Reverse Prefix of Word - Practice Coding Problems

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: "zyxxzxe"

💡 Note: First 'z' at index 2. Reverse prefix [0..2]: 'xyx' → 'yxx', then prepend 'z': 'zyxx', append 'xe' to get 'zyxxzxe'.

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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Understanding the Visualization

Input

String 'abcdefd' and target character 'd'

Find & Reverse

Locate first 'd' at index 3, reverse prefix 'abcd' → 'dcba'

Output

Concatenate reversed prefix with suffix: 'dcbaefd'

Key Takeaway

🎯 Key Insight: Only reverse the prefix up to the first occurrence of the target character, leaving the rest unchanged.

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

Approach	Time	Space	Notes
✓ Brute Force String Building	O(n²)	O(n)	Find the target character, then build result by concatenating characters in reverse order
Two Pointers In-Place Swap	O(n)	O(n)	Convert to char array and use two pointers to swap characters from both ends toward center
Built-in Find and Slice	O(n)	O(n)	Use built-in string functions to find character and slice/reverse efficiently

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

⚠ Quadratic 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

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force String Building — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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