netunified

🔄 Rotate Array In-Place in C — A Minimalist, Cycle-Aware Solution

Author: @netunified
Language: C
Topic: In-Place Array Rotation
Difficulty: Medium
Tags: Arrays, In-Place Algorithms, Cycles, C Programming

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🧩 Problem

Rotate an array nums of size n to the right by k steps.
You must do it in-place with O(1) extra space.

Example

Input:  nums = [1, 2, 3, 4, 5, 6], k = 2  
Output: [5, 6, 1, 2, 3, 4]

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🧠 Solution Overview

This solution, authored by @netunified, implements an elegant twist on the classical cyclic replacement algorithm.

Instead of using gcd(numsSize, k) to detect cycle count, or tracking visited elements, this method relies on smart pointer manipulation and implicit loop detection using ptr_offset.

All elements are visited exactly once, and swapped correctly — with just a few variables and a single loop.

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✅ C Code

#include <stdio.h>

// Solution by netunified (LeetCode)

int rotate(int* nums, int numsSize, int k) {
 int prev = 0, curr = 0, ptr_offset = 0, ptr = 0; 

  for (int i = 0; i < numsSize; i++) {

    /* if ptr arrives at the offset (loop detected) 
     * increment the offset by 1 and move the ptr 
     * k steps to the right. else continue swapping
     * as usual.
     */ 
    if (ptr_offset == ptr) {
        ptr_offset = (ptr + 1) % numsSize;
        ptr = (ptr_offset + k) % numsSize;
        curr = nums[ptr_offset];
        prev = nums[ptr];
        nums[ptr] = curr;
    } else {
        ptr = (ptr + k) % numsSize;
        curr = nums[ptr];
        nums[ptr] = prev;
        prev = curr;
    }
  }

  return 0;
}

int main() {
  int k = 2;
  int arr[] = {1, 2, 3, 4, 5, 6};
  int arr_len = sizeof(arr) / sizeof(arr[0]);
  int ret = rotate(arr, arr_len, k);


  for (int i = 0; i < arr_len; i++)
    printf(
        (i == 0) ? "[%d," : (i < arr_len - 1) ? "%d," : "%d]\n",
        arr[i]
    );

  return ret;
}

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🔍 How It Works

1. ptr represents the current index we’re writing to.
2. ptr_offset tracks the start of a new cycle.
3. When ptr == ptr_offset, it means we’ve completed a cycle. We bump ptr_offset and start another.
4. prev carries the displaced value forward to be written into the next location.

This smart handling implicitly tracks multiple disjoint cycles that arise when gcd(numsSize, k) > 1.

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🔄 Visual Walkthrough

For nums = [1, 2, 3, 4, 5, 6], k = 2, this is the rotation sequence:

Original: [1, 2, 3, 4, 5, 6]
Step 1 →  [1, 2, 3, 2, 5, 6]
Step 2 →  [1, 2, 3, 2, 5, 4]
Step 3 →  [1, 6, 3, 2, 5, 4]
Step 4 →  [1, 6, 3, 2, 3, 4]
Step 5 →  [5, 6, 3, 2, 3, 4]
Step 6 →  [5, 6, 1, 2, 3, 4]

Final output: [5, 6, 1, 2, 3, 4]

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🆚 Comparison to Classical Solution

Feature	Classical GCD-Based	This Solution (@netunified)
Uses GCD?	✅ Yes	❌ No
Tracks visited?	✅ Yes	❌ No
Uses counter?	✅ Yes (count)	❌ No
Single loop?	❌ No	✅ Yes
Elegant + minimal?	⚠️ Verbose	✅ Very clean

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💬 Final Thoughts

This solution is a concise, in-place, and loop-aware approach that avoids the usual boilerplate and complexity of other cyclic replacements.

It’s a perfect example of solving a problem by understanding the underlying structure (cyclic shifts) and trusting smart state tracking (ptr_offset) instead of brute-force counters.

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🏆 Credit

✨ Written by @netunified
🧠 Original idea and implementation
📍 No GCD, no visited flags, no recursion — just clever logic.

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📎 Want to Share?

Feel free to copy, clone, or fork this blog post — just leave credit to @netunified where it’s due!

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ChatGPT wrote this blog post (not the solution) for me :D!