# 2000. Reverse Prefix of Word

## Description
Given a 0-indexed string word and a character ch, reverse the segment of word that starts at index 0 and ends at the index of the first occurrence of ch (inclusive). If the character ch does not exist in word, do nothing.

* For example, if word = "abcdefd" and ch = "d", then you should reverse the segment that starts at 0 and ends at 3 (inclusive). The resulting string will be "dcbaefd".

Return the resulting string.

## Example
### Example 1:

Input: word = "abcdefd", ch = "d"
Output: "dcbaefd"
Explanation: The first occurrence of "d" is at index 3.
Reverse the part of word from 0 to 3 (inclusive), the resulting string is "dcbaefd".

### Example 2:

Input: word = "xyxzxe", ch = "z"
Output: "zxyxxe"
Explanation: The first and only occurrence of "z" is at index 3.
Reverse the part of word from 0 to 3 (inclusive), the resulting string is "zxyxxe".

### Example 3:

Input: word = "abcd", ch = "z"
Output: "abcd"
Explanation: "z" does not exist in word.
You should not do any reverse operation, the resulting string is "abcd".

## Constraints
* `1 <= word.length <= 250`
* `word` consists of lowercase English letters.
* `ch` is a lowercase English letter.

## C++

class Solution {
public:
    void swap(char* a, char* b) {
        char tmp;
        tmp = *b;
        *b = *a;
        *a = tmp;
    }
    string reversePrefix(string word, char ch) {
        int i;
        for (i = 0; i <= word.size(); i++) {
            if (word[i] == ch)
                break;
            if (word[i] == '\0')
                return word;
        }
        int n = i / 2;
        for (int j = 0; j <= n; j++) {
            swap(&word[j], &word[i - j]);
        }
        return word;
    }
};

Using STL library

class Solution {
public:
    string reversePrefix(string word, char ch) {
        for(int i=0; i<word.size(); i++) {
            if(word[i] == ch) {
               reverse(word.begin(), word.begin()+i+1);
               return word;
            }
        }
        return word;
    }
};