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# 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. Following implementation of above algorithm.

## C++

 `// C++ program for implementation of``// divide and conquer algorithm``// to find a given key in a row-wise``// and column-wise sorted 2D array``#include``#define ROW 4``#define COL 4``using` `namespace` `std;` `// A divide and conquer method to``// search a given key in mat[]``// in rows from fromRow to toRow``// and columns from fromCol to``// toCol``void` `search(``int` `mat[ROW][COL], ``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``        ``cout<<``"Found "``<< key << ``" at "``<< i <<``                            ``" "` `<< j<= fromCol)``                ``search(mat, fromRow, toRow, fromCol, j - 1, key);``            ``}``        ``}``    ``}` `// Driver code``int` `main()``{``    ``int` `mat[ROW][COL] = { {10, 20, 30, 40},``                            ``{15, 25, 35, 45},``                            ``{27, 29, 37, 48},``                            ``{32, 33, 39, 50}};``    ``int` `key = 50;``    ``for` `(``int` `i = 0; i < ROW; i++)``    ``for` `(``int` `j = 0; j < COL; j++)``        ``search(mat, 0, ROW - 1, 0, COL - 1, mat[i][j]);``    ``return` `0;``}` `// This code is contributed by Rajput-Ji`

## Java

 `// 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=fromCol)``                  ``search(mat, fromRow, toRow, fromCol, j-``1``, key);``            ``}``        ``}``    ``}``}`

## Python3

 `# Python3 program for implementation of``# divide and conquer algorithm to find``# a given key in a row-wise and column-wise``# sorted 2D array a divide and conquer method``# to search a given key in mat in rows from``# fromRow to toRow and columns from fromCol to``# toCol``def` `search(mat, fromRow, toRow, fromCol, toCol, key):` `    ``# Find middle and compare with middle``    ``i ``=` `fromRow ``+` `(toRow ``-` `fromRow) ``/``/` `2``;``    ``j ``=` `fromCol ``+` `(toCol ``-` `fromCol) ``/``/` `2``;``    ``if` `(mat[i][j] ``=``=` `key): ``# If key is present at middle``        ``print``(``"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 ``or` `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 ``and` `fromCol ``+` `1` `=``=` `toCol):``            ``if` `(mat[fromRow][toCol] ``=``=` `key):``                ``print``(``"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);` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``mat ``=` `[[ ``10``, ``20``, ``30``, ``40``],``           ``[``15``, ``25``, ``35``, ``45``],``           ``[``27``, ``29``, ``37``, ``48``],``           ``[``32``, ``33``, ``39``, ``50``]];``    ``rowcount ``=` `4``; colCount ``=` `4``; key ``=` `50``;``    ``for` `i ``in` `range``(rowcount):``        ``for` `j ``in` `range``(colCount):``            ``search(mat, ``0``, rowcount ``-` `1``, ``0``, colCount ``-` `1``, mat[i][j]);` `# This code is contributed by 29AjayKumar`

## C#

 `// 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`

## Javascript

 ``

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) `

Space Complexity: O(log(n))

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.

#### Approach :

The approach below  is a modified version of the binary search algorithm, called Binary Search on Rows algorithm.

1. Loop through each row of the matrix.
2. Check if the key falls within the range of the current row (i.e., if the first element of the row is less than or equal to the key and the last element of the row is greater than or equal to the key).
3. If the key falls within the range of the current row, perform binary search on that row to find the key.
4. If the key is found, print the row and column index of the key and return.

