Rotate Array

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

Tap to expand

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

✓ Brute Force - Move One by One

⏱️ Time: O(n × k) Space: O(1)

For each of the k steps, move every element one position to the right. The last element wraps around to the beginning.

Extra Array Approach

⏱️ Time: O(n) Space: O(n)

Create a new array and calculate where each element should go after rotation. Each element at index i moves to index (i + k) % n.

Three-Step Reversal (Optimal)

⏱️ Time: O(n) Space: O(1)

The key insight is that rotation can be achieved by three reversals: reverse the entire array, then reverse the first k elements, then reverse the remaining elements. This works in-place with O(1) space.

Brute Force - Move One by One — Algorithm Steps

Step 1: For each rotation step (k times)
Step 2: Save the last element
Step 3: Shift all elements one position right
Step 4: Place saved element at the beginning

Visualization

Tap to expand

Step-by-Step Walkthrough

Initial Array

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

After 1 Rotation

Last element moves to front [7,1,2,3,4,5,6]

After k Rotations

Final result after k=3 steps [5,6,7,1,2,3,4]

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
    
    for (int step = 0; step < k; step++) {
        // Save the last element
        int last = nums[numsSize - 1];
        // Shift all elements one position right
        for (int i = numsSize - 1; i > 0; i--) {
            nums[i] = nums[i - 1];
        }
        // Place saved element at beginning
        nums[0] = last;
    }
    
    return nums;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array using strtol
    int nums[1000];
    int numsSize = 0;
    char* p = line;
    
    // Skip to the opening bracket
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    // Parse numbers using strtol
    while (*p && *p != ']') {
        // Skip whitespace and commas
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        
        // Parse number
        nums[numsSize++] = (int)strtol(p, &p, 10);
    }
    
    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 × k)

k rotation steps, each step shifts n elements

✓ Linear Growth

Space Complexity

O(1)

Only one variable to store the last element

✓ Linear Space

78.0K Views

Very High Frequency

~15 min Avg. Time

2.2K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Move One by One — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler