Row with max 1s Python

PROGRAM TO FIND THE ROW WITH MAXIMUM NUMBER OF 1s

# Function to find the index

# of first index of 1 in a 

# boolean array arr[] 

def first( arr, low, high):

    if high >= low:

          

        # Get the middle index 

        mid = low + (high - low)//2

  

        # Check if the element at 

        # middle index is first 1

        if (mid == 0 or arr[mid - 1] == 0) and arr[mid] == 1:

            return mid

  

        # If the element is 0, 

        # recur for right side

        elif arr[mid] == 0:

            return first(arr, (mid + 1), high)

      

        # If element is not first 1, 

        # recur for left side

        else:

            return first(arr, low, (mid - 1))

    return -1

  

# Function that returns 

# index of row with maximum 

# number of 1s. 

def rowWithMax1s( mat):

      

    # Initialize max values

    R = len(mat)

    C = len(mat[0])

    max_row_index = 0

    max = -1

      

    # Traverse for each row and 

    # count number of 1s by finding

    #  the index of first 1

    for i in range(0, R):

        index = first (mat[i], 0, C - 1)

        if index != -1 and C - index > max:

            max = C - index

            max_row_index = i

  

    return max_row_index

  

# Driver Code

mat = [[0, 0, 0, 1],

       [0, 1, 1, 1],

       [1, 1, 1, 1],

       [0, 0, 0, 0]]

print ("Index of row with maximum 1s is", 

      rowWithMax1s(mat))







OUTPUT

Index of row with maximum 1s is 2

CREDITS: https://www.geeksforgeeks.org/find-the-row-with-maximum-number-1s/