Remove All Ones With Row And Column Flips - Leetcode Solution

Problem Statement:

You are given a 2D binary matrix A of size m x n consisting of only 0's and 1's. You can perform an operation on any row or column of the matrix which consists of toggling all elements in that row or column between 0 and 1.

Each operation can be any type (row toggle or column toggle), and can be used any number of times.

Given the matrix A, return the minimum number of operations needed to remove all 1's from the matrix. If the task is impossible, return -1.

Solution:

Approach:

Our goal is to remove all 1's from the given matrix, and we can do this by toggling rows and columns. We will traverse the matrix to find all 1's and then perform an operation on the row or column that contains that 1 to convert all the 1's present in that row or column to 0's.

Now, we will count the number of operations performed to do this. If at the end of this process, we still have 1's present in the matrix, it means that we cannot remove all 1's from the matrix, and we will return -1.

Algorithm:

1. We will traverse the matrix to find all the 1's.
2. For each 1 found, we will perform operations on the corresponding row and column to convert all the 1's present in that row and column to 0's.
3. We will keep track of the number of operations performed to remove all 1's.
4. We will check at the end if there are still 1's present in the matrix. If yes, we will return -1; otherwise, we will return the number of operations performed.

Let's go through the solution in detail.

Detailed Solution:

1. First, we will find all the 1's in the matrix A. We will maintain two arrays, rows and cols, to keep track of the row and column indices of all the 1's in the matrix as shown below:

    rows = []
    cols = []
   
    #Traverse the matrix to find the 1's and add their indices to rows and cols arrays
    for i in range(m):
        for j in range(n):
            if A[i][j]==1:
                rows.append(i)
                cols.append(j)
   

2. Next, we will perform operations on the rows and columns containing 1's to convert them to 0's. We will perform the operations that result in the minimum number of operations.

For example, if a row contains 4 ones and two columns contain one each, then performing the toggle operation on that row will minimize the number of operations required to convert all the ones in that row and columns to 0's.

We can perform operations on rows and columns simultaneously, and we will keep track of the number of operations performed while toggling rows and columns.

We will maintain separate variables, row_ops and col_ops, to keep track of the number of operations performed on rows and columns, respectively.

    row_ops = [0]*m
    col_ops = [0]*n
    
    #Traverse the rows and columns of the matrix that contains the 1's and perform the minimum number of operations
    for i in range(len(rows)):
        row_ops[rows[i]]+=1
        col_ops[cols[i]]+=1

    #Calculate the minimum number of operations
    min_ops = 0
    for i in range(m):
        if row_ops[i]%2!=0:
            min_ops+=1
    for j in range(n):
        if col_ops[j]%2!=0:
            min_ops+=1

3. Finally, we will check if there are still 1's in the matrix after performing the operations. If yes, we will return -1; otherwise, we will return the number of operations performed.

    #Check if there are still 1's in the matrix
    for i in range(m):
        for j in range(n):
            if A[i][j]==1:
                return -1

    #Return the number of operations performed
    return min_ops

Time Complexity:

The time complexity of the above algorithm is O(mn), where m and n are the number of rows and columns in the matrix, respectively.

Space Complexity:

The space complexity of the above algorithm is O(m+n), where m and n are the number of rows and columns in the matrix, respectively. We are using two arrays of size m and n to keep track of the row and column indices of all the 1's in the matrix, respectively.

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