Print all elements in sorted order from row and column wise sorted matrix

Given an n x n matrix, where every row and column is sorted in non-decreasing order. Print all elements of matrix in sorted order.

Example:

Input: mat[][]  =  { {10, 20, 30, 40},
                     {15, 25, 35, 45},
                     {27, 29, 37, 48},
                     {32, 33, 39, 50},
                   };

Output:
Elements of matrix in sorted order
10 15 20 25 27 29 30 32 33 35 37 39 40 45 48 50

We can use Young Tableau to solve the above problem. The idea is to consider given 2D array as Young Tableau and call extract minimum O(N)



C++

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// A C++ program to Print all elements in sorted order from row and
// column wise sorted matrix
#include<iostream>
#include<climits>
using namespace std;
  
#define INF INT_MAX
#define N 4
  
// A utility function to youngify a Young Tableau.  This is different
// from standard youngify.  It assumes that the value at mat[0][0] is 
// infinite.
void youngify(int mat[][N], int i, int j)
{
    // Find the values at down and right sides of mat[i][j]
    int downVal  = (i+1 < N)? mat[i+1][j]: INF;
    int rightVal = (j+1 < N)? mat[i][j+1]: INF;
  
    // If mat[i][j] is the down right corner element, return
    if (downVal==INF && rightVal==INF)
        return;
  
    // Move the smaller of two values (downVal and rightVal) to 
    // mat[i][j] and recur for smaller value
    if (downVal < rightVal)
    {
        mat[i][j] = downVal;
        mat[i+1][j] = INF;
        youngify(mat, i+1, j);
    }
    else
    {
        mat[i][j] = rightVal;
        mat[i][j+1] = INF;
        youngify(mat, i, j+1);
    }
}
  
// A utility function to extract minimum element from Young tableau
int extractMin(int mat[][N])
{
    int ret = mat[0][0];
    mat[0][0] = INF;
    youngify(mat, 0, 0);
    return ret;
}
  
// This function uses extractMin() to print elements in sorted order
void printSorted(int mat[][N])
{
   cout << "Elements of matrix in sorted order n";
   for (int i=0; i<N*N; i++)
     cout << extractMin(mat) << " ";
}
  
// driver program to test above function
int main()
{
  int mat[N][N] = { {10, 20, 30, 40},
                    {15, 25, 35, 45},
                    {27, 29, 37, 48},
                    {32, 33, 39, 50},
                  };
  printSorted(mat);
  return 0;
}

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Python 3

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# Python 3 program to Print all elements 
# in sorted order from row and column 
# wise sorted matrix
import sys
  
INF = sys.maxsize
N = 4
  
# A utility function to youngify a Young 
# Tableau. This is different from standard 
# youngify. It assumes that the value at 
# mat[0][0] is infinite.
def youngify(mat, i, j):
  
    # Find the values at down and
    # right sides of mat[i][j]
    downVal = mat[i + 1][j] if (i + 1 < N) else INF
    rightVal = mat[i][j + 1] if (j + 1 < N) else INF
  
    # If mat[i][j] is the down right
    # corner element, return
    if (downVal == INF and rightVal == INF):
        return
  
    # Move the smaller of two values 
    # (downVal and rightVal) to mat[i][j] 
    # and recur for smaller value
    if (downVal < rightVal):
        mat[i][j] = downVal
        mat[i + 1][j] = INF
        youngify(mat, i + 1, j)
      
    else:
        mat[i][j] = rightVal
        mat[i][j + 1] = INF
        youngify(mat, i, j + 1)
  
# A utility function to extract minimum 
# element from Young tableau
def extractMin(mat):
  
    ret = mat[0][0]
    mat[0][0] = INF
    youngify(mat, 0, 0)
    return ret
  
# This function uses extractMin() to 
# print elements in sorted order
def printSorted(mat):
          
    print("Elements of matrix in sorted order n")
    i = 0
    while i < N * N: 
        print(extractMin(mat), end = " ")
        i += 1
  
# Driver Code
if __name__ == "__main__":
      
    mat = [[10, 20, 30, 40],
           [15, 25, 35, 45],
           [27, 29, 37, 48],
           [32, 33, 39, 50]]
    printSorted(mat)
  
# This code is contributed by ita_c

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Output:

Elements of matrix in sorted order
10 15 20 25 27 29 30 32 33 35 37 39 40 45 48 50

Time complexity of extract minimum is O(N) and it is called O(N2) times. Therefore the overall time complexity is O(N3).