## C++

 `#include ``#define ROW 4``#define COL 4``using` `namespace` `std;` `bool` `binary_search(``int` `mat[ROW][COL], ``int` `row, ``int` `start,``                   ``int` `end, ``int` `key)``{``    ``while` `(start <= end) {``        ``int` `mid = start + (end - start) / 2;``        ``if` `(mat[row][mid] == key) {``            ``return` `true``;``        ``}``        ``else` `if` `(mat[row][mid] < key) {``            ``start = mid + 1;``        ``}``        ``else` `{``            ``end = mid - 1;``        ``}``    ``}``    ``return` `false``;``}` `void` `search(``int` `mat[ROW][COL], ``int` `key)``{``    ``// search in each row``    ``for` `(``int` `i = 0; i < ROW; i++) {``        ``// perform binary search in current row``        ``if` `(mat[i] <= key && mat[i][COL - 1] >= key) {``            ``if` `(binary_search(mat, i, 0, COL - 1, key)) {``                ``cout << ``"Found "` `<< key << ``" at "` `<< i``                     ``<< ``" "` `<< endl;``                ``return``;``            ``}``        ``}``    ``}``    ``cout << key << ``" not found"` `<< endl;``}` `// Driver code``int` `main()``{``    ``int` `mat[ROW][COL] = { { 10, 20, 30, 40 },``                          ``{ 15, 25, 35, 45 },``                          ``{ 27, 29, 37, 48 },``                          ``{ 32, 33, 39, 50 } };``    ``int` `key = 50;``    ``search(mat, key);``    ``return` `0;``}`

## Java

 `import` `java.util.*;` `class` `Main {``    ``static` `final` `int` `ROW = ``4``;``    ``static` `final` `int` `COL = ``4``;``    ``static` `boolean` `binarySearch(``int``[][] mat, ``int` `row,``                                ``int` `start, ``int` `end, ``int` `key)``    ``{``        ``while` `(start <= end) {``            ``int` `mid = start + (end - start) / ``2``;``            ``if` `(mat[row][mid] == key) {``                ``return` `true``;``            ``}``            ``else` `if` `(mat[row][mid] < key) {``                ``start = mid + ``1``;``            ``}``            ``else` `{``                ``end = mid - ``1``;``            ``}``        ``}``        ``return` `false``;``    ``}` `    ``static` `void` `search(``int``[][] mat, ``int` `key)``    ``{``        ``// search in each row``        ``for` `(``int` `i = ``0``; i < ROW; i++) {``            ``// perform binary search in current row``            ``if` `(mat[i][``0``] <= key``                ``&& mat[i][COL - ``1``] >= key) {``                ``if` `(binarySearch(mat, i, ``0``, COL - ``1``, key)) {``                    ``System.out.println(``"Found "` `+ key``                                       ``+ ``" at "` `+ i);``                    ``return``;``                ``}``            ``}``        ``}``        ``System.out.println(key + ``" not found"``);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[][] mat = { { ``10``, ``20``, ``30``, ``40` `},``                        ``{ ``15``, ``25``, ``35``, ``45` `},``                        ``{ ``27``, ``29``, ``37``, ``48` `},``                        ``{ ``32``, ``33``, ``39``, ``50` `} };``        ``int` `key = ``50``;``        ``search(mat, key);``    ``}``}`

## Python3

 `row ``=` `4``col ``=` `4` `# Function to perform binary search on a row of the matrix``def` `binary_search(mat, row, start, end, key):``    ``while` `start <``=` `end:``        ``mid ``=` `start ``+` `(end ``-` `start)` `        ``if` `mat[row][mid] ``=``=` `key:``            ``return` `True``        ``elif` `mat[row][mid] < key:``            ``start ``=` `mid ``+` `1``        ``else``:``            ``end ``=` `mid ``-` `1` `    ``return` `False`  `# Function to search for a given key in the matrix``def` `search(mat, key):``  ` `    ``# Search in each row``    ``for` `i ``in` `range``(row):``      ` `      ``# Perform binary search in current row``        ``if` `mat[i][``0``] <``=` `key <``=` `mat[i][col``-``1``]:``            ``if` `binary_search(mat, i, ``0``, col``-``1``, key):``                ``print``(f``"Found {key} at {i}"``)``                ``return` `True``    ``print``(f``"{key} not found"``)``    ``return` `False` `# Test the code with an example matrix and value of key``mat ``=` `[[``10``, ``20``, ``30``, ``40``],``       ``[``15``, ``25``, ``35``, ``45``],``       ``[``27``, ``29``, ``37``, ``48``],``       ``[``32``, ``33``, ``39``, ``50``]]``key ``=` `50``search(mat, key)`

