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Sort a 2D vector diagonally
  • Difficulty Level : Medium
  • Last Updated : 18 May, 2021

Given a 2D vector of NxM integers. The task is to sort the elements of the vectors diagonally from top-left to bottom-right in decreasing order.
Examples: 

Input: arr[][] = { { 10, 2, 3 }, { 4, 5, 6 }, {7, 8, 9 } } 
Output: 
10 2 0 
4 9 0 
0 0 5
Input: arr[][] = { { 10, 2, 43 }, { 40, 5, 16 }, { 71, 8, 29 }, {1, 100, 5} } 
Output: 
29 2 0 
8 10 0 
71 5 5 
0 0 0 

Approach: 
Observations: 
 

The above images show the difference between the column index and row index at each cell. The cells having the same difference from top-left to bottom-down cell forms a diagonal. 
Below are the steps to sort diagonal in decreasing order: 
 



  1. Store the diagonal element with a positive difference in one Array of Vectors(say Pos[]) such that elements at the cell having difference(say a) is stored at index an of Pos[] array.
  2. Store the diagonal element with the negative difference in another Array of Vectors(say Neg[]) such that elements at the cell having difference(say -b) is stored at index abs(-b) = b of Neg[] array.
  3. Sort both the Array of Vectors increasing order.
  4. Traverse the given 2D vector and updated the value at the current cell with the value stored in Pos[] and Neg[] array. 
    • If the difference between column and row index(say d) is positive, then updated the value from Pos[d] array and remove the last element as: 
       
d = i - j
arr[i][j] = Pos[d][Pos.size()-1]
Pos[d].pop_back()
  • If the difference between column and row index(say d) is negative, then updated the value from Neg[d] array and remove the last element as:
d = j - i
arr[i][j] = Neg[d][Neg.size()-1]
Neg[d].pop_back()

Below is the implementation of the above approach: 

CPP




// C++ program to sort the 2D vector
// diagonally in decreasing order
#include "bits/stdc++.h"
using namespace std;
 
// Function that sort the elements
// of 2D vector
void diagonalSort(vector<vector<int> >& mat)
{
 
    // Calculate the rows and column
    int row = mat.size();
    int col = mat[0].size();
 
    // Array of vectors to store the
    // diagonal elements
    vector<int> Neg[row];
    vector<int> Pos[col];
 
    // Traverse the 2D vector and put
    // element in Array of vectors at
    // index difference between indexes
    for (int i = 0; i < row; i++) {
        for (int j = 0; j < col; j++) {
 
            // If diff is negative, then
            // push element to Neg[]
            if (j < i) {
                Neg[i - j].push_back(mat[i][j]);
            }
 
            // If diff is positive, then
            // push element to Pos[]
            else if (j > i) {
                Pos[j - i].push_back(mat[i][j]);
            }
 
            // If diff is 0, then push
            // element to Pos[0]
            else {
                Pos[0].push_back(mat[i][j]);
            }
        }
    }
 
    // Sort the Array of vectors
    for (int i = 0; i < row; i++) {
        sort(Neg[i].begin(), Neg[i].end());
    }
    for (int i = 0; i < col; i++) {
        sort(Pos[i].begin(), Pos[i].end());
    }
 
    // Update the value to arr[][]
    // from the sorted Array of vectors
    for (int i = 0; i < row; i++) {
        for (int j = 0; j < col; j++) {
 
            // If diff is positive
            if (j < i) {
                int d = i - j;
                int l = Neg[d].size() - 1;
                mat[i][j] = Neg[d][l - 1];
                Neg[d].pop_back();
            }
 
            // If diff is negative
            else if (j > i) {
                int d = j - i;
                int l = Pos[d].size() - 1;
                mat[i][j] = Pos[d][l - 1];
                Pos[d].pop_back();
            }
 
            // If diff is 0
            else {
                int l = Pos[0].size();
                mat[i][j] = Pos[0][l - 1];
                Pos[0].pop_back();
            }
        }
    }
}
 
// Function to print element
void printElement(vector<vector<int> >& arr)
{
 
    // Traverse the 2D vector
    for (int i = 0; i < arr.size(); i++) {
        for (int j = 0; j < arr[0].size(); j++) {
            cout << arr[i][j] << ' ';
        }
        cout << endl;
    }
}
 
// Driver Code
int main()
{
    vector<vector<int> > arr
        = { { 10, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
    diagonalSort(arr);
 
    // Function call to print elements
    printElement(arr);
}

Python3




# Python program for the above approach
from collections import defaultdict
 
def diagonalSort(matrix, n, m):
   
    # make a dict of list, where we
    # wil store the diagonal elements
    to = defaultdict(list)
 
    # store the diagonal elements with
    # respect to their row-col value
    # remember every row-col value for
    # each diagonal will be different
    for row in range(n):
        for col in range(m):
            to[row-col].append(matrix[row][col])
 
    # sort the elements of each
    # diagonal as required
    for i in to:
       
        # upper triangle diagonals
        if i < 0:
            to[i].sort(reverse=True)
 
        # median
        elif i == 0:
            pass
 
        # lower triangle diagonals
        else:
            to[i].sort()
 
    # store the new diagonal elements to
    # their respective position in the matrix
    for row in range(n):
        for col in range(m):
            matrix[row][col] = to[row-col].pop(0)
 
    return matrix
 
# Driver Code
if __name__ == "__main__":
    matrix = [[10, 2, 3],
              [4, 5, 6],
              [7, 8, 9]]
 
    n = len(matrix)
    m = len(matrix[0])
    matrix = diagonalSort(matrix, n, m)
 
    for row in range(n):
        for col in range(m):
            print(matrix[row][col], end=' ')
        print()
Output: 
10 2 0 
4 9 0 
0 0 5

 

Time Complexity: O(N*M*log(min(N,M)))

Space Complexity: O(N*M)

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