Problem Description

You have a 2D matrix of size m x n containing non-negative integers, and you need to find the k-th largest value among all coordinate values in the matrix.

For any coordinate (a, b) in the matrix, its value is calculated as the XOR of all elements in the rectangular region from the top-left corner (0, 0) to (a, b) inclusive. In other words, the value at coordinate (a, b) is:

matrix[0][0] XOR matrix[0][1] XOR ... XOR matrix[0][b] XOR matrix[1][0] XOR ... XOR matrix[a][b]

This includes all elements where row index i satisfies 0 <= i <= a and column index j satisfies 0 <= j <= b.

Your task is to:

1. Calculate the XOR value for every coordinate in the matrix
2. Find the k-th largest value among all these calculated XOR values (where k is 1-indexed, meaning k=1 refers to the largest value)

For example, if you have a 2x2 matrix, you would calculate 4 different XOR values:

• Value at (0,0) = matrix[0][0]
• Value at (0,1) = matrix[0][0] XOR matrix[0][1]
• Value at (1,0) = matrix[0][0] XOR matrix[1][0]
• Value at (1,1) = matrix[0][0] XOR matrix[0][1] XOR matrix[1][0] XOR matrix[1][1]

Then return the k-th largest among these values.

Intuition

The naive approach would be to calculate the XOR value for each coordinate by iterating through all elements in its rectangle, but this would be inefficient with time complexity of O(m²n²).

The key insight is recognizing that we can use prefix XOR similar to how we use prefix sums. Just like prefix sums help us quickly calculate the sum of any subarray, prefix XOR can help us calculate the XOR of any subrectangle.

Let's think about how XOR operations behave. A crucial property of XOR is that a XOR a = 0 and a XOR 0 = a. This means if we XOR the same value twice, it cancels out.

Consider calculating the XOR value at coordinate (i, j). If we already know:

• The XOR of all elements from (0,0) to (i-1, j)
• The XOR of all elements from (0,0) to (i, j-1)
• The XOR of all elements from (0,0) to (i-1, j-1)

We can use these to compute the XOR at (i, j). When we XOR the first two values, we get all elements in the rectangles, but the overlap region from (0,0) to (i-1, j-1) gets XORed twice and cancels out. So we need to XOR it back once more, along with the current element matrix[i][j].

This gives us the formula: s[i+1][j+1] = s[i+1][j] XOR s[i][j+1] XOR s[i][j] XOR matrix[i][j]

By building up this 2D prefix XOR array, we can calculate all coordinate values in O(mn) time. Once we have all values, finding the k-th largest is a standard problem that can be solved by sorting them in O(mn log(mn)) or using a more efficient quick selection algorithm in O(mn) average time.

Learn more about Divide and Conquer, Prefix Sum, Sorting and Heap (Priority Queue) patterns.

Solution Approach

We implement the solution using a 2D prefix XOR array combined with sorting to find the k-th largest value.

Step 1: Initialize the Prefix XOR Array

We create a 2D array s of size (m+1) x (n+1) initialized with zeros. The extra row and column act as padding to handle boundary cases elegantly, avoiding the need for special checks when i=0 or j=0.

Step 2: Build the Prefix XOR Array

We iterate through each cell of the original matrix and calculate the prefix XOR value using the formula:

s[i+1][j+1] = s[i+1][j] XOR s[i][j+1] XOR s[i][j] XOR matrix[i][j]

Let's understand each component:

• s[i+1][j]: XOR of all elements from (0,0) to (i, j-1)
• s[i][j+1]: XOR of all elements from (0,0) to (i-1, j)
• s[i][j]: XOR of all elements from (0,0) to (i-1, j-1) (added back because it was XORed twice)
• matrix[i][j]: The current element

As we calculate each prefix XOR value, we append it to an ans list that will store all coordinate values.

Step 3: Find the k-th Largest Value

After calculating all m*n coordinate values, we need to find the k-th largest. The solution uses Python's nlargest(k, ans) function from the heapq module, which efficiently finds the k largest elements. The function returns these k elements in descending order, so nlargest(k, ans)[-1] gives us the k-th largest value.

Time Complexity: O(mn + mn*log(k)) where the first term is for building the prefix XOR array and the second term is for finding the k-th largest element using a heap.

Space Complexity: O(mn) for storing the prefix XOR array and the list of all coordinate values.

The beauty of this approach is that it transforms a potentially expensive repeated XOR calculation problem into an efficient dynamic programming solution using the prefix XOR technique.

Example Walkthrough

Let's walk through a small example with a 2×3 matrix and find the k=4th largest value.

Given matrix:

[5, 2, 1]
[1, 6, 3]

Step 1: Initialize prefix XOR array s (3×4 with padding):

[0, 0, 0, 0]
[0, ?, ?, ?]
[0, ?, ?, ?]

