LeetCode #1329 — MEDIUM

Sort the Matrix Diagonally

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

A matrix diagonal is a diagonal line of cells starting from some cell in either the topmost row or leftmost column and going in the bottom-right direction until reaching the matrix's end. For example, the matrix diagonal starting from mat[2][0], where mat is a 6 x 3 matrix, includes cells mat[2][0], mat[3][1], and mat[4][2].

Given an m x n matrix mat of integers, sort each matrix diagonal in ascending order and return the resulting matrix.

Example 1:

Input: mat = [[3,3,1,1],[2,2,1,2],[1,1,1,2]]
Output: [[1,1,1,1],[1,2,2,2],[1,2,3,3]]

Example 2:

Input: mat = [[11,25,66,1,69,7],[23,55,17,45,15,52],[75,31,36,44,58,8],[22,27,33,25,68,4],[84,28,14,11,5,50]]
Output: [[5,17,4,1,52,7],[11,11,25,45,8,69],[14,23,25,44,58,15],[22,27,31,36,50,66],[84,28,75,33,55,68]]

Constraints:

m == mat.length
n == mat[i].length
1 <= m, n <= 100
1 <= mat[i][j] <= 100

Step 01

Brute Force Baseline

Problem summary: A matrix diagonal is a diagonal line of cells starting from some cell in either the topmost row or leftmost column and going in the bottom-right direction until reaching the matrix's end. For example, the matrix diagonal starting from mat[2][0], where mat is a 6 x 3 matrix, includes cells mat[2][0], mat[3][1], and mat[4][2]. Given an m x n matrix mat of integers, sort each matrix diagonal in ascending order and return the resulting matrix.

Baseline thinking

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

Pattern signal: Array

Example 1

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

Example 2

[[11,25,66,1,69,7],[23,55,17,45,15,52],[75,31,36,44,58,8],[22,27,33,25,68,4],[84,28,14,11,5,50]]

Core Insight

What unlocks the optimal approach

Use a data structure to store all values of each diagonal.
How to index the data structure with the id of the diagonal?
All cells in the same diagonal (i,j) have the same difference so we can get the diagonal of a cell using the difference i-j.

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 #1329: Sort the Matrix Diagonally
class Solution {
    public int[][] diagonalSort(int[][] mat) {
        int m = mat.length, n = mat[0].length;
        List<Integer>[] g = new List[m + n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                g[m - i + j].add(mat[i][j]);
            }
        }
        for (var e : g) {
            Collections.sort(e, (a, b) -> b - a);
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                mat[i][j] = g[m - i + j].removeLast();
            }
        }
        return mat;
    }
}

// Accepted solution for LeetCode #1329: Sort the Matrix Diagonally
func diagonalSort(mat [][]int) [][]int {
	m, n := len(mat), len(mat[0])
	g := make([][]int, m+n)
	for i, row := range mat {
		for j, x := range row {
			g[m-i+j] = append(g[m-i+j], x)
		}
	}
	for _, e := range g {
		sort.Sort(sort.Reverse(sort.IntSlice(e)))
	}
	for i, row := range mat {
		for j := range row {
			k := len(g[m-i+j])
			mat[i][j] = g[m-i+j][k-1]
			g[m-i+j] = g[m-i+j][:k-1]
		}
	}
	return mat
}

# Accepted solution for LeetCode #1329: Sort the Matrix Diagonally
class Solution:
    def diagonalSort(self, mat: List[List[int]]) -> List[List[int]]:
        m, n = len(mat), len(mat[0])
        g = [[] for _ in range(m + n)]
        for i, row in enumerate(mat):
            for j, x in enumerate(row):
                g[m - i + j].append(x)
        for e in g:
            e.sort(reverse=True)
        for i in range(m):
            for j in range(n):
                mat[i][j] = g[m - i + j].pop()
        return mat

// Accepted solution for LeetCode #1329: Sort the Matrix Diagonally
impl Solution {
    pub fn diagonal_sort(mut mat: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        let m = mat.len();
        let n = mat[0].len();
        let mut g: Vec<Vec<i32>> = vec![vec![]; m + n];
        for i in 0..m {
            for j in 0..n {
                g[m - i + j].push(mat[i][j]);
            }
        }
        for e in &mut g {
            e.sort_by(|a, b| b.cmp(a));
        }
        for i in 0..m {
            for j in 0..n {
                mat[i][j] = g[m - i + j].pop().unwrap();
            }
        }
        mat
    }
}

// Accepted solution for LeetCode #1329: Sort the Matrix Diagonally
function diagonalSort(mat: number[][]): number[][] {
    const [m, n] = [mat.length, mat[0].length];
    const g: number[][] = Array.from({ length: m + n }, () => []);
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            g[m - i + j].push(mat[i][j]);
        }
    }
    for (const e of g) {
        e.sort((a, b) => b - a);
    }
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            mat[i][j] = g[m - i + j].pop()!;
        }
    }
    return mat;
}

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(m × n × log \min(m, n)

Space

O(m × n)

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.