[Coding Practice] Maximum number of zeros in NxN matrix

Question:

Input is a NxN matrix which contains only 0′s and 1′s. The condition is no 1 will occur in a row after 0. Find the index of the row which contains maximum number of zeros.

Example: lets say 5×5 matrix

1 0 0 0 0

1 1 0 0 0

1 1 1 1 1

0 0 0 0 0

1 1 1 0 0

For this input answer is 4th row.

solution should have time complexity less than N^2

http://www.geeksforgeeks.org/forums/topic/amazon-interview-question-for-software-engineerdeveloper-about-algorithms-24/

Solutions:

1. O(n²)Solution:

Start from a11 to a1n (go through the column) once 0 is met, return current index of row. If no 0 is met, go through the second column (a21 to a2n).

在最坏情况下需要遍历矩阵中的每个元素。

2. O(nlogn)Solution:

Search each row from top to down, in each row, use binary search to get the index of first 0. Use a variable to save the max index of 0. In worst case, O(nlogn) in time complexity.

3. O(n)Solution:
Start from the top right element a1n, if 0 is met, move left, if 1 is met or left corner is reached, record current index of row to variable T, then move down until 0 is met. Repeat these steps until move down to the bottom of matrix (anx) or left corner of matrix (ax1). The value of T is the answer. In worst case, time complexity O(2n) = O(n).

Code of Solution 3:

#include<stdio.h>
#include <ctime>
#include <cstdlib>
using namespace std;

int MaximumZeroRow(int** matrix, int size){
    if(matrix == NULL)
        return 0;
    int i = 0, j = size - 1, T = 0;
    while(1){
        if(!matrix[i][j]){
            j--;
        }
        if(matrix[i][j] || j < 0){
            T = i;
            i++;
        }
        if( i > (size - 1) || j < 0) break;
    }
    return T;
} 


int main(){
    srand(time(0));
    int **array;
    array = new int *[5];
    for(int i = 0; i < 5; i++){
        array[i] = new int[5];
        for(int j = 0; j < 5; j++){
            array[i][j] = (j == 0 ? (rand() & 1) : (rand() & 1) & array[i][j-1]);
            printf("%d ", array[i][j]);
        }
        printf("
");
    }
    printf("
");
    
    printf("Max 0 row index: %d
", MaximumZeroRow(array, 5));
    
    return 0;
}