## C#

 `using` `System;` `class` `Program``{``  ``const` `int` `ROW = 4;``  ``const` `int` `COL = 4;` `  ``static` `bool` `BinarySearch(``int``[,] mat, ``int` `row, ``int` `start, ``int` `end, ``int` `key)``  ``{``    ``while` `(start <= end)``    ``{``      ``int` `mid = start + (end - start) / 2;``      ``if` `(mat[row, mid] == key)``      ``{``        ``return` `true``;``      ``}``      ``else` `if` `(mat[row, mid] < key)``      ``{``        ``start = mid + 1;``      ``}``      ``else``      ``{``        ``end = mid - 1;``      ``}``    ``}``    ``return` `false``;``  ``}` `  ``static` `void` `Search(``int``[,] mat, ``int` `key)``  ``{``    ``// search in each row``    ``for` `(``int` `i = 0; i < ROW; i++)``    ``{``      ``// perform binary search in current row``      ``if` `(mat[i, 0] <= key && mat[i, COL - 1] >= key)``      ``{``        ``if` `(BinarySearch(mat, i, 0, COL - 1, key))``        ``{``          ``Console.WriteLine(``"Found "` `+ key + ``" at "` `+ i);``          ``return``;``        ``}``      ``}``    ``}``    ``Console.WriteLine(key + ``" not found"``);``  ``}` `  ``// Driver code``  ``static` `void` `Main()``  ``{``    ``int``[,] mat = { { 10, 20, 30, 40 },``                  ``{ 15, 25, 35, 45 },``                  ``{ 27, 29, 37, 48 },``                  ``{ 32, 33, 39, 50 } };``    ``int` `key = 50;``    ``Search(mat, key);``  ``}``}` `// This code is contributed by Prajwal Kandekar`

## Javascript

 `const ROW = 4;``const COL = 4;` `// Function to perform binary search on a row of the matrix``function` `binary_search(mat, row, start, end, key) {``    ``while` `(start <= end) {``        ``let mid = start + Math.floor((end - start) / 2);``        ``if` `(mat[row][mid] == key) {``            ``return` `true``;``        ``} ``else` `if` `(mat[row][mid] < key) {``            ``start = mid + 1;``        ``} ``else` `{``            ``end = mid - 1;``        ``}``    ``}``    ``return` `false``;``}` `// Function to search for a given key in the matrix``function` `search(mat, key) {``    ``// Search in each row``    ``for` `(let i = 0; i < ROW; i++) {``        ``// Perform binary search in current row``        ``if` `(mat[i] <= key && mat[i][COL-1] >= key) {``            ``if` `(binary_search(mat, i, 0, COL-1, key)) {``                ``console.log(``"Found "` `+ key + ``" at "` `+ i);``                ``return``;``            ``}``        ``}``    ``}``    ``console.log(key + ``" not found"``);``}` `// Test the code with an example matrix and value of key``let mat = [ [10, 20, 30, 40],``            ``[15, 25, 35, 45],``            ``[27, 29, 37, 48],``            ``[32, 33, 39, 50]];``    ``let key = 50;``    ``search(mat, key);`

Output

`Found 50 at 3 `

### Time complexity analysis:

In the worst case, we will perform binary search on each of the ROW rows of the matrix. The time complexity of binary search is O(log COL), and we perform it ROW times, so the overall time complexity of the Binary Search on Rows algorithm is O(ROW * log COL). Since the matrix is row-wise and column-wise sorted, we can assume that ROW and COL are roughly the same size, so the time complexity can be simplified to O(N log N), where N = ROW = COL.

Space Complexity Analysis: O(1), we only use a constant amount of extra space to store some variables and constants.