###### tags: `Weekly Contest`
# Weekly Contest 389

## [3083. Existence of a Substring in a String and Its Reverse](https://leetcode.com/problems/existence-of-a-substring-in-a-string-and-its-reverse/) (<font color=#00B8A3>Easy</font>)

限制 :

<ul>
<li><code>1 <= s.length <= 100</code></li>
<li><code>s consists only of lowercase English letters.</code></li>
</ul>

### Solution
屬於比較爛的雙指針寫法。

#### 時間複雜度: $O(n^2)$

#### 空間複雜度: $O(n)$

程式碼:

class Solution {
public:
    bool isSubstringPresent(string s) {
        for(int i=0;i<s.size()-1;i++)
        {
            for(int j=s.size()-1;j>0;j--)
            {
                if(s[j] == s[i] && s[i + 1] == s[j - 1])
                    return true;
            }
        }
        return false;
    }
};

## [3084. Count Substrings Starting and Ending with Given Character](https://leetcode.com/problems/count-substrings-starting-and-ending-with-given-character/) (<font color=#FFC011>Medium</font>)

限制 :

<ul>
<li><code>1 <= s.length <= 10⁵</code></li>
<li><code>s and c consist only of lowercase English letters.</code></li>
</ul>
### Solution

#### 時間複雜度: $O(n)$

#### 空間複雜度: $O(1)$

程式碼:

class Solution {
public:    
    long long countSubstrings(string s, char c) {
        int charCount = 0;
        long long result = 0;
        for(int i=0;i<s.size();i++)
        {
            if (c == s[i])
            {
                charCount++;
                result += charCount;
            }
        }
        
        return result;
    }
};

## [3]()(<font color=#FFC011>Medium</font>)

限制 :

<ul>
<li><code>10⁴</code></li>
</ul>
### Solution

#### 時間複雜度: $O()$

#### 空間複雜度: $O()$

程式碼:

## [4]()(<font color=#FF375F>Hard</font>)

限制 :

<ul>
<li><code>10⁴</code></li>
</ul>
### Solution

#### 時間複雜度: $O()$

#### 空間複雜度: $O()$

程式碼: