Kth Largest XOR Coordinate Value

🎯

Kth Largest XOR Coordinate Value

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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.

💡 This problem combines prefix XOR computations with a top-k selection using a heap. Beginners often struggle because it mixes 2D prefix computations with efficient data structures to track the kth largest value, which is not straightforward at first glance.

📋

Problem Statement

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

💡

Example

Input"matrix = [[5,2],[1,6]], k = 3"

Output4

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

• Single element matrix, k=1 → returns that element
• All elements zero → all XOR values zero, kth largest is zero
• k equals total number of elements → returns smallest XOR coordinate value
• Matrix with maximum size and large values → tests efficiency and memory

⚠️

Common Mistakes

Not using prefix XOR and recomputing XOR for each coordinate

Code runs too slowly and times out on large inputs

✅ Implement prefix XOR matrix to compute XOR coordinate values in O(1)

Sorting all XOR values even when k is small

Unnecessary overhead, slower than needed

✅ Use a min-heap of size k to track top k values dynamically

Incorrect prefix XOR formula causing wrong values

Wrong output or failing test cases

✅ Use inclusion-exclusion: prefix[i][j] = prefix[i-1][j] ^ prefix[i][j-1] ^ prefix[i-1][j-1] ^ matrix[i-1][j-1]

Off-by-one errors in prefix matrix indexing

Index out of bounds or incorrect results

✅ Use prefix matrix with size (m+1) x (n+1) and 1-based indexing for safe access

Using max-heap instead of min-heap for top k tracking

Incorrect kth largest value or inefficient heap operations

✅ Use min-heap to keep smallest of top k at root for easy replacement

🧠

Brute Force (Nested Loops for XOR Computation)

💡 This approach exists to build intuition on how XOR coordinate values are computed and why naive methods are inefficient. It helps understand the problem from first principles.

Intuition

Compute XOR for every coordinate by iterating over all submatrices from (0,0) to (a,b) directly, then collect all values and sort to find kth largest.

Algorithm

1. Initialize an empty list to store XOR coordinate values.
2. For each coordinate (a, b), compute XOR of all elements matrix[i][j] where 0 ≤ i ≤ a and 0 ≤ j ≤ b using nested loops.
3. Add the computed XOR value to the list.
4. Sort the list in descending order and return the kth element.

💡 The nested loops make it hard to see efficiency; this approach is straightforward but quickly becomes impractical as matrix size grows.

</>

Code

def kthLargestValue(matrix, k):
    m, n = len(matrix), len(matrix[0])
    xor_values = []
    for a in range(m):
        for b in range(n):
            val = 0
            for i in range(a + 1):
                for j in range(b + 1):
                    val ^= matrix[i][j]
            xor_values.append(val)
    xor_values.sort(reverse=True)
    return xor_values[k - 1]

# Example usage
if __name__ == '__main__':
    matrix = [[5,2],[1,6]]
    k = 3
    print(kthLargestValue(matrix, k))  # Output: 4

Line Notes

m, n = len(matrix), len(matrix[0])Determine matrix dimensions for iteration limits

for a in range(m):Iterate over all row indices for coordinates

for b in range(n):Iterate over all column indices for coordinates

for i in range(a + 1):Iterate over rows up to current coordinate to compute XOR

for j in range(b + 1):Iterate over columns up to current coordinate to compute XOR

val ^= matrix[i][j]Accumulate XOR of elements in the submatrix

xor_values.sort(reverse=True)Sort all XOR values descending to find kth largest

return xor_values[k - 1]Return kth largest XOR coordinate value

import java.util.*;
public class Solution {
    public int kthLargestValue(int[][] matrix, int k) {
        int m = matrix.length, n = matrix[0].length;
        List<Integer> xorValues = new ArrayList<>();
        for (int a = 0; a < m; a++) {
            for (int b = 0; b < n; b++) {
                int val = 0;
                for (int i = 0; i <= a; i++) {
                    for (int j = 0; j <= b; j++) {
                        val ^= matrix[i][j];
                    }
                }
                xorValues.add(val);
            }
        }
        Collections.sort(xorValues, Collections.reverseOrder());
        return xorValues.get(k - 1);
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        int[][] matrix = {{5,2},{1,6}};
        int k = 3;
        System.out.println(sol.kthLargestValue(matrix, k)); // Output: 4
    }
}

