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 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(n^{3})

** 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++ 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; }

Output:

5

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