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[Coding Challenge] Special Elements in Matrix

Aniruddha Chaudhari/28266/6
Code

This coding challenge was asked in Goldman Sachs coderpad coding interview round and also in the Syntel hackathon interview.

Problem Statement: Special Elements in Matrix

Given a matrix of size m*n, m denotes the row starting with index 0 and n denotes the column starting with index 0.

The matrix will hold distinct integers as elements.

We need to find a distinct number of positional elements which are either the minimum or maximum in their corresponding row or column.

Please return -1 if any row or any column has multiple minimum or maximum elements.

For example, given a matrix of size 3*3, the elements are stored as follows.

1  3  4
5  2  9
8  7  6

The expected output is 7.

In the above example, we identified the output as 7 based on below.

1 - minimum in row and column
4 - Maximum in row
2 - Minimum in column and row
9 - Maximum in row and column
8 - Maximum in row and column
7 - Maximum in column
6 - Minimum in row

Input:

m - integer - number of rows
n - integer - number of columns
m * n matrix

Output:

r - integer - result

Constraints:

0<m,n<100
Elements in the matrix are positive integers.

Programming Solution in Python:

Here is the code for special elements in matrix solved in Python.

# Complete the countSpecialElements function below.
  
def countSpecialElements(matrix):
  nRows= len(matrix)
  nCount=0
  
  for row in matrix:
    for indexCol, element in enumerate(row):
  
      if element==min(row) or element==max(row):
        if row.count(element)>1:
          return -1
        nCount=nCount+1
  
      else:
        listColumn=[]
  
        for indexRow in range(0, nRows):
          listColumn.append(matrix[indexRow][indexCol])
  
        if element==min(listColumn) or element==max(listColumn):
          if listColumn.count(element)>1:
            return -1
          nCount=nCount+1

  return nCount
  
if __name__ == '__main__':
  nCount = countSpecialElements([[1, 3, 4],[5, 2, 9],[8, 7, 6]])
  print(nCount)

Output:

7

In Python, matrix elements are stored as a nested list. To solve this kind of problem, you should have a good understanding of the list and its methods.

If you are solving this coding question in a competitive test round, consider all boundary cases.

There was one more question was asked in the Goldman Sachs coding round- Secure My Conversation.

Python Interview Questions eBook

coding challengePython
Aniruddha Chaudhari
I am complete Python Nut, love Linux and vim as an editor. I hold a Master of Computer Science from NIT Trichy. I dabble in C/C++, Java too. I keep sharing my coding knowledge and my own experience on CSEstack.org portal.

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Comments

  • Reply
    Ravi
    March 2, 2020 at 12:15 am

    The output of your code is 10, not 9.

    • Reply
      Aniruddha Chaudhari
      March 2, 2020 at 6:52 pm

      Edited. Thanks for the correction, Ravi!

  • Reply
    Moksha Jain
    May 9, 2020 at 1:58 pm

    Sir, there is a mistake in code…of indentation inside else block…

                    for indexRow in range(0, nRows):
                        listColumn.append(matrix[indexRow][indexCol])
    
                    if element == min(listColumn) or element == max(listColumn):
                            if listColumn.count(element) &gt; 1:
                                return -1
                            nCount = nCount + 1
    

    the correct output is 7.
    Thanks

    • Reply
      Aniruddha Chaudhari
      May 9, 2020 at 5:28 pm

      Thanks for the correction, Moksha! Indentation is fixed.

  • Reply
    Gordon
    March 21, 2022 at 7:40 pm

    Is there a more efficient way to solve this problem? Because at worst this code is using a triple nested for loop.

  • Reply
    LDrago
    August 10, 2022 at 9:28 pm

    One question is when will the -1 if any row or any column has multiple minimum or maximum elements occur. Since it is mentioned in the problem statement that The matrix will hold distinct integers as elements.

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Why?

Why Competitive Programming Important?

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