LeetCode #2946 — EASY

Matrix Similarity After Cyclic Shifts

Build confidence with an intuition-first walkthrough focused on array fundamentals.

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The Problem

Problem Statement

You are given an m x n integer matrix mat and an integer k. The matrix rows are 0-indexed.

The following proccess happens k times:

Even-indexed rows (0, 2, 4, ...) are cyclically shifted to the left.

Odd-indexed rows (1, 3, 5, ...) are cyclically shifted to the right.

Return true if the final modified matrix after k steps is identical to the original matrix, and false otherwise.

Example 1:

Input: mat = [[1,2,3],[4,5,6],[7,8,9]], k = 4

Output: false

Explanation:

In each step left shift is applied to rows 0 and 2 (even indices), and right shift to row 1 (odd index).

Example 2:

Input: mat = [[1,2,1,2],[5,5,5,5],[6,3,6,3]], k = 2

Output: true

Explanation:

Example 3:

Input: mat = [[2,2],[2,2]], k = 3

Output: true

Explanation:

As all the values are equal in the matrix, even after performing cyclic shifts the matrix will remain the same.

Constraints:

1 <= mat.length <= 25
1 <= mat[i].length <= 25
1 <= mat[i][j] <= 25
1 <= k <= 50

Step 01

Brute Force Baseline

Problem summary: You are given an m x n integer matrix mat and an integer k. The matrix rows are 0-indexed. The following proccess happens k times: Even-indexed rows (0, 2, 4, ...) are cyclically shifted to the left. Odd-indexed rows (1, 3, 5, ...) are cyclically shifted to the right. Return true if the final modified matrix after k steps is identical to the original matrix, and false otherwise.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[[1,2,3],[4,5,6],[7,8,9]]
4

Example 2

[[1,2,1,2],[5,5,5,5],[6,3,6,3]]
2

Example 3

[[2,2],[2,2]]
3

Step 02

Core Insight

What unlocks the optimal approach

You can reduce <code>k</code> shifts to <code>(k % n)</code> shifts as after <code>n</code> shifts the matrix will become similar to the initial matrix.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2946: Matrix Similarity After Cyclic Shifts
class Solution {
    public boolean areSimilar(int[][] mat, int k) {
        int m = mat.length, n = mat[0].length;
        k %= n;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (i % 2 == 1 && mat[i][j] != mat[i][(j + k) % n]) {
                    return false;
                }
                if (i % 2 == 0 && mat[i][j] != mat[i][(j - k + n) % n]) {
                    return false;
                }
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #2946: Matrix Similarity After Cyclic Shifts
func areSimilar(mat [][]int, k int) bool {
	n := len(mat[0])
	k %= n
	for i, row := range mat {
		for j, x := range row {
			if i%2 == 1 && x != mat[i][(j+k)%n] {
				return false
			}
			if i%2 == 0 && x != mat[i][(j-k+n)%n] {
				return false
			}
		}
	}
	return true
}

# Accepted solution for LeetCode #2946: Matrix Similarity After Cyclic Shifts
class Solution:
    def areSimilar(self, mat: List[List[int]], k: int) -> bool:
        n = len(mat[0])
        for i, row in enumerate(mat):
            for j, x in enumerate(row):
                if i % 2 == 1 and x != mat[i][(j + k) % n]:
                    return False
                if i % 2 == 0 and x != mat[i][(j - k + n) % n]:
                    return False
        return True

// Accepted solution for LeetCode #2946: Matrix Similarity After Cyclic Shifts
fn are_similar(mat: Vec<Vec<i32>>, k: i32) -> bool {
    let rows = mat.len();
    let cols = mat[0].len();
    let mut m = vec![vec![0; cols]; rows];

    for i in 0..rows {
        for j in 0..cols {
            let col = (j + k as usize) % cols;
            m[i][col] = mat[i][j];
        }
    }

    mat == m
}

fn main() {
    let mat = vec![vec![1, 2, 1, 2], vec![5, 5, 5, 5], vec![6, 3, 6, 3]];
    let ret = are_similar(mat, 2);
    println!("ret={ret}");
}

#[test]
fn test() {
    {
        let mat = vec![vec![1, 2, 1, 2], vec![5, 5, 5, 5], vec![6, 3, 6, 3]];
        let ret = are_similar(mat, 2);
        assert!(ret);
    }
    {
        let mat = vec![vec![2, 2], vec![2, 2]];
        let ret = are_similar(mat, 3);
        assert!(ret);
    }
    {
        let mat = vec![vec![1, 2]];
        let ret = are_similar(mat, 1);
        assert!(!ret);
    }
}

// Accepted solution for LeetCode #2946: Matrix Similarity After Cyclic Shifts
function areSimilar(mat: number[][], k: number): boolean {
    const m = mat.length;
    const n = mat[0].length;
    k %= n;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (i % 2 === 1 && mat[i][j] !== mat[i][(j + k) % n]) {
                return false;
            }
            if (i % 2 === 0 && mat[i][j] !== mat[i][(j - k + n) % n]) {
                return false;
            }
        }
    }
    return true;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.