Given an integers **N** such that there is a chessboard of size **N*N** and an array **pos[][]** of **K** pairs of integers which represent the positions of placed rooked in the given chessboard. The task is to find the maximum number of rooks with their positions that can be placed on the given chessboard such that no rook attacks some other rook. Print the positions in lexicographical order.**Examples:**

Input:N = 4, K = 2, pos[][] = {{1, 4}, {2, 2}}Output:

2

3 1

4 3Explanation:

Only 2 more rooks can be placed on the given chessboard and their positions are (3, 1) and (4, 3).Input:N = 5, K = 0, pos[][] = {}Output:

5

1 1

2 2

3 3

4 4

5 5Explanation:

Since the chessboard is empty we can place 5 rooks the given chessboard and their positions are (1, 1), (2, 2), (3, 3), (4, 4) and (5, 5).

**Naive Approach:** The simplest approach is to try to place a rook at every empty position of the chessboard and check if it attacks the already placed rooks or not. Below are the steps:

- Initialize a 2D matrix
**M[][]**of size N*N to represent the chessboard and place the already given rooks in it. - Transverse the complete matrix
**M[][]**and check if the ith row and jth column contains any rook - If the ith row and jth column both don’t contain any rook, then a rook is placed there and this cell is added to the result.
- Otherwise, move to the next empty cell on the chessboard.

**Time Complexity:** O(N^{3}) **Auxiliary Space:** O(N^{2})**Efficient Approach:** The approach is based on the idea that a maximum of **(N – K)** rooks can be placed on the chessboard according to the Pigeonhole Principle. Below are the steps:

- Since no two of the given rooks attack each other, all the rows given in the input must be unique. Similarly, all the columns given in the input must be unique.
- So, place the rooks only in
**N – K**unused rows and**N – K**unused columns. - Therefore, lexicographically minimum configuration can be achieved by pairing the smallest unused row with the smallest unused column, the second smallest unused row with the second smallest unused column, and so on.

Below is the implementation of the above approach:

## C++

`// C++ Program to implement` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to print the maximum rooks` `// and their positions` `void` `countRooks(` `int` `n, ` `int` `k,` ` ` `int` `pos[2][2])` ` ` `{` ` ` `int` `row[n] = {0};` ` ` `int` `col[n] = {0};` ` ` `// Initialize row and col array` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `{` ` ` `row[i] = 0;` ` ` `col[i] = 0;` ` ` `}` ` ` `// Marking the location of` ` ` `// already placed rooks` ` ` `for` `(` `int` `i = 0; i < k; i++) ` ` ` `{` ` ` `row[pos[i][0] - 1] = 1;` ` ` `col[pos[i][1] - 1] = 1;` ` ` `}` ` ` `int` `res = n - k;` ` ` `// Print number of non-attacking` ` ` `// rooks that can be placed` ` ` `cout << res << ` `" "` `<< endl;` ` ` `// To store the placed rook` ` ` `// location` ` ` `int` `ri = 0, ci = 0;` ` ` `while` `(res-- > 0) ` ` ` `{` ` ` `// Print lexographically` ` ` `// smallest order` ` ` `while` `(row[ri] == 1) ` ` ` `{` ` ` `ri++;` ` ` `}` ` ` `while` `(col[ci] == 1) ` ` ` `{` ` ` `ci++;` ` ` `}` ` ` `cout << (ri + 1) << ` `" "` `<< ` ` ` `(ci + 1) << ` `" "` `<<endl;` ` ` `ri++;` ` ` `ci++;` ` ` `}` ` ` `}` `// Driver Code` `int` `main()` `{` ` ` `// Size of board` ` ` `int` `N = 4;` ` ` `// Number of rooks already placed` ` ` `int` `K = 2;` ` ` `// Position of rooks` ` ` `int` `pos[2][2] = {{1, 4}, {2, 2}};` ` ` `// Function call` ` ` `countRooks(N, K, pos);` `}` `// This code is contributed by shikhasingrajput` |

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## Java

`// Java program for the above approach` `public` `class` `GFG {` ` ` `// Function to print the maximum rooks` ` ` `// and their positions` ` ` `private` `static` `void` `countRooks(` `int` `n, ` `int` `k,` ` ` `int` `pos[][])` ` ` `{` ` ` `int` `row[] = ` `new` `int` `[n];` ` ` `int` `col[] = ` `new` `int` `[n];` ` ` `// Initialize row and col array` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) {` ` ` `row[i] = ` `0` `;` ` ` `col[i] = ` `0` `;` ` ` `}` ` ` `// Marking the location of` ` ` `// already placed rooks` ` ` `for` `(` `int` `i = ` `0` `; i < k; i++) {` ` ` `row[pos[i][` `0` `] - ` `1` `] = ` `1` `;` ` ` `col[pos[i][` `1` `] - ` `1` `] = ` `1` `;` ` ` `}` ` ` `int` `res = n - k;` ` ` `// Print number of non-attacking` ` ` `// rooks that can be placed` ` ` `System.out.println(res + ` `" "` `);` ` ` `// To store the placed rook` ` ` `// location` ` ` `int` `ri = ` `0` `, ci = ` `0` `;` ` ` `while` `(res-- > ` `0` `) {` ` ` `// Print lexographically` ` ` `// smallest order` ` ` `while` `(row[ri] == ` `1` `) {` ` ` `ri++;` ` ` `}` ` ` `while` `(col[ci] == ` `1` `) {` ` ` `ci++;` ` ` `}` ` ` `System.out.println(` ` ` `(ri + ` `1` `)` ` ` `+ ` `" "` `+ (ci + ` `1` `)` ` ` `+ ` `" "` `);` ` ` `ri++;` ` ` `ci++;` ` ` `}` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `// Size of board` ` ` `int` `N = ` `4` `;` ` ` `// Number of rooks already placed` ` ` `int` `K = ` `2` `;` ` ` `// Position of rooks` ` ` `int` `pos[][] = { { ` `1` `, ` `4` `}, { ` `2` `, ` `2` `} };` ` ` `// Function call` ` ` `countRooks(N, K, pos);` ` ` `}` `}` |

