100101" -> "110101" 
"110101" -> "110011" 
"110011" -> "111011"

Example 2:

Input: binary = "01"
Output: "01"
Explanation: "01" cannot be transformed any further.

 

Constraints:

• 1 <= binary.length <= 105
• binary consist of '0' and '1'.

Solutions

Solution 1: Quick Thinking

We observe that operation 2 can move all $1$s to the end of the string, and operation 1 can change all 0000..000 strings to 111..110.

Therefore, to get the maximum binary string, we should move all $1$s that are not at the beginning to the end of the string, making the string in the form of 111..11...00..11, and then use operation 1 to change the middle 000..00 to 111..10. In this way, we can finally get a binary string that contains at most one $0$, which is the maximum binary string we are looking for.

The time complexity is $O(n)$, where $n$ is the length of the string binary. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

class Solution:
  def maximumBinaryString(self, binary: str) -> str:
    k = binary.find('0')
    if k == -1:
      return binary
    k += binary[k + 1 :].count('0')
    return '1' * k + '0' + '1' * (len(binary) - k - 1)

class Solution {
  public String maximumBinaryString(String binary) {
    int k = binary.indexOf('0');
    if (k == -1) {
      return binary;
    }
    int n = binary.length();
    for (int i = k + 1; i < n; ++i) {
      if (binary.charAt(i) == '0') {
        ++k;
      }
    }
    char[] ans = binary.toCharArray();
    Arrays.fill(ans, '1');
    ans[k] = '0';
    return String.valueOf(ans);
  }
}

class Solution {
public:
  string maximumBinaryString(string binary) {
    int k = binary.find('0');
    if (k == binary.npos) return binary;
    int n = binary.size();
    for (int i = k + 1; i < n; ++i) {
      if (binary[i] == '0') {
        ++k;
      }
    }
    return string(k, '1') + '0' + string(n - k - 1, '1');
  }
};

func maximumBinaryString(binary string) string {
  k := strings.IndexByte(binary, '0')
  if k == -1 {
    return binary
  }
  for _, c := range binary[k+1:] {
    if c == '0' {
      k++
    }
  }
  ans := []byte(binary)
  for i := range ans {
    ans[i] = '1'
  }
  ans[k] = '0'
  return string(ans)
}