A better solution is to use the approach used for merging k sorted arrays. The idea is to use a Min Heap of size N which stores elements of first column. The do extract minimum. In extract minimum, replace the minimum element with the next element of the row from which the element is extracted. Time complexity of this solution is O(N2LogN).

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// C++ program to merge k sorted arrays of size n each.
#include<iostream>
#include<climits>
using namespace std;
  
#define N 4
  
// A min heap node
struct MinHeapNode
{
    int element; // The element to be stored
    int i; // index of the row from which the element is taken
    int j; // index of the next element to be picked from row
};
  
// Prototype of a utility function to swap two min heap nodes
void swap(MinHeapNode *x, MinHeapNode *y);
  
// A class for Min Heap
class MinHeap
{
    MinHeapNode *harr; // pointer to array of elements in heap
    int heap_size; // size of min heap
public:
    // Constructor: creates a min heap of given size
    MinHeap(MinHeapNode a[], int size);
  
    // to heapify a subtree with root at given index
    void MinHeapify(int );
  
    // to get index of left child of node at index i
    int left(int i) { return (2*i + 1); }
  
    // to get index of right child of node at index i
    int right(int i) { return (2*i + 2); }
  
    // to get the root
    MinHeapNode getMin() { return harr[0]; }
  
    // to replace root with new node x and heapify() new root
    void replaceMin(MinHeapNode x) { harr[0] = x;  MinHeapify(0); }
};
  
// This function prints elements of a given matrix in non-decreasing
//  order. It assumes that ma[][] is sorted row wise sorted.
void printSorted(int mat[][N])
{
    // Create a min heap with k heap nodes.  Every heap node
    // has first element of an array
    MinHeapNode *harr = new MinHeapNode[N];
    for (int i = 0; i < N; i++)
    {
        harr[i].element = mat[i][0]; // Store the first element
        harr[i].i = i;  // index of row
        harr[i].j = 1;  // Index of next element to be stored from row
    }
    MinHeap hp(harr, N); // Create the min heap
  
    // Now one by one get the minimum element from min
    // heap and replace it with next element of its array
    for (int count = 0; count < N*N; count++)
    {
        // Get the minimum element and store it in output
        MinHeapNode root = hp.getMin();
  
        cout << root.element << " ";
  
        // Find the next elelement that will replace current
        // root of heap. The next element belongs to same
        // array as the current root.
        if (root.j < N)
        {
            root.element = mat[root.i][root.j];
            root.j += 1;
        }
        // If root was the last element of its array
        else root.element =  INT_MAX; //INT_MAX is for infinite
  
        // Replace root with next element of array
        hp.replaceMin(root);
    }
}
  
// FOLLOWING ARE IMPLEMENTATIONS OF STANDARD MIN HEAP METHODS
// FROM CORMEN BOOK
// Constructor: Builds a heap from a given array a[] of given size
MinHeap::MinHeap(MinHeapNode a[], int size)
{
    heap_size = size;
    harr = a;  // store address of array
    int i = (heap_size - 1)/2;
    while (i >= 0)
    {
        MinHeapify(i);
        i--;
    }
}
  
// A recursive method to heapify a subtree with root at given index
// This method assumes that the subtrees are already heapified
void MinHeap::MinHeapify(int i)
{
    int l = left(i);
    int r = right(i);
    int smallest = i;
    if (l < heap_size && harr[l].element < harr[i].element)
        smallest = l;
    if (r < heap_size && harr[r].element < harr[smallest].element)
        smallest = r;
    if (smallest != i)
    {
        swap(&harr[i], &harr[smallest]);
        MinHeapify(smallest);
    }
}
  
// A utility function to swap two elements
void swap(MinHeapNode *x, MinHeapNode *y)
{
    MinHeapNode temp = *x;  *x = *y;  *y = temp;
}
  
// driver program to test above function
int main()
{
  int mat[N][N] = { {10, 20, 30, 40},
                    {15, 25, 35, 45},
                    {27, 29, 37, 48},
                    {32, 33, 39, 50},
                  };
  printSorted(mat);
  return 0;
}

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Output:

10 15 20 25 27 29 30 32 33 35 37 39 40 45 48 50

Exercise:
Above solutions work for a square matrix. Extend the above solutions to work for an M*N rectangular matrix.

This article is contributed by Varun. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



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Improved By : Ita_c



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