Reverse String - Practice Coding Problems

Reverse String - Problem

The Challenge: You're given a string as an array of characters and need to reverse it in-place. This means you must modify the original array without creating a new one, using only O(1) extra memory space.

What you get: An array of characters like ['h','e','l','l','o']
What you return: The same array, but reversed: ['o','l','l','e','h']

This is a fundamental string manipulation problem that tests your understanding of array indexing and space-efficient algorithms. The constraint of in-place modification makes it more challenging than simply creating a new reversed string.

Input & Output

example_1.py — Basic String

$ Input: s = ['h','e','l','l','o']

› Output: ['o','l','l','e','h']

💡 Note: We reverse the string 'hello' by swapping characters from both ends moving inward. First swap h↔o, then e↔l, leaving the middle 'l' unchanged.

example_2.py — Even Length

$ Input: s = ['H','a','n','n','a','h']

› Output: ['h','a','n','n','a','H']

💡 Note: For even-length strings, we swap all pairs: H↔h, a↔a, n↔n. Each character finds its mirror position.

example_3.py — Single Character

$ Input: s = ['A']

› Output: ['A']

💡 Note: Edge case: A single character string remains unchanged since there are no other characters to swap with.

Constraints

1 ≤ s.length ≤ 10⁵
s[i] is a printable ASCII character
Must modify the input array in-place with O(1) extra memory

Visualization

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Understanding the Visualization

Place Pointers

Position left pointer at start (index 0) and right pointer at end (index n-1)

Swap & Move

Swap characters at pointer positions, then move left pointer right and right pointer left

Meet in Middle

Continue until pointers meet or cross - the string is now reversed!

Key Takeaway

🎯 Key Insight: By using two pointers that converge toward the center, we can reverse any string in-place with optimal space efficiency, requiring only constant extra memory regardless of input size.

Asked in

G Google 25 a Amazon 20 f Meta 15 ⊞ Microsoft 18

The optimal solution uses the two pointers technique to achieve O(1) space complexity. By placing pointers at both ends and swapping elements while moving inward, we reverse the string in-place with only n/2 swaps.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Extra Array)	O(n)	O(n)	Create a new array and copy elements in reverse order
Two Pointers (Optimal)	O(n)	O(1)	Use two pointers from both ends, swapping elements as they meet in the middle

Brute Force (Extra Array) — Algorithm Steps

Create a new array of the same size
Iterate through original array from end to beginning
Copy each element to the new array
Copy the new array back to original (or return new array)

Visualization

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Step-by-Step Walkthrough

Create temp array

Allocate new array of same size

Copy in reverse

Copy each element from end to beginning

Copy back

Replace original array with reversed copy

Code -

solution.c — C

void reverseString(char* s, int sSize) {
    // This violates O(1) space constraint
    char* temp = (char*)malloc(sSize * sizeof(char));
    
    // Copy elements in reverse order
    for (int i = 0; i < sSize; i++) {
        temp[i] = s[sSize - 1 - i];
    }
    
    // Copy back to original array
    for (int i = 0; i < sSize; i++) {
        s[i] = temp[i];
    }
    
    free(temp);
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We visit each element once to copy it

✓ Linear Growth

Space Complexity

O(n)

We create an additional array of size n

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Extra Array) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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