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## Python3

`# Python3 program for the above approach` `# Function to prthe maximum rooks` `# and their positions` `def` `countRooks(n, k, pos):` ` ` ` ` `row ` `=` `[` `0` `for` `i ` `in` `range` `(n)]` ` ` `col ` `=` `[` `0` `for` `i ` `in` `range` `(n)]` ` ` `# Marking the location of` ` ` `# already placed rooks` ` ` `for` `i ` `in` `range` `(k):` ` ` `row[pos[i][` `0` `] ` `-` `1` `] ` `=` `1` ` ` `col[pos[i][` `1` `] ` `-` `1` `] ` `=` `1` ` ` `res ` `=` `n ` `-` `k` ` ` `# Print number of non-attacking` ` ` `# rooks that can be placed` ` ` `print` `(res)` ` ` `# To store the placed rook` ` ` `# location` ` ` `ri ` `=` `0` ` ` `ci ` `=` `0` ` ` ` ` `while` `(res > ` `0` `):` ` ` `# Print lexographically` ` ` `# smallest order` ` ` `while` `(row[ri] ` `=` `=` `1` `):` ` ` `ri ` `+` `=` `1` ` ` ` ` `while` `(col[ci] ` `=` `=` `1` `):` ` ` `ci ` `+` `=` `1` ` ` ` ` `print` `((ri ` `+` `1` `), (ci ` `+` `1` `))` ` ` ` ` `ri ` `+` `=` `1` ` ` `ci ` `+` `=` `1` ` ` `res ` `-` `=` `1` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `# Size of board` ` ` `N ` `=` `4` ` ` `# Number of rooks already placed` ` ` `K ` `=` `2` ` ` `# Position of rooks` ` ` `pos` `=` `[ [ ` `1` `, ` `4` `], [ ` `2` `, ` `2` `] ]` ` ` `# Function call` ` ` `countRooks(N, K, pos)` `# This code is contributed by mohit kumar 29` |

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## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` ` ` `// Function to print the maximum rooks` ` ` `// and their positions` ` ` `private` `static` `void` `countRooks(` `int` `n, ` `int` `k,` ` ` `int` `[, ]pos)` ` ` `{` ` ` `int` `[]row = ` `new` `int` `[n];` ` ` `int` `[]col = ` `new` `int` `[n];` ` ` `// Initialize row and col array` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `{` ` ` `row[i] = 0;` ` ` `col[i] = 0;` ` ` `}` ` ` `// Marking the location of` ` ` `// already placed rooks` ` ` `for` `(` `int` `i = 0; i < k; i++) ` ` ` `{` ` ` `row[pos[i, 0] - 1] = 1;` ` ` `col[pos[i, 1] - 1] = 1;` ` ` `}` ` ` `int` `res = n - k;` ` ` `// Print number of non-attacking` ` ` `// rooks that can be placed` ` ` `Console.WriteLine(res + ` `" "` `);` ` ` `// To store the placed rook` ` ` `// location` ` ` `int` `ri = 0, ci = 0;` ` ` `while` `(res -- > 0) ` ` ` `{` ` ` `// Print lexographically` ` ` `// smallest order` ` ` `while` `(row[ri] == 1) ` ` ` `{` ` ` `ri++;` ` ` `}` ` ` `while` `(col[ci] == 1) ` ` ` `{` ` ` `ci++;` ` ` `}` ` ` `Console.WriteLine((ri + 1) + ` `" "` `+ ` ` ` `(ci + 1) + ` `" "` `);` ` ` `ri++;` ` ` `ci++;` ` ` `}` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main(String[] args)` ` ` `{` ` ` `// Size of board` ` ` `int` `N = 4;` ` ` `// Number of rooks already placed` ` ` `int` `K = 2;` ` ` `// Position of rooks` ` ` `int` `[, ]pos = {{1, 4}, {2, 2}};` ` ` `// Function call` ` ` `countRooks(N, K, pos);` ` ` `}` `}` `// This code is contributed by Rajput-Ji` |

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**Output:**

2 3 1 4 3

**Time Complexity:** O(N^{2}) **Auxiliary Space:** O(N^{2})

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