Rotate Image

Rotate Image - Problem

You are given an n x n 2D matrix representing an image. Rotate the image by 90 degrees clockwise.

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Input & Output

Example 1 — 3×3 Matrix

$ Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]

› Output: [[7,4,1],[8,5,2],[9,6,3]]

💡 Note: After 90° clockwise rotation: first row [1,2,3] becomes last column [1,2,3], second row [4,5,6] becomes middle column [4,5,6], third row [7,8,9] becomes first column [7,8,9]

Example 2 — 4×4 Matrix

$ Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]

› Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]

💡 Note: Each element moves to its rotated position: element at (0,0) which is 5 moves to (0,3), element at (0,1) which is 1 moves to (1,3), etc.

Example 3 — 1×1 Matrix

$ Input: matrix = [[1]]

› Output: [[1]]

💡 Note: Single element matrix remains unchanged after rotation

Constraints

1 ≤ matrix.length ≤ 20
matrix.length == matrix[i].length
-1000 ≤ matrix[i][j] ≤ 1000

Visualization

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Asked in

🍎 Apple 25 Ⓜ Microsoft 20 a Amazon 18 G Google 15

The key insight is that 90° clockwise rotation can be achieved by transpose + reverse rows, or by rotating 4-element cycles in concentric layers. The optimal approach uses layer-by-layer rotation with O(n²) time and O(1) space.

Common Approaches

✓ Element by Element with Temp Array

⏱️ Time: O(n²) Space: O(n²)

Create a temporary matrix and copy each element from position (i,j) to its new rotated position (j, n-1-i). Then copy back to original matrix.

Layer by Layer Rotation

⏱️ Time: O(n²) Space: O(1)

Process the matrix in concentric square layers from outside to inside. For each layer, rotate groups of 4 elements cyclically using a single temporary variable.

Transpose + Reverse Rows

⏱️ Time: O(n²) Space: O(1)

A 90-degree clockwise rotation can be achieved by combining two simpler operations: transpose (flip across main diagonal) and then reverse each row horizontally.

Element by Element with Temp Array — Algorithm Steps

Create temporary n×n matrix
Copy each element to rotated position
Copy temp matrix back to original

Visualization

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Step-by-Step Walkthrough

Original Matrix

3x3 matrix with elements 1-9

Copy to Temp

Each (i,j) goes to (j,n-1-i)

Copy Back

Temp matrix copied back to original

Code -

solution.c — C

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

static int matrix[1000][1000];
static int temp[1000][1000];

void solution(int n) {
    // Copy to rotated positions
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            temp[j][n-1-i] = matrix[i][j];
        }
    }
    
    // Copy back to original
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            matrix[i][j] = temp[i][j];
        }
    }
}

int main() {
    char input[100000];
    fgets(input, sizeof(input), stdin);
    
    // Parse 2D array [[a,b],[c,d]]
    char *p = input;
    int n = 0;
    
    // Count rows by counting '[' after initial '['
    int bracket_count = 0;
    for (char *q = input; *q; q++) {
        if (*q == '[') {
            bracket_count++;
            if (bracket_count > 1) n++;
        }
    }
    
    // Reset and parse values
    p = input;
    while (*p && *p != '[') p++; // Find first '['
    p++; // Skip it
    
    for (int i = 0; i < n; i++) {
        while (*p && *p != '[') p++; // Find row start '['
        p++; // Skip '['
        
        for (int j = 0; j < n; j++) {
            while (*p && !isdigit(*p) && *p != '-') p++; // Find number
            
            char *endptr;
            matrix[i][j] = strtol(p, &endptr, 10);
            p = endptr;
            
            while (*p && *p != ',' && *p != ']') p++; // Skip to next delimiter
            if (*p == ',') p++; // Skip comma
        }
        
        while (*p && *p != ']') p++; // Skip to row end ']'
        p++; // Skip ']'
    }
    
    solution(n);
    
    // Print result
    printf("[");
    for (int i = 0; i < n; i++) {
        printf("[");
        for (int j = 0; j < n; j++) {
            printf("%d", matrix[i][j]);
            if (j < n-1) printf(",");
        }
        printf("]");
        if (i < n-1) printf(",");
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Visit each element once in n×n matrix

⚠ Quadratic Growth

Space Complexity

O(n²)

Temporary matrix stores all n×n elements

✓ Linear Space

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Ln 1, Col 1

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Input & Output

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Common Approaches

Element by Element with Temp Array — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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