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Find any possible two coordinates of Rectangle whose two coordinates are given

  • Last Updated : 16 Jul, 2021

Given a matrix mat[][] of size N×N where two elements of the matrix are ‘1’ denoting the coordinate of the rectangle and ‘0’ denotes the empty space, the task is to find the other two coordinates of the rectangle.
Note: There can be multiple answers possible for this problem, print any one of them.

Examples:

 Input: mat[][] = {{0, 0, 1, 0}, {0, 0, 1, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}}
Output: {{0, 0, 1, 1}, {0, 0, 1, 1}, {0, 0, 0, 0}, {0, 0, 0, 0}}
Explanation: 
0 0 1 1 
0 0 1 1
0 0 0 0
0 0 0 0
The coordinates {{0, 2}, {0, 3}, {1, 2}, {1, 3}} forms the rectangle

Input: mat[][] = {{0, 0, 1, 0}, {0, 0, 0, 0}, {1, 0, 0, 0}, {0, 0, 0, 0}} 
Output: {{1, 0, 1, 0}, {0, 0, 0, 0}, {1, 0, 1, 0}, {0, 0, 0, 0}}

Approach: The remaining coordinate can be found using these given coordinates because some points may have a common row and some might have a common column. Follow the steps below to solve the problem:



  • Initialize two pairs, say p1 and p2 to store the position of 1 in the initial matrix mat[].
  • Initialize two pairs, say p3 and p4 to store position where new 1 is to be inserted to make it a rectangle.
  • Traverse through the matrix using two nested loops and find the pairs p1 and p2.
  • Now there are three possible cases:
    • If p1.first and p2.first are same in this case adding 1 to p1.first and p2.first gives us p3.first and p4.first while p3.second and p4.second remain the same as p1.second and p2.second respectively.
    • If p1.second and p2.second are the same in this case adding 1 to p1.second and p2.second gives us p3.second and p4.second while p3.first and p4.first remains the same as p1.first and p2.first 
    • If no coordinates are same, then p3.first = p2.first, p3.second = p1.second, p4.first = p1.first and p4.second = p2.second.
  • Replace the coordinates of p3 and p4 with 1 and print the matrix.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the remaining
// two rectangle coordinates
void Create_Rectangle(vector<string> arr, int n)
{
 
    // Pairs to store the position of given
    // two coordinates of the rectangle.
    pair<int, int> p1 = { -1, -1 };
    pair<int, int> p2 = { -1, -1 };
 
    // Pairs to store the remaining two
    // coordinates of the rectangle.
    pair<int, int> p3;
    pair<int, int> p4;
 
    // Traverse through matrix and
    // find pairs p1 and p2
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (arr[i][j] == '1')
                if (p1.first == -1)
                    p1 = { i, j };
                else
                    p2 = { i, j };
        }
    }
 
    p3 = p1;
    p4 = p2;
 
    // First Case
    if (p1.first == p2.first) {
        p3.first = (p1.first + 1) % n;
        p4.first = (p2.first + 1) % n;
    }
    // Second Case
    else if (p1.second == p2.second) {
        p3.second = (p1.second + 1) % n;
        p4.second = (p2.second + 1) % n;
    }
    // Third Case
    else {
        swap(p3.first, p4.first);
    }
 
    arr[p3.first][p3.second] = '1';
    arr[p4.first][p4.second] = '1';
 
    // Print the matrix
    for (int i = 0; i < n; i++) {
        cout << arr[i] << endl;
    }
}
 
// Driver code
int main()
{
    // Given Input
    int n = 4;
    vector<string> arr{ "0010", "0000", "1000", "0000" };
 
    // Function Call
    Create_Rectangle(arr, n);
    return 0;
}

Java




// Java program for above approach
import java.awt.*;
import java.util.*;
class GFG{
    static class pair<T, V>{
        T first;
        V second;
    }
   
    // Function to find the remaining
    // two rectangle coordinates
    static void Create_Rectangle(ArrayList<String> arr, int n)
    {
 
        // Pairs to store the position of given
        // two coordinates of the rectangle.
        pair<Integer, Integer> p1 = new pair<>();
        p1.first = -1;
        p1.second= -1;
        pair<Integer, Integer> p2 = new pair<>();
        p2.first = -1;
        p2.second = -1;
 
        // Pairs to store the remaining two
        // coordinates of the rectangle.
        pair<Integer,Integer> p3 = new pair<>();
        pair<Integer, Integer> p4 = new pair<>();
 
        // Traverse through matrix and
        // find pairs p1 and p2
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (arr.get(i).charAt(j) == '1')
                    if (p1.first == -1) {
                        p1.first =i;
                        p1.second = j;
                    }
                    else {
                        p2.first = i;
                        p2.second = j;
                    }
            }
        }
 
        p3 = p1;
        p4 = p2;
 