Step 2: Build the prefix XOR array using the formula: s[i+1][j+1] = s[i+1][j] XOR s[i][j+1] XOR s[i][j] XOR matrix[i][j]

For position (0,0): matrix value = 5

• s[1][1] = s[1][0] XOR s[0][1] XOR s[0][0] XOR 5
• s[1][1] = 0 XOR 0 XOR 0 XOR 5 = 5
• Add 5 to our answer list: ans = [5]

For position (0,1): matrix value = 2

• s[1][2] = s[1][1] XOR s[0][2] XOR s[0][1] XOR 2
• s[1][2] = 5 XOR 0 XOR 0 XOR 2 = 7
• Add 7 to answer list: ans = [5, 7]

For position (0,2): matrix value = 1

• s[1][3] = s[1][2] XOR s[0][3] XOR s[0][2] XOR 1
• s[1][3] = 7 XOR 0 XOR 0 XOR 1 = 6
• Add 6 to answer list: ans = [5, 7, 6]

For position (1,0): matrix value = 1

• s[2][1] = s[2][0] XOR s[1][1] XOR s[1][0] XOR 1
• s[2][1] = 0 XOR 5 XOR 0 XOR 1 = 4
• Add 4 to answer list: ans = [5, 7, 6, 4]

For position (1,1): matrix value = 6

• s[2][2] = s[2][1] XOR s[1][2] XOR s[1][1] XOR 6
• s[2][2] = 4 XOR 7 XOR 5 XOR 6 = 0
• Add 0 to answer list: ans = [5, 7, 6, 4, 0]

For position (1,2): matrix value = 3

• s[2][3] = s[2][2] XOR s[1][3] XOR s[1][2] XOR 3
• s[2][3] = 0 XOR 6 XOR 7 XOR 3 = 2
• Add 2 to answer list: ans = [5, 7, 6, 4, 0, 2]

Final prefix XOR array:

[0, 0, 0, 0]
[0, 5, 7, 6]
[0, 4, 0, 2]

Step 3: Find the 4th largest value All coordinate values: [5, 7, 6, 4, 0, 2] Sorted in descending order: [7, 6, 5, 4, 2, 0] The 4th largest value is 4.

Verification of some values:

• Value at (0,1) = 5 XOR 2 = 7 ✓
• Value at (1,1) = 5 XOR 2 XOR 1 XOR 6 = 0 ✓
• Value at (1,2) = 5 XOR 2 XOR 1 XOR 1 XOR 6 XOR 3 = 2 ✓

Solution Implementation

Time and Space Complexity

Time Complexity: O(m × n × log(m × n)) or O(m × n) depending on the implementation of nlargest.

The algorithm consists of two main parts:

1. Computing XOR prefix sums: The nested loops iterate through all m × n elements of the matrix. For each element, we perform constant time XOR operations to compute s[i + 1][j + 1]. This takes O(m × n) time.

2. Finding the k-th largest value: The nlargest(k, ans) function is called on a list of m × n elements.

   • If nlargest uses a heap-based approach (which is typical in Python's heapq implementation), it maintains a min-heap of size k and processes all m × n elements, resulting in O(m × n × log k) time. When k is close to m × n, this becomes O(m × n × log(m × n)).
   • If nlargest uses a quickselect-based approach, the average case would be O(m × n).

Therefore, the overall time complexity is O(m × n × log(m × n)) in the worst case or O(m × n) with an optimal selection algorithm.

Space Complexity: O(m × n)

The space usage includes:

1. The 2D prefix sum array s of size (m + 1) × (n + 1), which is O(m × n).
2. The list ans that stores all m × n XOR values, which is O(m × n).
3. The space used by nlargest which could be O(k) for a heap-based approach or O(m × n) if it creates a copy of the list.

The dominant space usage is O(m × n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Misunderstanding XOR Properties in Prefix Calculation

The Problem: A common mistake is treating XOR operations like regular addition when building the prefix array. Developers might incorrectly write:

# INCORRECT - treating XOR like addition
prefix_xor[i+1][j+1] = prefix_xor[i+1][j] ^ prefix_xor[i][j+1] ^ matrix[i][j]

This misses the crucial prefix_xor[i][j] term. Unlike addition where we subtract the overlapping region, with XOR we need to XOR it back because:

• The region (0,0) to (i-1, j-1) is included in both prefix_xor[i+1][j] and prefix_xor[i][j+1]
• XORing the same value twice cancels it out: a ^ a = 0
• We need to XOR it once more to restore it

The Solution: Always include all four terms in the prefix XOR calculation:

prefix_xor[i+1][j+1] = (prefix_xor[i+1][j] ^ 
                       prefix_xor[i][j+1] ^ 
                       prefix_xor[i][j] ^     # Don't forget this!
                       matrix[i][j])

Pitfall 2: Using Wrong Method to Find k-th Largest

The Problem: Developers might sort the entire array and access the k-th element:

# INEFFICIENT - sorting entire array
all_xor_values.sort(reverse=True)
return all_xor_values[k-1]

This has O(mn*log(mn)) time complexity, which is unnecessary when k is small.

The Solution: Use heapq.nlargest(k, all_xor_values)[-1] which maintains a min-heap of size k, giving O(mn*log(k)) complexity. This is significantly better when k << mn.

Pitfall 3: Off-by-One Errors with Array Indexing

The Problem: Confusion between the padded prefix array indices and original matrix indices:

# INCORRECT - using wrong indices
prefix_xor[i][j] = prefix_xor[i-1][j] ^ prefix_xor[i][j-1] ^ matrix[i][j]

This fails when i=0 or j=0 due to negative indices.

The Solution: Consistently use the padding strategy where prefix_xor[i+1][j+1] corresponds to matrix[i][j]. This eliminates boundary condition checks and makes the code cleaner.