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.

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Example 1:

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


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Constraints:


	1 &lt;= word.length &lt;= 250
	word consists of lowercase English letters.
	ch is a lowercase English letter.


class Solution:
    def reversePrefix(self, word: str, ch: str) -> str:
        idx=word.find(ch)
        if idx:
            return word[:idx+1][::-1]+ word[idx+1:]
        return word
class Solution {
    public String reversePrefix(String word, char ch) {
        char[] c = word.toCharArray();
        int locate = 0;
        for (int i = 0; i < word.length(); i++) { //first occurrence of ch
            if (ch == c[i]) {
                locate = i;
                break;
            }
        }
        char[] res = new char[word.length()];
        for (int i = 0; i <= locate; i++) {
            res[i] = c[locate - i];
        }
        for (int i = locate + 1; i < word.length(); i++) {
            res[i] = c[i];
        }
        return String.valueOf(res);
    }
}
class Solution {
public:
    string reversePrefix(string word, char ch) {
        string ans = "";
        int tr = true;
        for(auto w : word){
            ans.push_back(w);
            if(tr && w == ch){
                tr=false;
                reverse(ans.begin(), ans.end());
            }
        }
        return ans;
    }
};
var reversePrefix = function(word, ch) {
    return word.indexOf(ch) !== -1 ? word.split("").slice(0, word.indexOf(ch) + 1).reverse().join("") + word.slice(word.indexOf(ch) + 1) : word;
};
Reverse Prefix of Word