Daily LeetCode - by Vítor Belim

Description

Given two n x n binary matrices mat and target, return true if it is possible to make mat equal to target by rotating mat in 90-degree increments, or false otherwise.

Example 1:

Input: mat = [[0,1],[1,0]], target = [[1,0],[0,1]]
Output: true
Explanation: We can rotate mat 90 degrees clockwise to make mat equal target.

Example 2:

Input: mat = [[0,1],[1,1]], target = [[1,0],[0,1]]
Output: false
Explanation: It is impossible to make mat equal to target by rotating mat.

Example 3:

Input: mat = [[0,0,0],[0,1,0],[1,1,1]], target = [[1,1,1],[0,1,0],[0,0,0]]
Output: true
Explanation: We can rotate mat 90 degrees clockwise two times to make mat equal target.

Constraints:

n == mat.length == target.length
n == mat[i].length == target[i].length
1 <= n <= 10
mat[i][j] and target[i][j] are either 0 or 1.

Solutions

This function determines whether a target matrix is a rotated version of the input matrix by checking all four possible rotations (0°, 90°, 180°, and 270°). It maintains four boolean flags—rot0, rot90, rot180, and rot270—and iterates through every position [r, c] in the matrix, comparing elements from the source matrix at rotation-adjusted coordinates to the corresponding position in the target. For each rotation, it uses the formula mat[adjusted_row][adjusted_col] === target[r][c] to verify if that rotation produces the target; for example, a 90° rotation reads from mat[c][n-1-r], while a 180° rotation reads from mat[n-1-r][n-1-c]. If all four rotations are ruled out at any point (all flags become false), the function returns early for efficiency. Finally, it returns true if any of the four rotations produced a match with the target matrix.

Language: javascript(2026-03-22 10:40)DONE

CPU Performance100.00%

Memory Performance80.88%

/**
 * @param {number[][]} mat
 * @param {number[][]} target
 * @return {boolean}
 */
var findRotation = function(mat, target) {
    const n = mat.length;

    let rot0 = true;
    let rot90 = true;
    let rot180 = true;
    let rot270 = true;

    for (let r = 0; r < n; r++) {
        for (let c = 0; c < n; c++) {
            const inv_r = n - 1 - r;
            const inv_c = n - 1 - c;

            rot0 &= mat[r][c] === target[r][c];
            rot90 &= mat[c][inv_r] === target[r][c];
            rot180 &= mat[inv_r][inv_c] === target[r][c];
            rot270 &= mat[inv_c][r] === target[r][c];

            if (!rot0 && !rot90 && !rot180 && !rot270) {
                return false;
            }
        }
    }

    return rot0 || rot90 || rot180 || rot270;
};

Description

Solutions

AI Explanation