Reverse Prefix of Word

Reverse Prefix of Word - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: word = "abcdefd", ch = "d"

› Output: "dcbaefd"

💡 Note: The first 'd' appears at index 3. Reverse segment [0..3]: 'abcd' → 'dcba', then append 'efd' to get 'dcbaefd'.

Example 2 — Character Not Found

$ Input: word = "xyxzxe", ch = "z"

› Output: "zxyxze"

💡 Note: First 'z' at index 3. Reverse prefix [0..3]: 'xyxz' → 'zxyx', then append 'xe' to get 'zxyxze'.

Example 3 — Character at Start

$ Input: word = "abcd", ch = "a"

› Output: "abcd"

💡 Note: Target 'a' is at index 0. Reversing segment [0..0] gives 'a', so result remains 'abcd'.

Constraints

1 ≤ word.length ≤ 250
word consists of lowercase English letters
ch is a lowercase English letter

Visualization

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Asked in

G Google 12 a Amazon 8 m Microsoft 6

The key insight is to find the first occurrence of the target character and reverse only the prefix up to that point. The optimal approach uses built-in string functions for O(n) time complexity. Time: O(n), Space: O(n).

Common Approaches

✓ Brute Force String Building

⏱️ Time: O(n²) Space: O(n)

First scan the string to find the index of the target character. Then build the result string by concatenating characters from the prefix in reverse order, followed by the remaining suffix.

Built-in Find and Slice

⏱️ Time: O(n) Space: O(n)

Leverage language built-in functions for finding character index and string slicing. Reverse the prefix using built-in reverse functions and concatenate with the suffix.

Two Pointers In-Place Swap

⏱️ Time: O(n) Space: O(n)

First find the target character index, then convert string to character array for in-place modification. Use two pointers starting from both ends of the prefix and swap characters while moving toward center.

Brute Force String Building — Algorithm Steps

Find index of first occurrence of ch
Build reversed prefix by concatenating chars backwards
Append remaining suffix to result

Visualization

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Step-by-Step Walkthrough

Find Target

Scan string to find first occurrence of 'd' at index 3

Reverse Prefix

Concatenate chars from index 3 to 0: 'd'+'c'+'b'+'a'

Append Suffix

Add remaining characters: 'efd'

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* solution(char* word, char* ch) {
    int len = strlen(word);
    char* result = (char*)malloc((len + 1) * sizeof(char));
    
    // Find first occurrence of ch
    int targetIndex = -1;
    char target = ch[0];
    for (int i = 0; i < len; i++) {
        if (word[i] == target) {
            targetIndex = i;
            break;
        }
    }
    
    // If ch not found, return original word
    if (targetIndex == -1) {
        strcpy(result, word);
        return result;
    }
    
    // Build result by reversing prefix
    int pos = 0;
    for (int i = targetIndex; i >= 0; i--) {
        result[pos++] = word[i];
    }
    
    // Append remaining suffix
    for (int i = targetIndex + 1; i < len; i++) {
        result[pos++] = word[i];
    }
    
    result[pos] = '\0';
    return result;
}

int main() {
    char word[1001], ch[2];
    fgets(word, sizeof(word), stdin);
    word[strcspn(word, "\n")] = 0;
    fgets(ch, sizeof(ch), stdin);
    ch[strcspn(ch, "\n")] = 0;
    
    char* result = solution(word, ch);
    printf("%s\n", result);
    free(result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Finding character takes O(n), then string concatenation in loop takes O(n²) due to immutable strings

✓ Linear Growth

Space Complexity

O(n)

Result string and temporary strings during concatenation

⚡ Linearithmic Space

35.6K Views

Medium Frequency

~10 min Avg. Time

890 Likes

Ln 1, Col 1

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force String Building — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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