1329 - Sort the Matrix Diagonally (Medium)

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

Approach 1: Priority Queue

Written by @wkw

class Solution {
public:
    vector<vector<int>> diagonalSort(vector<vector<int>>& mat) {
        int n = mat.size(), m = mat[0].size();
        // observation:
        // for each cell mat[i][j] in the same diagonal,
        // they will have the same value i - j

        // if the diagonal line is pointing to upper right direction,
        // then they will have the same value i + j (see https://leetcode.com/problems/diagonal-traverse/)

        // the idea is to use priority queue for each diagonal
        // as priority queue could sort it internally when a value is pushed / popped
        // we use greater<int> as we want the smallest go first
        map<int, priority_queue<int, vector<int>, greater<int>>> diag;
        // iterate each row
        for (int i = 0; i < n; i++) {
            // iterate each col
            for (int j = 0; j < m; j++) {
                // for each mat[i][j]
                // add it to corresponding priority queue
                diag[i - j].push(mat[i][j]);
            }
        }
        // iterate each row
        for (int i = 0; i < n; i++) {
            // iterate each col
            for (int j = 0; j < m; j++) {
                // diag[i - j].top() would return the smallest number in the current queue
                // we can just perform in-place replacement here
                mat[i][j] = diag[i - j].top();
                // since we don't need this value anymore, pop it out
                // so that next cell would get the smallest value
                diag[i - j].pop();
            }
        }
        return mat;
    }
};