5484. Find Kth Bit in Nth Binary String

Given two positive integers n and k, the binary string  Sn is formed as follows:

• S1 = "0"
• Si = Si-1 + "1" + reverse(invert(Si-1)) for i > 1

Where + denotes the concatenation operation, reverse(x) returns the reversed string x, and invert(x) inverts all the bits in x (0 changes to 1 and 1 changes to 0).

For example, the first 4 strings in the above sequence are:

• S1 = "0"
• S2 = "011"
• S3 = "0111001"
• S4 = "011100110110001"

Return the kth bit in Sn. It is guaranteed that k is valid for the given n.

Example 1:

Input: n = 3, k = 1
Output: "0"
Explanation: S3 is "0111001". The first bit is "0".

Example 2:

Input: n = 4, k = 11
Output: "1"
Explanation: S4 is "011100110110001". The 11th bit is "1".

Example 3:

Input: n = 1, k = 1
Output: "0"

Example 4:

Input: n = 2, k = 3
Output: "1"

Constraints:

• 1 <= n <= 20
• 1 <= k <= 2n - 1

---

简单题，直接字符串模拟操作即可，完整代码如下：

class Solution {
private:
	string InvertStr(string &s) {
		string ans = "";
		for (int i = 0; i < s.size(); ++i) {
			if (s[i] == '0')ans.push_back('1');
			else ans.push_back('0');
		}
		return ans;
	}
public:
	char findKthBit(int n, int k) {
		string s = "0";
		for (int i = 2; i <= n; ++i) {
			string inv_str = InvertStr(s);
			reverse(inv_str.begin(), inv_str.end());
			s = s + "1" + inv_str;
		}
		return s[k - 1];
	}
};

本代码提交AC。

5483. Make The String Great

Given a string s of lower and upper case English letters.

A good string is a string which doesn’t have two adjacent characters s[i] and s[i + 1] where:

• 0 <= i <= s.length - 2
• s[i] is a lower-case letter and s[i + 1] is the same letter but in upper-case or vice-versa.

To make the string good, you can choose two adjacent characters that make the string bad and remove them. You can keep doing this until the string becomes good.

Return the string after making it good. The answer is guaranteed to be unique under the given constraints.

Notice that an empty string is also good.

Example 1:

Input: s = "leEeetcode"
Output: "leetcode"
Explanation: In the first step, either you choose i = 1 or i = 2, both will result "leEeetcode" to be reduced to "leetcode".

Example 2:

Input: s = "abBAcC"
Output: ""
Explanation: We have many possible scenarios, and all lead to the same answer. For example:
"abBAcC" --> "aAcC" --> "cC" --> ""
"abBAcC" --> "abBA" --> "aA" --> ""

Example 3:

Input: s = "s"
Output: "s"

Constraints:

• 1 <= s.length <= 100
• s contains only lower and upper case English letters.

---

给定一个字符串，如果相邻两个字符是同一字母的大小写形式，则需要删除这两个字母，问最终的字符串形式是什么。

简单题，使用栈，完整代码如下：

class Solution {
public:
    string makeGood(string s) {
        stack<char> stk;
        for(int i = 0; i < s.size(); ++i) {
            if(!stk.empty() && abs(stk.top() - s[i]) == 32) {
                stk.pop();
            } else {
                stk.push(s[i]);
            }
        }
        string ans="";
        while(!stk.empty()) {
            ans = string(1, stk.top()) + ans;
            stk.pop();
        }
        return ans;
    }
};

本代码提交AC。