# [Coding Challenge] Special Elements in Matrix

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.

1. Ravi says:

The output of your code is 10, not 9.

2. Moksha Jain says:

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

3. Gordon says:

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

4. LDrago says:

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.