How do you solve Find Kth Largest XOR Coordinate Value on LeetCode?

Find Kth Largest XOR Coordinate Value (LeetCode #1738) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n log(m * n)) time complexity and O(m * n) space complexity. Key insight: Using prefix XOR allows for efficient submatrix XOR calculations.

What is the time complexity of Find Kth Largest XOR Coordinate Value?

The optimal solution for Find Kth Largest XOR Coordinate Value runs in O(m * n log(m * n)) time and uses O(m * n) space. The time complexity is dominated by the sorting step after calculating the prefix XOR values, which is O(m * n log(m * n)).

What is the brute force solution for Find Kth Largest XOR Coordinate Value?

The brute force approach for Find Kth Largest XOR Coordinate Value is Brute Force, with O(m * n * (m + n)) time complexity and O(m * n) space complexity. The brute force approach calculates the XOR value for every coordinate in the matrix directly. This is straightforward but inefficient for larger matrices. The optimal Optimal Solution approach improves this to O(m * n log(m * n)).

What are the interview tips for Find Kth Largest XOR Coordinate Value?

Explain your thought process clearly while coding. Consider edge cases, such as small matrices or matrices with identical values. Practice calculating prefix sums or XORs to become comfortable with the concept.

What are common mistakes when solving Find Kth Largest XOR Coordinate Value?

Forgetting to handle the boundaries when calculating prefix XOR. Not sorting the XOR values before retrieving the k-th largest.

#1738

Find Kth Largest XOR Coordinate Value

Medium

ArrayDivide and ConquerBit ManipulationSortingHeap (Priority Queue)MatrixPrefix SumQuickselectPrefix SumBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m * n * (m + n))	O(m * n log(m * n))
Space	O(m * n)	O(m * n)

💡

Intuition

Time O(m * n log(m * n))Space O(m * n)

Using a prefix XOR array allows us to compute the XOR for any submatrix in constant time after an initial setup. This significantly reduces the time complexity.

⚙️

Algorithm

5 steps

1Step 1: Create a prefix XOR matrix where each cell (i, j) contains the XOR of all elements from (0, 0) to (i, j).
2Step 2: Iterate through the original matrix to fill the prefix XOR matrix.
3Step 3: Use the prefix XOR matrix to compute the XOR value for each coordinate (a, b) in constant time.
4Step 4: Store the XOR values in a list and sort it.
5Step 5: Return the k-th largest value from the sorted list.

solution.py15 lines

1def kthLargestValue(matrix, k):
2    m, n = len(matrix), len(matrix[0])
3    prefix_xor = [[0] * n for _ in range(m)]
4    for i in range(m):
5        for j in range(n):
6            prefix_xor[i][j] = matrix[i][j]
7            if i > 0:
8                prefix_xor[i][j] ^= prefix_xor[i - 1][j]
9            if j > 0:
10                prefix_xor[i][j] ^= prefix_xor[i][j - 1]
11            if i > 0 and j > 0:
12                prefix_xor[i][j] ^= prefix_xor[i - 1][j - 1]
13    xor_values = [prefix_xor[i][j] for i in range(m) for j in range(n)]
14    xor_values.sort(reverse=True)
15    return xor_values[k - 1]

ℹ

Complexity note: The time complexity is dominated by the sorting step after calculating the prefix XOR values, which is O(m * n log(m * n)).

1Using prefix XOR allows for efficient submatrix XOR calculations.
2Sorting the results helps in easily retrieving the k-th largest value.

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