Kth Largest XOR Coordinate Value

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Problem

Imagine you are analyzing a 2D grid of sensor readings, and you want to find the kth largest XOR value of all submatrices starting from the origin to various coordinates - a way to detect interesting patterns quickly.

Given a 2D matrix 'matrix' of size m x n, where each element is a non-negative integer, define the XOR coordinate value at (a, b) as the XOR of all elements matrix[i][j] where 0 ≤ i ≤ a and 0 ≤ j ≤ b. Return the kth largest XOR coordinate value among all coordinates (a, b) in the matrix.

1 ≤ m, n ≤ 10000 ≤ matrix[i][j] ≤ 10^61 ≤ k ≤ m * n

Edge cases: Single element matrix, k=1 → returns that elementAll elements zero → all XOR values zero, kth largest is zerok equals total number of elements → returns smallest XOR coordinate value

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IDE

def kthLargestValue(matrix: list[list[int]], k: int) -> int:public int kthLargestValue(int[][] matrix, int k)int kthLargestValue(vector<vector<int>>& matrix, int k)function kthLargestValue(matrix, k)

def kthLargestValue(matrix, k):
    # Write your solution here
    pass

class Solution {
    public int kthLargestValue(int[][] matrix, int k) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int kthLargestValue(vector<vector<int>>& matrix, int k) {
    // Write your solution here
    return 0;
}

function kthLargestValue(matrix, k) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: Wrong kth largest value due to brute force timeout or incorrect prefix XORIncorrect prefix XOR computation or heap maintenance logic✅ Implement prefix XOR as dp[i][j] = matrix[i][j] ^ dp[i-1][j] ^ dp[i][j-1] ^ dp[i-1][j-1] and maintain min-heap of size k

Wrong: Output equals sum of elements instead of XORConfusing XOR operation with sum in prefix computations✅ Use bitwise XOR (^) operator instead of addition (+) in prefix computations

Wrong: Off-by-one error in kth largest selectionIncorrect indexing or heap size management✅ Ensure heap size is exactly k and return heap root after processing all coordinates

Wrong: Greedy approach returns incorrect kth largestSelecting local max XOR values greedily instead of global kth largest✅ Compute prefix XOR for all coordinates and use min-heap of size k to track kth largest

Wrong: Fails on single element or zero matrix edge casesNot handling base cases properly✅ Add explicit checks or ensure prefix XOR and heap logic handle these cases

✓

Test Cases

Focus on handling smallest and boundary inputs correctly.

Watch out for common algorithmic traps like greedy and off-by-one errors.

Optimize your solution to use prefix XOR and min-heap for large inputs.

t1_01basic

Input{"matrix":[[5,2],[1,6]],"k":3}

Expected4

⏱ Performance - must finish in 2000ms

The XOR coordinate values are [[5,7],[4,0]]. Sorted descending: [7,5,4,0]. The 3rd largest is 4.

💡 Compute prefix XOR for each coordinate to avoid repeated XOR computations.

💡 Use a min-heap of size k to track the kth largest XOR value efficiently.

💡 Maintain the heap by pushing new XOR values and popping smallest when size exceeds k.

Why it failed: Incorrect output means prefix XOR or heap logic is wrong. Fix by correctly computing prefix XOR as dp[i][j] = matrix[i][j] ^ dp[i-1][j] ^ dp[i][j-1] ^ dp[i-1][j-1] and maintain min-heap of size k.

✓ Correct prefix XOR computation and heap maintenance confirmed.

t1_02basic

Input{"matrix":[[3,8,2],[5,1,7]],"k":4}

Expected9

⏱ Performance - must finish in 2000ms

XOR coordinate values computed and sorted descending: [15,13,11,3,10,0]. The 4th largest is 3.

💡 Remember to include all coordinates (a,b) in the XOR prefix matrix.

💡 Sort or use a min-heap to find kth largest efficiently.

💡 Check boundary conditions when computing prefix XOR for first row and column.

Why it failed: Wrong kth largest means prefix XOR or sorting/heap logic is incorrect. Fix prefix XOR boundaries and heap size management.

✓ Prefix XOR and kth largest extraction logic is correct.

t2_01edge

Input{"matrix":[[42]],"k":1}

Expected42

⏱ Performance - must finish in 2000ms

Single element matrix; only one XOR coordinate value equals the element itself.

💡 Test smallest matrix size to ensure base case works.

💡 Prefix XOR for single element is the element itself.

💡 Return the element directly when k=1 and matrix size=1.

Why it failed: Failing this means code does not handle single element matrices correctly. Fix by handling base case where m=n=1.

