Solved! Leetcode 2000. Reverse Prefix of Word

Description: Reverse Prefix of Word

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 1

<strong>Input:</strong> word = "abcdefd", ch = "d"
<strong>Output:</strong> "dcbaefd"
<strong>Explanation:</strong>&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".

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

<strong>Input:</strong> word = "xyxzxe", ch = "z"
<strong>Output:</strong> "zxyxxe"
<strong>Explanation:</strong>&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".

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

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

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.

Solution

class Solution {
    public String reversePrefix(String word, char ch) {
        if(word==null || word.isBlank()){
            return word;
        }

        //find the first index of ch in the word
        int end = word.indexOf(ch);
        if(end!=-1){
            char[] c = word.toCharArray();
            int left = 0;
            int right = end;

            //reverse the portion between 0 and the index of ch
            while(left<right){
                char t = c[left];
                c[left] = c[right];
                c[right] = t;
                left++;
                right--;
            }
            word = new String(c);
        }
        return word;
    }
}

class Solution {

public String reversePrefix(String word, char ch) {

if(word==null || word.isBlank()){

return word;

}

//find the first index of ch in the word

int end = word.indexOf(ch);

if(end!=-1){

char[] c = word.toCharArray();

int left = 0;

int right = end;

//reverse the portion between 0 and the index of ch

while(left<right){

char t = c[left];

c[left] = c[right];

c[right] = t;

left++;

right--;

}

word = new String(c);

}

return word;

}

Time Complexity

O(n), where n is the number of characters in the word (specifically O(n/2))

Space Complexity

O(n), where n is the number of characters in the word

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