Build confidence with an intuition-first walkthrough focused on two pointers fundamentals.
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.
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:
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 <= 250word consists of lowercase English letters.ch is a lowercase English letter.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers · Stack
"abcdefd" "d"
"xyxzxe" "z"
"abcd" "z"
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2000: Reverse Prefix of Word
class Solution {
public String reversePrefix(String word, char ch) {
int j = word.indexOf(ch);
if (j == -1) {
return word;
}
char[] cs = word.toCharArray();
for (int i = 0; i < j; ++i, --j) {
char t = cs[i];
cs[i] = cs[j];
cs[j] = t;
}
return String.valueOf(cs);
}
}
// Accepted solution for LeetCode #2000: Reverse Prefix of Word
func reversePrefix(word string, ch byte) string {
j := strings.IndexByte(word, ch)
if j < 0 {
return word
}
s := []byte(word)
for i := 0; i < j; i++ {
s[i], s[j] = s[j], s[i]
j--
}
return string(s)
}
# Accepted solution for LeetCode #2000: Reverse Prefix of Word
class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
i = word.find(ch)
return word if i == -1 else word[i::-1] + word[i + 1 :]
// Accepted solution for LeetCode #2000: Reverse Prefix of Word
impl Solution {
pub fn reverse_prefix(word: String, ch: char) -> String {
match word.find(ch) {
Some(i) => word[..=i].chars().rev().collect::<String>() + &word[i + 1..],
None => word,
}
}
}
// Accepted solution for LeetCode #2000: Reverse Prefix of Word
function reversePrefix(word: string, ch: string): string {
const i = word.indexOf(ch) + 1;
if (!i) {
return word;
}
return [...word.slice(0, i)].reverse().join('') + word.slice(i);
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.
Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.
Usually fails on: Indices point to blocked elements and outputs shift.
Fix: Pop while invariant is violated before pushing current element.