# The Matrix Find Algorithm in Python

Challenge: How to find an element in a sorted matrix where row and column values increase monotonically?

What is a matrix? A matrix is a table of values consisting of rows and columns. Here, we represent the matrix as a list of integer lists. Hence, we can access matrix values with the indexing and slicing notation.

## How to Find an Element in a Matrix in Python?

The naive algorithm to find an element in a Python matrix is to iterate over all rows and values in that rows and compare these elements to the searched value:

```def matrix_find(matrix, value):
for row in matrix:
for element in row:
if element == value:
return True
return False

matrix = [[3, 4, 4, 6],
[6, 8, 11, 12],
[6, 8, 11, 15],
[9, 11, 12, 17]]

print(matrix_find(matrix, 7))
# False

print(matrix_find(matrix, 17))
# True```

If the matrix has n rows and m columns, the runtime complexity of the algorithm is O(n*m) because you have to perform n*m comparisons. This is not optimal for a sorted matrix (see next)!

## How to Find an Element in a Sorted Matrix in Python?

What is a sorted matrix? The matrix is sorted as the integers in the rows and columns increase monotonically with the row and column number.

The matrix-find algorithm is a beautiful way to search a value in a sorted matrix without visiting all values in the sorted matrix.

Here’s the algorithm:

```def matrix_find(matrix, value):
if not matrix or not matrix:
return False

j = len(matrix) - 1
for row in matrix:
while row[j] > value:
j = j - 1
if j == -1:
return False
if row[j] == value:
return True
return False

matrix = [[3, 4, 4, 6],
[6, 8, 11, 12],
[6, 8, 11, 15],
[9, 11, 12, 17]]
print(matrix_find(matrix=matrix, value=7))```

The function `matrixFind` takes a sorted integer matrix and an integer value. It returns `True` if the matrix contains the integer value.

### Explanation

In the first few lines, the algorithm checks whether the matrix is empty and returns `False` if this is the case.

Then, the while loop iterates over rows `i` and column `j` of the matrix starting with the first row `i=0` and the last column `j=n-1`.

But instead of searching the whole matrix, the algorithm uses a smarter strategy. It skips whole rows and columns at a time with the following method.

Check the last value of the first row (top right matrix value). We denote this value as `matrix[i][j]`. There are three cases.

1. The top right matrix value `matrix[i][j]` is equal to the searched value. In this case, the algorithm returns `True`.
2. The top right matrix value `matrix[i][j]` is smaller than the searched value. Because the matrix is sorted, the top right matrix value is the largest element in the row i. Thus, we can skip row i completely by going to the next row `i = i+1`. Next, we repeat this procedure with a smaller matrix that has one row less (i.e., row i).
3. The top right matrix value `matrix[i][j]` is larger than the searched value. This means that the whole column j has only elements that are larger than the searched value. Thus, we are sure that our searched value is not in column j and we can skip this column completely by decreasing `j = j-1`. Next, we repeat this procedure with a smaller matrix that has one column less (i.e., row j).

### Runtime Complexity Matrix-Find

In summary, the idea of this great algorithm from Keith Schwartz reduces one row or one column in each iteration. The runtime is only `O(2n)` instead of `O(n^2)` for a squared matrix with n rows and columns.

### Puzzle Matrix-Find

Test your understanding by solving the following Python puzzle on our Finxter app:

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## Where to Go From Here?

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