Given a binary N x N matrix, we need to find the total number of matrix positions from which there is an endless path. Any position (i, j) is said to have an endless path if and only if the position (i,j) has the value 1 and all of the next positions in its row(i) and its column(j) should have value 1. If any position next to (i,j) either in row(i) or in column(j) will have 0 then position (i,j) doesn’t have any endless path.
Examples:
Input : 0 1 0
1 1 1
0 1 1
Output : 4
Endless points are (1, 1), (1, 2),
(2, 1) and (2, 2). For all other
points path to some corner is
blocked at some point.
Input : 0 1 1
1 1 0
0 1 0
Output : 1
Endless point is (0, 2).
Naive Approach :
We traverse all positions, for every position, we check that does this position has endless path or not. If yes then count it otherwise ignore it. But as usual its time complexity seems to be high.
Time complexity : O(n3)
Advance Approach (Dynamic programming):
We can easily say that if there is a zero at any position, then it will block path for all the positions left to it and top of it.
Also, we can say that any position (i,j) will have an endless row if (i,j+1) will have an endless row and value of (i,j) is 1.
Similarly, we can say that any position (i,j) will have an endless column if (i+1,j) will have an endless column and value of (i,j) is 1.
So we should maintain two matrices one for row and one for column. Always start from right most position for row and bottom most position for column and only check for next position whether it has endless path or not.
And Finally, if any position will have an endless path in both row and column matrix then that position is said to have an endless path.
C++
// C++ program to find count of endless points #include<bits/stdc++.h> using namespace std; const int MAX = 100; // Returns count of endless points int countEndless(bool input[][MAX], int n) { bool row[n][n], col[n][n]; // Fills column matrix. For every column, start // from every last row and fill every entry as // blockage after a 0 is found. for (int j=0; j<n; j++) { // flag which will be zero once we get a '0' // and it will be 1 otherwise bool isEndless = 1; for (int i=n-1; i>=0; i--) { // encountered a '0', set the isEndless // variable to false if (input[i][j] == 0) isEndless = 0; col[i][j] = isEndless; } } // Similarly, fill row matrix for (int i=0; i<n; i++) { bool isEndless = 1; for (int j= n-1; j>=0; j--) { if (input[i][j] == 0) isEndless = 0; row[i][j] = isEndless; } } // Calculate total count of endless points int ans = 0; for (int i=0; i<n; i++) for (int j=1; j<n; j++) // If there is NO blockage in row // or column after this point, // increment result. if (row[i][j] && col[i][j]) ans++; return ans; } // Driver code int main() { bool input[][MAX] = { {1, 0, 1, 1}, {0, 1, 1, 1}, {1, 1, 1, 1}, {0, 1, 1, 0}}; int n = 4; cout << countEndless(input, n); return 0; } |
Java
// Java program to find count of endless points class GFG { static final int MAX = 100; // Returns count of endless points static int countEndless(boolean input[][], int n) { boolean row[][] = new boolean[n][n]; boolean col[][] = new boolean[n][n]; // Fills column matrix. For every column, // start from every last row and fill every // entry as blockage after a 0 is found. for (int j = 0; j < n; j++) { // flag which will be zero once we get // a '0' and it will be 1 otherwise boolean isEndless = true; for (int i = n-1; i >= 0; i--) { // encountered a '0', set the // isEndless variable to false if (input[i][j] == false) isEndless = false; col[i][j] = isEndless; } } // Similarly, fill row matrix for (int i = 0; i < n; i++) { boolean isEndless = true; for (int j = n-1; j >= 0; j--) { if (input[i][j] == false) isEndless = false; row[i][j] = isEndless; } } // Calculate total count of endless points int ans = 0; for (int i = 0; i < n; i++) for (int j = 1; j < n; j++) // If there is NO blockage in row // or column after this point, // increment result. if (row[i][j] && col[i][j]) ans++; return ans; } //driver code public static void main(String arg[]) { boolean input[][] = { {true, false, true, true}, {false, true, true, true}, {true, true, true, true}, {false, true, true, false}}; int n = 4; System.out.print(countEndless(input, n)); } } // This code is contributed by Anant Agarwal. |
