xyz Interview Question for Developer Program Engineers

Country: United States
Interview Type: Phone Interview

Comment hidden because of low score. Click to expand.
of 3 vote

There are surprisingly 4 solutions to this problem! We can start with the brute force and improve from O(n^2) -> O(nlogn) -> O(n) -> O(lgn). I have outlined the 4 solutions below with explanations.
PS:- I would appreciate if you can plz upvote :P, I put in some effort to clearly explain the solution. :P

Python solution:

# Brute Force - Look at each element one by one
# O(n^2) solution
def searchVersion1(matrix, target):
  for i, row in enumerate(matrix):
    for j, sq in enumerate(row):
      if sq == target:
        return True
  return False

# Since each row is sorted, perform binary search
# Since there are n rows, the runtime is O(n*lgn)
def searchVersion2(matrix, target):
  def binarySearchHelper(row, target, low, high):
    while low <= high:
      mid = (low + high) // 2
      if row[mid] == target:
        return True
      elif row[mid] < target:
        low = mid + 1
      elif row[mid] > target:
        high = mid - 1
    return False

  for i, row in enumerate(matrix):
    if binarySearchHelper(row, target, 0, len(row)-1):
      return True
  return False

# O(n) solution
# We can start from top-right corner of the matrix
# and move up to bottom right. If current element < target, move down
# else if element > target, move left. If we go out of bounds, we can
# return False since the element is not found.
def searchVersion3(matrix, target):
  x, y = 0, len(matrix[0])-1

  while 0 <= x < len(matrix) and 0 <= y < len(matrix[0]):
    currentElement = matrix[x][y]
    if currentElement == target:
      return True
    elif currentElement < target:
      x += 1 # Move down
    elif currentElement > target:
      y -= 1 # Move left
  return False

# O(lgn) solution
# Treat the 2-d matrix as a 1-d sorted array
# and just map the indices using math while applying binary search.
def searchVersion4(matrix, target):
  rows, cols = len(matrix), len(matrix[0])
  low, high = 0, rows * cols - 1

  while low <= high:
    mid = (low + high) // 2
    curRow = mid // cols
    curCol = mid % cols
    currentElement = matrix[curRow][curCol]
    if currentElement == target:
      return True
    elif currentElement < target:
      low = mid + 1
      high = mid - 1
  return False

matrix =  \
  [1 , 3, 5, 7, 9] ,

print(searchVersion1(matrix, 23)) # True
print(searchVersion1(matrix, 28)) # False
print(searchVersion1(matrix, 5)) # True
print(searchVersion1(matrix, 10))# False
print('#' * 20)
print(searchVersion2(matrix, 23))# True
print(searchVersion2(matrix, 28))# False
print(searchVersion2(matrix, 5))# True
print(searchVersion2(matrix, 10))# False
print('#' * 20)
print(searchVersion3(matrix, 23))# True
print(searchVersion3(matrix, 28))# False
print(searchVersion3(matrix, 5))# True
print(searchVersion3(matrix, 10))# False
print('#' * 20)
print(searchVersion4(matrix, 23))# True
print(searchVersion4(matrix, 28))# False
print(searchVersion4(matrix, 5))# True
print(searchVersion4(matrix, 10))# False

- prudent_programmer March 06, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
of 1 vote

Binary search to find the row and binary search to find the element in the row, log(n) + log(n). Complexity should be log(n).

- Biswanath March 06, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
of 0 vote

Also, you can do a single binary search, if you can mange the index conversion from two dimensional to single dimensional. As in (row, column) (2,3) means index 2n+3. So you can run in binary search on from 1 and n*n.

- BIswanath March 06, 2018 | Flag Reply

Add a Comment

Writing Code? Surround your code with {{{ and }}} to preserve whitespace.


is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.

Learn More


CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance.

Learn More

Resume Review

Most engineers make critical mistakes on their resumes -- we can fix your resume with our custom resume review service. And, we use fellow engineers as our resume reviewers, so you can be sure that we "get" what you're saying.

Learn More

Mock Interviews

Our Mock Interviews will be conducted "in character" just like a real interview, and can focus on whatever topics you want. All our interviewers have worked for Microsoft, Google or Amazon, you know you'll get a true-to-life experience.

Learn More