Reverse Words in a String II

Reverse Words in a String II - Problem

Given a character array s, reverse the order of the words.

A word is defined as a sequence of non-space characters. The words in s will be separated by a single space.

Your code must solve the problem in-place, i.e. without allocating extra space.

Input & Output

Example 1 — Basic Case

$ Input: s = ["t","h","e"," ","s","k","y"," ","i","s"," ","b","l","u","e"]

› Output: ["b","l","u","e"," ","i","s"," ","s","k","y"," ","t","h","e"]

💡 Note: The words are "the", "sky", "is", "blue". Reversing their order gives "blue", "is", "sky", "the".

Example 2 — Single Word

$ Input: s = ["a"]

› Output: ["a"]

💡 Note: Only one word, so it remains the same after reversal.

Example 3 — Two Words

$ Input: s = ["h","i"," ","b","y","e"]

› Output: ["b","y","e"," ","h","i"]

💡 Note: Two words: "hi" and "bye". After reversal: "bye hi".

Constraints

1 ≤ s.length ≤ 10⁴
s[i] is either a lowercase English letter or a space ' '
There is at least one word in s
s does not contain leading or trailing spaces
All words are separated by exactly one space

Visualization

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Asked in

M Microsoft 35 a Amazon 28 G Google 22 f Facebook 18

The key insight is using a two-step reversal: first reverse the entire array, then reverse each individual word. This achieves in-place word reversal with O(1) space. Best approach is the two-pointer technique. Time: O(n), Space: O(1)

Common Approaches

✓ Extra Space Approach

⏱️ Time: O(n) Space: O(n)

Convert array to string, split by spaces to get words, reverse the word list, join back to string, and convert back to character array. This violates the in-place constraint but shows the basic logic.

Two-Step Reversal (Optimal)

⏱️ Time: O(n) Space: O(1)

First reverse the entire character array, which puts words in correct order but each word is backwards. Then iterate through and reverse each individual word to restore correct spelling.

Extra Space Approach — Algorithm Steps

Convert char array to string
Split string into words by spaces
Reverse the order of words
Join words back with spaces
Convert back to character array

Visualization

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Step-by-Step Walkthrough

Split

Convert to string and split by spaces

Reverse

Reverse the word array

Join

Join words back with spaces

Code -

solution.c — C

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

void solution(char** s, int size) {
    // Convert char array to string
    char* str = (char*)malloc(size + 1);
    str[0] = '\0';
    for (int i = 0; i < size; i++) {
        strcat(str, s[i]);
    }
    
    // Split into words
    char* words[100];
    int wordCount = 0;
    char* token = strtok(str, " ");
    while (token != NULL) {
        words[wordCount] = (char*)malloc(strlen(token) + 1);
        strcpy(words[wordCount], token);
        wordCount++;
        token = strtok(NULL, " ");
    }
    
    // Reverse the order of words
    for (int i = 0; i < wordCount / 2; i++) {
        char* temp = words[i];
        words[i] = words[wordCount - 1 - i];
        words[wordCount - 1 - i] = temp;
    }
    
    // Join back
    strcpy(str, "");
    for (int i = 0; i < wordCount; i++) {
        if (i > 0) strcat(str, " ");
        strcat(str, words[i]);
    }
    
    // Update original array in-place
    for (int i = 0; i < (int)strlen(str); i++) {
        s[i][0] = str[i];
        s[i][1] = '\0';
    }
    
    free(str);
    for (int i = 0; i < wordCount; i++) {
        free(words[i]);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse JSON array of strings
    char** s = (char**)malloc(1000 * sizeof(char*));
    int size = 0;
    
    char* p = line;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',' || *p == '\n' || *p == '\t') p++;
        if (*p == ']' || *p == '\0') break;
        
        if (*p == '"') {
            p++; // skip opening quote
            s[size] = (char*)malloc(2);
            s[size][0] = *p;
            s[size][1] = '\0';
            size++;
            p++; // skip the character
            if (*p == '"') p++; // skip closing quote
        }
    }
    
    solution(s, size);
    
    // Print result
    printf("[");
    for (int i = 0; i < size; i++) {
        if (i > 0) printf(",");
        printf("\"%s\"", s[i]);
    }
    printf("]\n");
    
    for (int i = 0; i < size; i++) {
        free(s[i]);
    }
    free(s);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass to convert array to string, split, reverse, and join

✓ Linear Growth

Space Complexity

O(n)

Creates new strings and word arrays during processing

✓ Linear Space

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Input & Output

Constraints

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Common Approaches

Extra Space Approach — Algorithm Steps

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Code -

Time & Space Complexity

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