3446. Sort Matrix by Diagonals.cpp

[PrashamMehta-04]

You are given an n x n square matrix of integers grid. Return the matrix such that: -> The diagonals in the bottom-left triangle (including the middle diagonal) are sorted in non-increasing order. -> The diagonals in the top-right triangle are sorted in non-decreasing order.

Intuition

The problem requires sorting matrix diagonals differently based on their position:

Bottom-left triangle (including the main diagonal) → sort in non-increasing order
Top-right triangle → sort in non-decreasing order

Each diagonal can be uniquely identified by its starting position: either from the first column (bottom-left) or from the first row (top-right). Once we know the starting point of a diagonal, we can extract all its elements, sort them according to the required order, and place them back.

By processing one diagonal at a time, we ensure that sorting is localized and does not interfere with other diagonals. This approach leverages the small constraint on 𝑛 ≤ 10 to keep sorting simple and efficient.

Approach

Identify diagonals:
Bottom-left diagonals start from the first column (rows from n-1 to 0).
Top-right diagonals start from the first row (columns from 1 to n-1).
Extract elements of each diagonal into a temporary array.
Sort the diagonal:
Bottom-left → descending order (non-increasing)
Top-right → ascending order (non-decreasing)
Place elements back into the matrix along the same diagonal path.
Repeat for all diagonals until the entire matrix is updated.
Return the updated matrix.

Code Solution (C++)

class Solution {
public:
    vector<vector<int>> sortMatrix(vector<vector<int>>& grid) {
        vector<vector<int>> ans;
        vector<int> t;
        int cnt = 1;
        int j=0;
        for(int i=grid.size()-1;i>=0;i--){
            int p=i;
            while(j < grid.size() && p < grid.size()){
                t.push_back(grid[p][j]);
                p++;
                j++;
            }
            sort(t.begin(), t.end());
            reverse(t.begin(), t.end());
            ans.push_back(t);
            t.clear();
            j=0;
        }
        vector<vector<int>> a1;
        int r = 1;
        for(int k=1;k<grid.size();k++){
            int q = 0;
            int d = r;
            while(q < grid.size() && d < grid.size()){
                t.push_back(grid[q][d]);
                q++;
                d++;
            }
            sort(t.begin(), t.end());
            a1.push_back(t);
            t.clear();
            r++;
        }
        for(int i=0;i<a1.size();i++){
            ans.push_back(a1[i]);
        }
        j = grid.size()-1;
        int p = 0,q=0;
        for(int i=grid.size()-1;i>=0;i--){
            int r = i;
            while(q < ans[p].size()){
                grid[r][j] = ans[p][q];
                q++;
                r++;
                j++;
            } 
            j=0;
            q=0;
            p++;
        }
        q=0;
        j=1;
        r=0;
        for(int i=1;i<grid.size();i++){
            while(q < ans[p].size()){
                grid[r][j] = ans[p][q];
                q++;
                r++;
                j++;
            }
            j=i+1;
            r=0;
            q=0;
            p++;
        }
        return grid;
    }
};

By submitting this PR, I confirm that:

• This is my original work not totally AI genearted
• I have tested the solution thoroughly on leetcode
• I have maintained proper PR description format
• This is a meaningful contribution, not spam

[github-actions]

Thanks for raising the PR, the owner will be review it soon' keep patience, keep contributing>>>!!! make sure you have star ⭐ the repo

[SjxSubham]

@PrashamMehta-04 Star the repo as well...