it is the question of programming and i do not have enough knowledge to solve it so i am asking to you and i hope i will get best answer

Question

Take a Boolean 2D array of 10X 10, where each row is sorted from the user, Find the row with the

maximum number of 1s.

Example:

Input matrix

0 1 1 1

0 0 1 1

1 1 1 1 // this row has maximum 1s

0 0 0 0

Output: 2

Answer 1

The actual answer will depend on what language you are using, but a very crude approach to this in Python would be something like the code below. Note that there are many ways to actually do this, some of which are a little hard to understand if you are a new programmer, so this is a "brute force" approach that should be fairly clear as far as the logic goes:

# Our input matrix

matrix = [[0, 1, 1, 1], [0, 0, 1, 1], [1, 1, 1, 1], [0, 0, 0, 0]]

# Initialize our variables we'll use to track the answer

longest_row_count = 0

longest_row_index = 0

# Iterate through our matrix, and store the current index (i) for reference

for i, row in enumerate(matrix):

counter = 0

# Now examine each element in the row to see if it is True/1

for index in row:

if index:

counter += 1

# Compare the current row to our previous "maximum" and update if needed

if counter > longest_row_count:

longest_row_index = i

longest_row_count = counter

# Print our our answer

print(f'The most True/1s were in index {longest_row_index}. It had {longest_row_count}')

By the way, adding this for if you want a more advanced solution (very Python-specific):

matrix = [[0, 1, 1, 1], [0, 0, 1, 1], [1, 1, 1, 1], [0, 0, 0, 0]]

counts = []

for row in matrix:

counts.append(row.count(1))

print(f'The most True/1s were in index {counts.index(max(counts))}. It had {max(counts)}')