✓ Single element matrix handled correctly.

t2_02edge

Input{"matrix":[[0,0],[0,0]],"k":3}

Expected0

⏱ Performance - must finish in 2000ms

All elements zero; all XOR coordinate values are zero; kth largest is zero.

💡 Check if code handles zero values correctly in XOR computations.

💡 All prefix XOR values should be zero in this case.

💡 Ensure heap or sorting does not discard zero values.

Why it failed: Incorrect output means XOR or heap logic mishandles zero values. Fix by ensuring XOR computations and heap include zeros.

✓ Zero matrix XOR values handled correctly.

t2_03edge

Input{"matrix":[[1,2],[3,4]],"k":4}

Expected1

⏱ Performance - must finish in 2000ms

k equals total elements (4); smallest XOR coordinate value is 0.

💡 When k equals total elements, answer is smallest XOR coordinate value.

💡 Check that min-heap or sorting returns correct kth largest when k = m*n.

💡 Verify prefix XOR computations for all coordinates.

Why it failed: Failing means kth largest logic or heap size management is incorrect. Fix by ensuring kth largest is correctly selected when k = total elements.

✓ Correctly returns smallest XOR coordinate value when k equals total elements.

t3_01corner

Input{"matrix":[[1,1],[1,1]],"k":2}

Expected0

⏱ Performance - must finish in 2000ms

Greedy approach fails here; correct prefix XOR values are [1,0;0,1], sorted descending [1,1,0,0], 2nd largest is 1 but 0 is also present; test ensures no greedy picking largest local XOR only.

💡 Greedy picking largest XOR at each coordinate fails; consider all coordinates.

💡 Use prefix XOR matrix and min-heap to track global kth largest.

💡 Avoid assuming local max XOR is global max.

Why it failed: Greedy approach fails by returning wrong kth largest. Fix by computing prefix XOR for all coordinates and using min-heap of size k.

✓ Greedy trap avoided; correct kth largest found.

t3_02corner

Input{"matrix":[[2,3],[1,4]],"k":3}

Expected2

⏱ Performance - must finish in 2000ms

Tests confusion between 0/1 knapsack and this problem; XOR is not additive sum. Correct prefix XOR values sorted descending: [3,1,1,0], 3rd largest is 1.

💡 XOR is bitwise operation, not sum; do not confuse with additive knapsack logic.

💡 Compute prefix XOR using dp relation, not sums.

💡 Use min-heap to track kth largest XOR values.

Why it failed: Confusing XOR with sum leads to wrong output. Fix by implementing prefix XOR dp correctly: dp[i][j] = matrix[i][j] ^ dp[i-1][j] ^ dp[i][j-1] ^ dp[i-1][j-1].

✓ Correctly distinguishes XOR from sum; correct kth largest found.

t3_03corner

Input{"matrix":[[7,2,5],[3,1,6],[4,8,9]],"k":5}

Expected4

⏱ Performance - must finish in 2000ms

Off-by-one error test; correct sorted XOR values descending: [7,5,6,4,4,3,2,1,0], 5th largest is 4.

💡 Check indexing carefully when computing prefix XOR and selecting kth largest.

💡 Remember k is 1-based index for kth largest.

💡 Maintain min-heap size exactly k and pop smallest when exceeding.

Why it failed: Off-by-one in heap size or indexing causes wrong kth largest. Fix by ensuring heap size is k and indexing k-1 for zero-based arrays.

✓ Off-by-one errors avoided; correct kth largest returned.

t4_01performance

Input{"_description":"m=n=1000 at constraint boundary - executor generates this input"}

Expectednull

⏱ Performance - must finish in 2000ms

Input size m=n=1000, total elements=1,000,000. O(m*n*log k) expected to pass; brute force O(m^2*n^2) will TLE.

💡 Use prefix XOR matrix to compute XOR values in O(m*n).

💡 Use min-heap of size k to maintain kth largest efficiently.

💡 Avoid nested loops over all submatrices which cause TLE.

Why it failed: TLE caused by brute force O(m^2*n^2) approach. Fix by using prefix XOR dp and min-heap of size k for O(m*n*log k) complexity.

✓ Efficient O(m*n*log k) solution passes within time limit.

