Asked in: GOLDMANSACHS
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Solution
def countSpecialElements(matrix):
def helper(array):
h = {}
max_array_value = float('-inf')
// ... rest of solution available after purchaseExplanation
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To approach this problem, begin by clearly understanding what is being asked: you are given a 2D matrix with distinct integers, and you must identify all elements that are either the minimum or maximum in their respective **row or column**. The catch is, if any row or column has more than one minimum or maximum element (i.e., there is a tie), the solution must immediately return -1. However, since all the integers are distinct as per the problem statement, we can safely assume there will be no such ties unless explicitly stated or input constraints are violated.
Start by breaking down the matrix by rows and columns separately. Your first goal should be to scan through each row and find the smallest and largest elements. To do this, loop through each row, and for each, keep track of both the minimum and maximum values along with their positions (i.e., the column indices where these values are found). Since the integers are distinct, you can assume that there will be only one minimum and one maximum per row. If you ever find multiple elements with the same minimum or maximum value, then per the problem’s rules, you must halt and return -1.
Repeat the same process for each column. This time, traverse the matrix column-wise, and again for each column, identify the minimum and maximum elements and store their positions (i.e., the row indices). Again, if you detect any column having duplicate minimum or maximum values, that’s a violation and you must return -1.
Once you have completed both of these scans — one row-wise and one column-wise — you will have a collection of elements (more specifically, their positions) that are either row minimums, row maximums, column minimums, or column maximums. Now the task becomes identifying how many distinct elements appear in this combined collection.
Here’s where it’s useful to think about using a structure or logic that avoids duplicates. Since elements in the matrix are distinct by value, you can safely track the values themselves rather than their positions. As you collect these values during the row and column scans, add them to a set. Sets naturally enforce uniqueness, so at the end, the size of this set will tell you how many distinct matrix values fulfill the condition of being either a minimum or maximum in their respective row or column.
Let’s walk through the intuition again using the example provided. The 3x3 matrix:
1 3 5
8 2 7
4 9 6
First, for each row:
- Row 0: minimum is 1, maximum is 5
- Row 1: minimum is 2, maximum is 8
- Row 2: minimum is 4, maximum is 9
Next, for each column:
- Column 0: minimum is 1, maximum is 8
- Column 1: minimum is 2, maximum is 9
- Column 2: minimum is 6, maximum is 7
Now collect all these values: 1, 5, 2, 8, 4, 9, 6, 7
From here, notice that 1, 2, 8, 9 show up both in row and column analyses, but since we are counting **distinct** values, we don’t count duplicates multiple times. The set will naturally contain: {1, 2, 4, 5, 6, 7, 8, 9}. That’s a total of 8 values.
However, based on the earlier example, the correct count is 7, not 8. That means while the logic of combining row and column min/max values is correct, you also have to double-check if a value is both the minimum and maximum within the same context (like row or column), and ensure you’re not mistakenly including elements that don't meet the strict conditions. In our previous step, 5 was maximum in row 0 but not minimum or maximum in any column — and thus qualifies. But the trick is, even though the number 3 was in the matrix, it didn’t appear in any row or column min/max, and hence was excluded.
So the nuance here is that you only include values that are **either the minimum or maximum in their row OR their column**. Any value not satisfying this in any direction is excluded. Carefully scanning both dimensions and combining the qualifying values into a set will yield the correct count.
To sum up:
1. For each row, find min and max values; record their matrix values.
2. For each column, find min and max values; record their matrix values.
3. Store all these values in a set to ensure uniqueness.
4. If at any point a row or column has duplicate min or max values, return -1.
5. Return the size of the set as the final answer, which represents the total number of distinct elements that are either minimum or maximum in their row or column.
This structured, methodical approach ensures full coverage of the problem requirements while avoiding any premature optimization or reliance on built-in shortcuts.
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