Python3
# Python3 program to find count of # endless points import numpy as np # Returns count of endless points def countEndless(input_mat, n) : row = np.zeros((n, n)) col = np.zeros((n, n)) # Fills column matrix. For every column, # start from every last row and fill # every entry as blockage after a 0 is found. for j in range(n) : # flag which will be zero once we # get a '0' and it will be 1 otherwise isEndless = 1 for i in range(n - 1, -1, -1) : # encountered a '0', set the # isEndless variable to false if (input_mat[i][j] == 0) : isEndless = 0 col[i][j] = isEndless # Similarly, fill row matrix for i in range(n) : isEndless = 1 for j in range(n - 1, -1, -1) : if (input_mat[i][j] == 0) : isEndless = 0 row[i][j] = isEndless # Calculate total count of endless points ans = 0 for i in range(n) : for j in range(1, n) : # If there is NO blockage in row # or column after this point, # increment result. #print(row[i][j] , col[i][j]) if (row[i][j] and col[i][j]) : ans += 1 #print(ans) return ans # Driver code if __name__ == "__main__" : input_mat = [[1, 0, 1, 1], [0, 1, 1, 1], [1, 1, 1, 1], [0, 1, 1, 0]] n = 4 print(countEndless(input_mat, n)) # This code is contributed by Ryuga |
C#
// C# program to find count of // endless points using System; public class GFG { // Returns count of endless points static int countEndless(bool [,]input, int n) { bool [,]row = new bool[n,n]; bool [,]col = new bool[n,n]; // Fills column matrix. For every // column, start from every last // row and fill every entry as // blockage after a 0 is found. for (int j = 0; j < n; j++) { // flag which will be zero // once we get a '0' and it // will be 1 otherwise bool isEndless = true; for (int i = n - 1; i >= 0; i--) { // encountered a '0', set // the isEndless variable // to false if (input[i,j] == false) isEndless = false; col[i,j] = isEndless; } } // Similarly, fill row matrix for (int i = 0; i < n; i++) { bool isEndless = true; for (int j = n - 1; j >= 0; j--) { if (input[i,j] == false) isEndless = false; row[i,j] = isEndless; } } // Calculate total count of // endless points int ans = 0; for (int i = 0; i < n; i++) for (int j = 1; j < n; j++) // If there is NO blockage // in row or column after // this point, increment // result. if (row[i,j] && col[i,j]) ans++; return ans; } //Driver code public static void Main() { bool [,]input = { {true, false, true, true}, {false, true, true, true}, {true, true, true, true}, {false, true, true, false}}; int n = 4; Console.Write(countEndless(input, n)); } } // This code is contributed by Sam007. |
PHP
<?php // PHP program to find // count of endless points // Returns count of // endless points function countEndless($input, $n) { // Fills column matrix. For // every column, start from // every last row and fill // every entry as blockage // after a 0 is found. for ($j = 0; $j < $n; $j++) { // flag which will be zero // once we get a '0' and // it will be 1 otherwise $isEndless = 1; for ($i = $n - 1; $i >= 0; $i--) { // encountered a '0', // set the isEndless // variable to false if ($input[$i][$j] == 0) $isEndless = 0; $col[$i][$j] = $isEndless; } } // Similarly, fill row matrix for ($i = 0; $i < $n; $i++) { $isEndless = 1; for ($j = $n - 1; $j >= 0; $j--) { if ($input[$i][$j] == 0) $isEndless = 0; $row[$i][$j] = $isEndless; } } // Calculate total count // of endless points $ans = 0; for ($i = 0; $i < $n; $i++) for ($j = 1; $j < $n; $j++) // If there is NO blockage // or column after this point, // increment result. if ($row[$i][$j] && $col[$i][$j]) $ans++; return $ans; } // Driver code $input = array(array(1, 0, 1, 1), array(0, 1, 1, 1), array(1, 1, 1, 1), array(0, 1, 1, 0)); $n = 4; echo countEndless($input, $n); // This code is contributed // by shiv_bhakt. ?> |
Output:
5
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