LeetCode #566 — EASY

Reshape the Matrix

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

The Problem

Problem Statement

In MATLAB, there is a handy function called reshape which can reshape an m x n matrix into a new one with a different size r x c keeping its original data.

You are given an m x n matrix mat and two integers r and c representing the number of rows and the number of columns of the wanted reshaped matrix.

The reshaped matrix should be filled with all the elements of the original matrix in the same row-traversing order as they were.

If the reshape operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.

Example 1:

Input: mat = [[1,2],[3,4]], r = 1, c = 4
Output: [[1,2,3,4]]

Example 2:

Input: mat = [[1,2],[3,4]], r = 2, c = 4
Output: [[1,2],[3,4]]

Constraints:

m == mat.length
n == mat[i].length
1 <= m, n <= 100
-1000 <= mat[i][j] <= 1000
1 <= r, c <= 300

Step 01

Brute Force Baseline

Problem summary: In MATLAB, there is a handy function called reshape which can reshape an m x n matrix into a new one with a different size r x c keeping its original data. You are given an m x n matrix mat and two integers r and c representing the number of rows and the number of columns of the wanted reshaped matrix. The reshaped matrix should be filled with all the elements of the original matrix in the same row-traversing order as they were. If the reshape operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.

Baseline thinking

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

Pattern signal: Array

Example 1

[[1,2],[3,4]]
1
4

Example 2

[[1,2],[3,4]]
2
4

Core Insight

What unlocks the optimal approach

Do you know how 2d matrix is stored in 1d memory? Try to map 2-dimensions into one.
M[i][j]=M[n*i+j] , where n is the number of cols. This is the one way of converting 2-d indices into one 1-d index. Now, how will you convert 1-d index into 2-d indices?
Try to use division and modulus to convert 1-d index into 2-d indices.
M[i] => M[i/n][i%n] Will it result in right mapping? Take some example and check this formula.

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 #566: Reshape the Matrix
class Solution {
    public int[][] matrixReshape(int[][] mat, int r, int c) {
        int m = mat.length, n = mat[0].length;
        if (m * n != r * c) {
            return mat;
        }
        int[][] ans = new int[r][c];
        for (int i = 0; i < m * n; ++i) {
            ans[i / c][i % c] = mat[i / n][i % n];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #566: Reshape the Matrix
func matrixReshape(mat [][]int, r int, c int) [][]int {
	m, n := len(mat), len(mat[0])
	if m*n != r*c {
		return mat
	}
	ans := make([][]int, r)
	for i := range ans {
		ans[i] = make([]int, c)
	}
	for i := 0; i < m*n; i++ {
		ans[i/c][i%c] = mat[i/n][i%n]
	}
	return ans
}

# Accepted solution for LeetCode #566: Reshape the Matrix
class Solution:
    def matrixReshape(self, mat: List[List[int]], r: int, c: int) -> List[List[int]]:
        m, n = len(mat), len(mat[0])
        if m * n != r * c:
            return mat
        ans = [[0] * c for _ in range(r)]
        for i in range(m * n):
            ans[i // c][i % c] = mat[i // n][i % n]
        return ans

// Accepted solution for LeetCode #566: Reshape the Matrix
impl Solution {
    pub fn matrix_reshape(mat: Vec<Vec<i32>>, r: i32, c: i32) -> Vec<Vec<i32>> {
        let r = r as usize;
        let c = c as usize;
        let m = mat.len();
        let n = mat[0].len();
        if m * n != r * c {
            return mat;
        }
        let mut i = 0;
        let mut j = 0;
        (0..r)
            .into_iter()
            .map(|_| {
                (0..c)
                    .into_iter()
                    .map(|_| {
                        let res = mat[i][j];
                        j += 1;
                        if j == n {
                            j = 0;
                            i += 1;
                        }
                        res
                    })
                    .collect()
            })
            .collect()
    }
}

// Accepted solution for LeetCode #566: Reshape the Matrix
function matrixReshape(mat: number[][], r: number, c: number): number[][] {
    let m = mat.length,
        n = mat[0].length;
    if (m * n != r * c) return mat;
    let ans = Array.from({ length: r }, v => new Array(c).fill(0));
    let k = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            ans[Math.floor(k / c)][k % c] = mat[i][j];
            ++k;
        }
    }
    return ans;
}

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.