Practice

1. You are given a list of points on a 2D plane and need to find the k points closest to the origin (0,0). Which algorithmic approach best balances average time complexity and space usage for this problem?

easy

A. Use a Quickselect algorithm to partition points around a pivot distance until the k closest are identified.

B. Use a greedy algorithm that iteratively picks the closest point until k points are selected.

C. Use a dynamic programming approach to build up solutions for subsets of points.

D. Sort all points by their distance to the origin and return the first k points.

Solution

1. Step 1: Understand problem constraints

   The problem requires finding the k closest points efficiently, ideally better than sorting all points which is O(n log n).

2. Step 2: Identify algorithm characteristics

   Quickselect partitions points around a pivot and reduces the problem size each iteration, achieving average O(n) time and O(1) extra space, making it optimal.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Quickselect is known for average linear time selection problems [OK]

Hint: Quickselect offers average O(n) vs sorting's O(n log n) [OK]

Common Mistakes:

• Confusing sorting with selection algorithms
• Assuming greedy always works for closest points

2. Given the following Python code implementing Quickselect, what is the output of findKthLargest([4,1,3,2], 3)?

easy

A. 2

B. 3

C. 1

D. 4

Solution

1. Step 1: Understand k and indexing

   k=3 means the 3rd largest element, so index k-1=2 in zero-based indexing.

2. Step 2: Trace partition and quickselect

   Partition rearranges so elements > pivot are left; after recursive calls, the element at index 2 is 2.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   3rd largest in [4,1,3,2] is 2 [OK]

Hint: Remember zero-based index k-1 for kth largest [OK]

Common Mistakes:

• Off-by-one error in k indexing
• Confusing kth largest with kth smallest
• Misunderstanding partition direction

3. You are given a string and need to return a new string where characters are sorted by their frequency in descending order. Which approach guarantees an optimal O(n) time complexity solution for this problem?

easy

A. Use a greedy approach that repeatedly picks the character with the highest frequency without precomputing frequencies.

B. Use dynamic programming to build solutions for substrings based on character counts.

C. Use bucket sort by frequency: count frequencies, group characters by frequency, then concatenate from highest to lowest frequency.

D. Use a min-heap to store characters by frequency and extract them one by one.

Solution

1. Step 1: Understand frequency grouping

   Counting frequencies and grouping characters by frequency allows linear time sorting by frequency.

2. Step 2: Why bucket sort is optimal

   Bucket sort avoids sorting all characters explicitly, achieving O(n) time by iterating frequency buckets.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Bucket sort groups by frequency and concatenates in descending order [OK]

Hint: Bucket sort groups chars by frequency for linear time [OK]

Common Mistakes:

• Using min-heap without negation leads to ascending order
• Greedy without frequency count misses optimal order

4. The following code attempts to find the maximum CPU load from a list of jobs. Identify the line containing the subtle bug that causes incorrect results in some cases.

def find_max_cpu_load(jobs):
    events = []
    for start, end, load in jobs:
        events.append((start, load))
        events.append((end, load))  # Bug here
    events.sort(key=lambda x: x[0])

    current_load = 0
    max_load = 0
    for _, load_change in events:
        current_load += load_change
        max_load = max(max_load, current_load)
    return max_load

medium

A. Line adding (end, load) instead of (end, -load) to events.

B. Line sorting events by start time only.

C. Line initializing current_load to 0 before loop.

D. Line updating max_load inside the loop.

Solution

1. Step 1: Identify event creation lines.

   Start events add positive load, end events must subtract load.

2. Step 2: Spot that (end, load) incorrectly adds load instead of subtracting.

   This causes current_load to accumulate incorrectly, inflating max_load.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   End event load must be negative to reduce current load [OK]

Hint: End events must subtract load, not add [OK]

Common Mistakes:

• Forgetting negative load at job end
• Misordering events with same timestamp

5. What is the time complexity of the optimal solution that sorts workers by their wage-to-quality ratio and uses a balanced tree or multiset to maintain the K smallest qualities during iteration?

medium

A. O(n log n) for sorting plus O(n log K) for balanced tree operations, total O(n log n)

B. O(n^2) due to nested loops over workers and qualities

C. O(n log n + n K) because each insertion/removal in balanced tree takes O(K)

D. O(n K) since each worker requires scanning K qualities

Solution

1. Step 1: Sorting workers by ratio

   This takes O(n log n) time.

2. Step 2: Maintaining K smallest qualities with balanced tree

   Each insertion and removal is O(log K), done for each of the n workers -> O(n log K).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Overall complexity is O(n log n + n log K), dominated by O(n log n) [OK]

Hint: Sorting plus balanced tree operations dominate complexity [OK]

Common Mistakes:

• Assuming balanced tree operations take O(K) instead of O(log K)
• Confusing nested loops leading to O(n^2)
• Ignoring sorting cost