        // First Case
        if (p1.first.intValue() == (p2.first).intValue()) {
            p3.first = (p1.first + 1) % n;
            p4.first = (p2.first + 1) % n;
        }
        // Second Case
        else if (p1.second.intValue()==(p2.second).intValue()) {
            p3.second = (p1.second + 1) % n;
            p4.second = (p2.second + 1) % n;
        }
        // Third Case
        else {
            int temp = p3.first;
            p3.first = p4.first;
            p4.first = temp;
        }
 
        // Print the matrix
        for (int i = 0; i < n; i++) {
            if(i==p3.first){
                for (int j = 0;j<n;j++){
                    if(j==p3.second)
                        System.out.print('1');
                    else
                        System.out.print(arr.get(i).charAt(j));
                }
            }
            else if(i==p4.first){
                for (int j = 0;j<n;j++){
                    if(j==p4.second)
                        System.out.print('1');
                    else
                        System.out.print(arr.get(i).charAt(j));
                }
            }
            else
                System.out.print(arr.get(i));
            System.out.println();
        }
    }
 
 
    // Driver Code
    public static void main(String[] args)
    {
       
        // Given Input
        int n = 4;
        ArrayList<String> arr = new ArrayList<>();
        arr.add("0010");
        arr.add("0000");
        arr.add("1000");
        arr.add("0000");
        //{ , "0000", "1000", "0000" };
 
        // Function Call
        Create_Rectangle(arr, n);
    }
}
 
// This code is contributed by hritikrommie.

Python3




# Python program for the above approach
 
# Function to find the remaining
# two rectangle coordinates
def Create_Rectangle(arr, n):
 
    for i in range(n):
        arr[i] = [i for i in arr[i]]
 
    # Pairs to store the position of given
    # two coordinates of the rectangle.
    p1 = [-1, -1]
    p2 = [-1, -1]
 
    # Pairs to store the remaining two
    # coordinates of the rectangle.
    p3 = []
    p4 = []
 
    # Traverse through matrix and
    # find pairs p1 and p2
    for i in range(n):
        for j in range(n):
            if (arr[i][j] == '1'):
                if (p1[0] == -1):
                    p1 = [i, j]
                else:
                    p2 = [i, j]
 
    p3 = p1
    p4 = p2
 
    # First Case
    if (p1[0] == p2[0]):
        p3[0] = (p1[0] + 1) % n
        p4[0] = (p2[0] + 1) % n
    # Second Case
    elif (p1[1] == p2[1]):
        p3[1] = (p1[1] + 1) % n
        p4[1] = (p2[1] + 1) % n
    # Third Case
    else:
        p3[0], p4[0] = p4[0],p3[0]
 
    arr[p3[0]][p3[1]] = '1'
    arr[p4[0]][p4[1]] = '1'
 
    # Print the matrix
    for i in range(n):
        print("".join(arr[i]))
 
 
# Driver code
if __name__ == '__main__':
    # Given Input
    n = 4
    arr = ["0010", "0000", "1000", "0000"]
 
    # Function Call
    Create_Rectangle(arr, n)
 
# This code is contributed by mohit kumar 29.

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to find the remaining
// two rectangle coordinates
function Create_Rectangle(arr, n) {
 
    for (let i = 0; i < n; i++) {
        arr[i] = arr[i].split("")
    }
    // Pairs to store the position of given
    // two coordinates of the rectangle.
    let p1 = [-1, -1];
    let p2 = [-1, -1];
 
    // Pairs to store the remaining two
    // coordinates of the rectangle.
    let p3 = [];
    let p4 = [];
 
    // Traverse through matrix and
    // find pairs p1 and p2
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++) {
            if (arr[i][j] == '1')
                if (p1[0] == -1)
                    p1 = [i, j];
                else
                    p2 = [i, j];
        }
    }
 
    p3 = p1;
    p4 = p2;
 
    // First Case
    if (p1[0] == p2[0]) {
        p3[0] = (p1[0] + 1) % n;
        p4[0] = (p2[0] + 1) % n;
    }
    // Second Case
    else if (p1[1] == p2[1]) {
        p3[1] = (p1[1] + 1) % n;
        p4[1] = (p2[1] + 1) % n;
    }
    // Third Case
    else {
        let temp = p3[0];
        p3[0] = p4[0];
        p4[0] = temp;
    }
 
    arr[p3[0]][p3[1]] = '1';
    arr[p4[0]][p4[1]] = '1';
 
    // Print the matrix
    for (let i = 0; i < n; i++) {
        document.write(arr[i].join("") + "<br>");
    }
}
 
// Driver code
 
// Given Input
let n = 4;
let arr = ["0010", "0000", "1000", "0000"];
 
// Function Call
Create_Rectangle(arr, n);
 
</script>

 
 

Output
1010
0000
1010
0000

 

Time Complexity: O(N2)  
Auxiliary Space: O(1)

 

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