Given a Boolean Matrix, find k such that all elements in k'th row are 0 and k'th column are 1. - GeeksforGeeks

20 Jan, 2022
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 1, them this column cannot be a
.........solution decrement j.

3) If result is -1, return it. 

4) Else check validity of result by checking all row and column
          elements of result
Below is the implementation based on above idea. 
C++
Java
Python3
C#
PHP
Javascript
// 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;}
// Java program to find i such that all entries in i'th row are 0// and all entries in i't column are 1class GFG {     static int n = 5;     static int find(boolean arr[][]) {        // 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 false, then this row may be a solution            if (arr[i][j] == false) {                // Check for all elements in this row                while (j >= 0 && (arr[i][j] == false || i == j)) {                    j--;                }                 // If all values are false, 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] == true || 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 k = 0; k < n; k++) {            if (res != k && arr[k][res] != true) {                return -1;            }        }        for (int l = 0; l < n; l++) {            if (res != l && arr[res][l] != false) {                return -1;            }        }         return res;    }     /* Driver program to test above functions */    public static void main(String[] args) {        boolean mat[][] = {{false, false, true, true, false},        {false, false, false, true, false},        {true, true, true, true, false},        {false, false, false, false, false},        {true, true, true, true, true}};        System.out.println(find(mat));    }} /* This Java code is contributed by PrinciRaj1992*/
''' Python program to find k such that all elements in k'th row    are 0 and k'th column are 1''' def find(arr):     # store length of the array    n = len(arr)     # start from top right-most corner    i = 0    j = n - 1         # initialise result    res = -1     # find the index (This loop runs at most 2n times, we    # either increment row number or decrement column number)    while i < n and j >= 0:         # if the current element is 0, then this row may be a solution        if arr[i][j] == 0:             # check for all the elements in this row            while j >= 0 and (arr[i][j] == 0 or i == j):                j -= 1             # if all values are 0, update result as row number               if j == -1:                res = i                break             # if found a 1 in current row, the row can't be a            # solution, increment row number            else: i += 1         # if the current element is 1        else:             #check for all the elements in this column            while i < n and (arr[i][j] == 1 or i == j):                i +=1             # if all elements are 1, update result as col number            if i == n:                res = j                break             # if found a 0 in current column, the column can't be a            # solution, decrement column number            else: j -= 1     # if we couldn't find result in above loop, result doesn't exist    if res == -1:        return res     # check if the above computed res value is valid    for i in range(0, n):        if res != i and arr[i][res] != 1:            return -1    for j in range(0, j):        if res != j and arr[res][j] != 0:            return -1;      return res; # test find(arr) functionarr = [ [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] ] print (find(arr))
// C# program to find i such that all entries in i'th row are 0// and all entries in i't column are 1 using System;public class GFG{     static int n = 5;     static int find(bool [,]arr) {        // 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 false, then this row may be a solution            if (arr[i,j] == false) {                // Check for all elements in this row                while (j >= 0 && (arr[i,j] == false || i == j)) {                    j--;                }                 // If all values are false, 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] == true || 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 k = 0; k < n; k++) {            if (res != k && arr[k,res] != true) {                return -1;            }        }        for (int l = 0; l < n; l++) {            if (res != l && arr[res,l] != false) {                return -1;            }        }         return res;    }     /* Driver program to test above functions */    public static void Main() {        bool [,]mat = {{false, false, true, true, false},        {false, false, false, true, false},        {true, true, true, true, false},        {false, false, false, false, false},        {true, true, true, true, true}};        Console.WriteLine(find(mat));    }} // This code is contributed by PrinciRaj1992
<?php// PHP program to find i such that all// entries in i'th row are 0 and all// entries in i'th column are 1 function find(&$arr){    $n = 5;         // Start from top-most rightmost corner    // (We could start from other corners also)    $i = 0;    $j = $n - 1;     // Initialize result    $res = -1;     // Find the index (This loop runs at most    // 2n times, we either increment row number    // or decrement column number)    while ($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 $res;     // Check if above computed res is valid    for ($i = 0; $i < $n; $i++)    if ($res != $i && $arr[$i][$res] != 1)        return -1;    for ($j = 0; $j < $n; $j++)    if ($res != $j && $arr[$res][$j] != 0)        return -1;     return $res;}  // Driver Code$mat = array(array(0, 0, 1, 1, 0),             array(0, 0, 0, 1, 0),             array(1, 1, 1, 1, 0),             array(0, 0, 0, 0, 0),             array(1, 1, 1, 1, 1));echo (find($mat)); // This code is contributed by Shivi_Aggarwal?>
<script> // JavaScript program to find i such that// all entries in i'th row are 0// and all entries in i't column are 1var n = 5;function find(arr) {    // Start from top-most rightmost corner    // (We could start from other corners also)    var i = 0, j = n - 1;    // Initialize result    var 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 false,        // then this row may be a solution        if (arr[i][j] == false) {            // Check for all elements in this row            while (j >= 0 && (arr[i][j] == false || i == j)) {                j--;            }            // If all values are false,            // 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] == true || 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 (var k = 0; k < n; k++) {        if (res != k && arr[k][res] != true) {            return -1;        }    }    for (var l = 0; l < n; l++) {        if (res != l && arr[res][l] != false) {            return -1;        }    }    return res;} /* Driver program to test above functions */var mat = [[false, false, true, true, false],[false, false, false, true, false],[true, true, true, true, false],[false, false, false, false, false],[true, true, true, true, true]];document.write(find(mat)); </script>
Output: 
3
Time complexity of this solution is O(n). Note that we traverse at most 2n elements in the main while loop.This article is contributed by Ashish Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above 
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