Find k such that all elements in k’th row are 0 and k’th column are 1 in a boolean matrix

Given a square boolean matrix mat[n][n], find k such that all elements in k’th row are 0 and all elements in k’th column are 1. The value of mat[k][k] can be anything (either 0 or 1). If no such k exists, return -1.

Examples:

Input: bool mat[n][n] = { {1, 0, 0, 0},
                          {1, 1, 1, 0},
                          {1, 1, 0, 0},
                          {1, 1, 1, 0},
                        };
Output: 0
All elements in 0'th row are 0 and all elements in 
0'th column are 1.  mat[0][0] is 1 (can be any value)


Input: bool mat[n][n] = {{0, 1, 1, 0, 1},
                         {0, 0, 0, 0, 0},
                         {1, 1, 1, 0, 0},
                         {1, 1, 1, 1, 0},
                         {1, 1, 1, 1, 1}};
Output: 1
All elements in 1'st row are 0 and all elements in 
1'st column are 1.  mat[1][1] is 0 (can be any value)


Input: bool mat[n][n] = {{0, 1, 1, 0, 1},
                         {0, 0, 0, 0, 0},
                         {1, 1, 1, 0, 0},
                         {1, 0, 1, 1, 0},
                         {1, 1, 1, 1, 1}};
Output: -1
There is no k such that k'th row elements are 0 and
k'th column elements are 1.

Expected time complexity is O(n)

We strongly recommend you to minimize your browser and try this yourself first.

A Simple Solution is check all rows one by one. If we find a row ‘i’ such that all elements of this row are 0 except mat[i][i] which may be either 0 or 1, then we check all values in column ‘i’. If all values are 1 in the column, then we return i. Time complexity of this solution is O(n2).

An Efficient Solution can solve this problem in O(n) time. The solution is based on below facts.
1) There can be at most one k that can be qualified to be an answer (Why? Note that if k’th row has all 0’s probably except mat[k][k], then no column can have all 1′)s.
2) If we traverse the given matrix from a corner (preferably from top right and bottom left), we can quickly discard complete row or complete column based on below rules.
….a) If mat[i][j] is 0 and i != j, then column j cannot be the solution.
….b) If mat[i][j] is 1 and i != j, then row i cannot be the solution.

Below is complete algorithm based on above observations.

1) Start from top right corner, i.e., i = 0, j = n-1.  
   Initialize result as -1.

2) Do following until we find the result or reach outside the matrix.

......a) If mat[i][j] is 0, then check all elements on left of j in current row. 
.........If all elements on left of j are also 0, then set result as i. Note 
.........that i may not be result, but if there is a result, then it must be i 
.........(Why? we reach mat[i][j] after discarding all rows above it and all 
.........columns on right of it)

.........If all left side elements of i'th row are not 0, them this row cannot 
.........be a solution, increment i.

......b) If mat[i][j] is 1, then check all elements below i in current column. 
.........If all elements below i are 1, then set result as j. Note that j may
......... not be result, but if there is a result, then it must be j 

.........If all elements of j'th column are not 0, them this row cannot be a,
.........solution increment i.

3) If result is -1, return it. 

4) Else check validity of result by checking all row and column
          elements of result

Below is C++ implementation based on above idea.

// C++ program to find i such that all entries in i'th row are 0 // and all entries in i't column are 1 #include <iostream> using namespace std; #define n 5 int find(bool arr[n][n]) {     // Start from top-most rightmost corner     // (We could start from other corners also)     int i=0, j=n-1;     // Initialize result     int res = -1;     // Find the index (This loop runs at most 2n times, we either     // increment row number or decrement column number)     while (i<n && j>=0)     {         // If current element is 0, then this row may be a solution         if (arr[i][j] == 0)         {             // Check for all elements in this row             while (j >= 0 && (arr[i][j] == 0 || i == j))                 j--;             // If all values are 0, then store this row as result             if (j == -1)             {                 res = i;                 break;             }             // We reach here if we found a 1 in current row, so this             //  row cannot be a solution, increment row number             else i++;         }         else // If current element is 1         {             // Check for all elements in this column             while (i<n && (arr[i][j] == 1 || i == j))                 i++;             // If all elements are 1             if (i == n)             {                 res = j;                 break;             }             // We reach here if we found a 0 in current column, so this             // column cannot be a solution, increment column number             else j--;         }     }     // If we could not find result in above loop, then result doesn't exist     if (res == -1)        return res;     // Check if above computed res is valid     for (int i=0; i<n; i++)        if (res != i && arr[i][res] != 1)           return -1;     for (int j=0; j<n; j++)        if (res != j && arr[res][j] != 0)           return -1;     return res; } /* Driver program to test above functions */ int main() {     bool mat[n][n] = {{0, 0, 1, 1, 0},                       {0, 0, 0, 1, 0},                       {1, 1, 1, 1, 0},                       {0, 0, 0, 0, 0},                       {1, 1, 1, 1, 1}};     cout << find(mat);     return 0; }

Output:

3

Time complexity of this solution is O(n). Note that we traverse at most 2n elements in the main while loop.