Rotate Array - Practice Coding Problems

Rotate Array - Problem

Given an integer array nums, rotate the array to the right by k steps, where k is non-negative.

Example: If nums = [1,2,3,4,5,6,7] and k = 3, after rotation the array becomes [5,6,7,1,2,3,4].

Note: Try to solve this problem in-place with O(1) extra memory.

Input & Output

Example 1 — Basic Rotation

$ Input: nums = [1,2,3,4,5,6,7], k = 3

› Output: [5,6,7,1,2,3,4]

💡 Note: Rotate right by 3: the last 3 elements [5,6,7] move to the front, and [1,2,3,4] shift to the back

Example 2 — Small Array

$ Input: nums = [-1,-100,3,99], k = 2

› Output: [3,99,-1,-100]

💡 Note: Rotate right by 2: [3,99] moves to front, [-1,-100] moves to back

Example 3 — No Change

$ Input: nums = [1,2], k = 2

› Output: [1,2]

💡 Note: k equals array length, so after full rotation the array remains unchanged

Constraints

1 ≤ nums.length ≤ 10⁵
-2³¹ ≤ nums[i] ≤ 2³¹ - 1
0 ≤ k ≤ 10⁵

Visualization

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Understanding the Visualization

Input

Original array [1,2,3,4,5,6,7] and rotation steps k=3

Process

Elements move k positions to the right, with wraparound

Output

Rotated array [5,6,7,1,2,3,4]

Key Takeaway

🎯 Key Insight: Three array reversals can achieve any rotation in O(1) space

Asked in

M Microsoft 45 A Amazon 38 G Google 32 🍎 Apple 28

The key insight is that rotating an array can be achieved through three strategic reversals. The optimal approach uses the three-step reversal method: reverse the entire array, then reverse the first k elements, then reverse the remaining elements. Best time complexity: O(n), space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Extra Array Approach	O(n)	O(n)	Use extra space to place elements directly in their final positions
Brute Force - Move One by One	O(n × k)	O(1)	Move each element one position at a time, k times
Three-Step Reversal (Optimal)	O(n)	O(1)	Reverse entire array, then reverse first k and last n-k elements

Extra Array Approach — Algorithm Steps

Step 1: Create a new array of same size
Step 2: For each element, calculate its new position using (i + k) % n
Step 3: Place element in new position
Step 4: Copy result back to original array

Visualization

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Step-by-Step Walkthrough

Calculate Positions

Each element at index i goes to (i + k) % n

Place Elements

Copy elements to their calculated positions

Result

New array has all elements in rotated positions

Code -

solution.c — C

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

int* solution(int* nums, int numsSize, int k) {
    k = k % numsSize;  // Handle k > numsSize
    
    // Create new array
    int* rotated = (int*)malloc(numsSize * sizeof(int));
    
    // Place each element at its new position
    for (int i = 0; i < numsSize; i++) {
        int newPos = (i + k) % numsSize;
        rotated[newPos] = nums[i];
    }
    
    // Copy back to original array
    for (int i = 0; i < numsSize; i++) {
        nums[i] = rotated[i];
    }
    
    free(rotated);
    return nums;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int k;
    scanf("%d", &k);
    
    int* result = solution(nums, numsSize, k);
    printf("[");
    for (int i = 0; i < numsSize; i++) {
        if (i > 0) printf(",");
        printf("%d", result[i]);
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass to copy each element to its new position

✓ Linear Growth

Space Complexity

O(n)

Extra array to store the rotated elements

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Extra Array Approach — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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