Search in a Row-wise and Column-wise Sorted 2D Array using Divide and Conquer algorithm

Given an n x n matrix, where every row and column is sorted in increasing order. Given a key, how to decide whether this key is in the matrix.
A linear time complexity is discussed in the previous post. This problem can also be a very good example for divide and conquer algorithms. Following is divide and conquer algorithm.

1) Find the middle element.
2) If middle element is same as key return.
3) If middle element is lesser than key then
….3a) search submatrix on lower side of middle element
….3b) Search submatrix on right hand side.of middle element
4) If middle element is greater than key then
….4a) search vertical submatrix on left side of middle element
….4b) search submatrix on right hand side.

DaCMat3



Following Java implementation of above algorithm.

Java

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// Java program for implementation of divide and conquer algorithm 
// to find a given key in a row-wise and column-wise sorted 2D array
class SearchInMatrix
{
    public static void main(String[] args)
    {
        int[][] mat = new int[][] { {10, 20, 30, 40}, 
                                    {15, 25, 35, 45},
                                    {27, 29, 37, 48},
                                    {32, 33, 39, 50}};
        int rowcount = 4,colCount=4,key=50;
        for (int i=0; i<rowcount; i++)
          for (int j=0; j<colCount; j++)
             search(mat, 0, rowcount-1, 0, colCount-1, mat[i][j]);
    }
  
    // A divide and conquer method to search a given key in mat[]
    // in rows from fromRow to toRow and columns from fromCol to
    // toCol
    public static void search(int[][] mat, int fromRow, int toRow, 
                              int fromCol, int toCol, int key)
    {
        // Find middle and compare with middle 
        int i = fromRow + (toRow-fromRow )/2;
        int j = fromCol + (toCol-fromCol )/2;
        if (mat[i][j] == key) // If key is present at middle
          System.out.println("Found "+ key + " at "+ i + 
                               " " + j);
        else
        {
            // right-up quarter of matrix is searched in all cases.
            // Provided it is different from current call
            if (i!=toRow || j!=fromCol)
             search(mat,fromRow,i,j,toCol,key);
  
            // Special case for iteration with 1*2 matrix
            // mat[i][j] and mat[i][j+1] are only two elements.
            // So just check second element
            if (fromRow == toRow && fromCol + 1 == toCol)
              if (mat[fromRow][toCol] == key)
                System.out.println("Found "+ key+ " at "
                                   fromRow + " " + toCol);
  
            // If middle key is lesser then search lower horizontal 
            // matrix and right hand side matrix
            if (mat[i][j] < key)
            {
                // search lower horizontal if such matrix exists
                if (i+1<=toRow)
                  search(mat, i+1, toRow, fromCol, toCol, key);
            }
  
            // If middle key is greater then search left vertical 
            // matrix and right hand side matrix
            else
            {
                // search left vertical if such matrix exists
                if (j-1>=fromCol)
                  search(mat, fromRow, toRow, fromCol, j-1, key);
            }
        }
    }
}

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C#

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// C# program for implementation of
// divide and conquer algorithm 
// to find a given key in a row-wise 
// and column-wise sorted 2D array
using System;
  
public class SearchInMatrix
{
    public static void Main(String[] args)
    {
        int[,] mat = new int[,] { {10, 20, 30, 40}, 
                                    {15, 25, 35, 45},
                                    {27, 29, 37, 48},
                                    {32, 33, 39, 50}};
        int rowcount = 4, colCount = 4, key = 50;
        for (int i = 0; i < rowcount; i++)
            for (int j = 0; j < colCount; j++)
            search(mat, 0, rowcount - 1, 0, colCount - 1, mat[i, j]);
    }
  
    // A divide and conquer method to 
    // search a given key in mat[]
    // in rows from fromRow to toRow 
    // and columns from fromCol to
    // toCol
    public static void search(int[,] mat, int fromRow, int toRow, 
                            int fromCol, int toCol, int key)
    {
        // Find middle and compare with middle 
        int i = fromRow + (toRow-fromRow )/2;
        int j = fromCol + (toCol-fromCol )/2;
        if (mat[i, j] == key) // If key is present at middle
        Console.WriteLine("Found "+ key + " at "+ i + 
                            " " + j);
        else
        {
            // right-up quarter of matrix is searched in all cases.
            // Provided it is different from current call
            if (i != toRow || j != fromCol)
            search(mat, fromRow, i, j, toCol, key);
  
            // Special case for iteration with 1*2 matrix
            // mat[i][j] and mat[i][j+1] are only two elements.
            // So just check second element
            if (fromRow == toRow && fromCol + 1 == toCol)
            if (mat[fromRow,toCol] == key)
                Console.WriteLine("Found "+ key + " at "
                                fromRow + " " + toCol);
  
            // If middle key is lesser then search lower horizontal 
            // matrix and right hand side matrix
            if (mat[i, j] < key)
            {
                // search lower horizontal if such matrix exists
                if (i + 1 <= toRow)
                search(mat, i + 1, toRow, fromCol, toCol, key);
            }
  
            // If middle key is greater then search left vertical 
            // matrix and right hand side matrix
            else
            {
                // search left vertical if such matrix exists
                if (j - 1 >= fromCol)
                search(mat, fromRow, toRow, fromCol, j - 1, key);
            }
        }
    }
}
  
// This code has been contributed by 29AjayKumar

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Output:

Found 10 at 0 0
Found 20 at 0 1
Found 30 at 0 2
Found 40 at 0 3
Found 15 at 1 0
Found 25 at 1 1
Found 35 at 1 2
Found 45 at 1 3
Found 27 at 2 0
Found 29 at 2 1
Found 37 at 2 2
Found 48 at 2 3
Found 32 at 3 0
Found 33 at 3 1
Found 39 at 3 2
Found 50 at 3 3

Time complexity:
We are given a n*n matrix, the algorithm can be seen as recurring for 3 matrices of size n/2 x n/2. Following is recurrence for time complexity

 T(n) = 3T(n/2) + O(1) 

The solution of recurrence is O(n1.58) using Master Method.
But the actual implementation calls for one submatrix of size n x n/2 or n/2 x n, and other submatrix of size n/2 x n/2.

This article is contributed by Kaushik Lele. 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 : 29AjayKumar



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