Line Notes

int m = matrix.length, n = matrix[0].length;Get matrix dimensions for loops

for (int a = 0; a < m; a++) {Loop over rows for coordinate a

for (int b = 0; b < n; b++) {Loop over columns for coordinate b

for (int i = 0; i <= a; i++) {Iterate rows up to a for XOR calculation

for (int j = 0; j <= b; j++) {Iterate columns up to b for XOR calculation

val ^= matrix[i][j];XOR accumulate elements in submatrix

Collections.sort(xorValues, Collections.reverseOrder());Sort XOR values descending

return xorValues.get(k - 1);Return kth largest XOR value

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int kthLargestValue(vector<vector<int>>& matrix, int k) {
    int m = matrix.size(), n = matrix[0].size();
    vector<int> xorValues;
    for (int a = 0; a < m; a++) {
        for (int b = 0; b < n; b++) {
            int val = 0;
            for (int i = 0; i <= a; i++) {
                for (int j = 0; j <= b; j++) {
                    val ^= matrix[i][j];
                }
            }
            xorValues.push_back(val);
        }
    }
    sort(xorValues.rbegin(), xorValues.rend());
    return xorValues[k - 1];
}

int main() {
    vector<vector<int>> matrix = {{5,2},{1,6}};
    int k = 3;
    cout << kthLargestValue(matrix, k) << endl; // Output: 4
    return 0;
}

Line Notes

int m = matrix.size(), n = matrix[0].size();Get matrix dimensions for iteration

for (int a = 0; a < m; a++) {Loop over rows for coordinate a

for (int b = 0; b < n; b++) {Loop over columns for coordinate b

for (int i = 0; i <= a; i++) {Iterate rows up to a for XOR calculation

for (int j = 0; j <= b; j++) {Iterate columns up to b for XOR calculation

val ^= matrix[i][j];XOR accumulate elements in submatrix

sort(xorValues.rbegin(), xorValues.rend());Sort XOR values descending

return xorValues[k - 1];Return kth largest XOR value

function kthLargestValue(matrix, k) {
    const m = matrix.length, n = matrix[0].length;
    const xorValues = [];
    for (let a = 0; a < m; a++) {
        for (let b = 0; b < n; b++) {
            let val = 0;
            for (let i = 0; i <= a; i++) {
                for (let j = 0; j <= b; j++) {
                    val ^= matrix[i][j];
                }
            }
            xorValues.push(val);
        }
    }
    xorValues.sort((x, y) => y - x);
    return xorValues[k - 1];
}

// Example usage
const matrix = [[5,2],[1,6]];
const k = 3;
console.log(kthLargestValue(matrix, k)); // Output: 4

Line Notes

const m = matrix.length, n = matrix[0].length;Get matrix dimensions for loops

for (let a = 0; a < m; a++) {Loop over rows for coordinate a

for (let b = 0; b < n; b++) {Loop over columns for coordinate b

for (let i = 0; i <= a; i++) {Iterate rows up to a for XOR calculation

for (let j = 0; j <= b; j++) {Iterate columns up to b for XOR calculation

val ^= matrix[i][j];XOR accumulate elements in submatrix

xorValues.sort((x, y) => y - x);Sort XOR values descending

return xorValues[k - 1];Return kth largest XOR value

⏱

Complexity

TimeO(m^2 * n^2) due to nested loops over all submatrices

SpaceO(m * n) to store all XOR coordinate values

For each of the m*n coordinates, we compute XOR over up to (a+1)*(b+1) elements, leading to O(m^2 * n^2) time.

💡 For a 10x10 matrix, this means 10,000 operations per coordinate, totaling 1,000,000 operations, which is slow but manageable for small inputs.

Interview Verdict: TLE

This approach is too slow for large inputs but is useful to understand the problem and correctness before optimizing.

🧠

Prefix XOR Matrix + Sorting All Values

💡 This approach introduces prefix XOR to compute XOR coordinate values efficiently, reducing redundant computations and improving performance significantly.

Intuition

Use a prefix XOR matrix where each cell stores XOR of all elements from (0,0) to (i,j). Then XOR coordinate values can be computed in O(1) using prefix XOR properties.

Algorithm

1. Create a prefix XOR matrix of same size as input.
2. Fill prefix XOR matrix using XOR of previous prefix values and current element.
3. Collect all prefix XOR values into a list.
4. Sort the list descending and return the kth element.

💡 Prefix XOR reduces nested loops to single pass, but sorting all values still costs O(m*n log(m*n)).

</>

Code

def kthLargestValue(matrix, k):
    m, n = len(matrix), len(matrix[0])
    prefix = [[0] * (n + 1) for _ in range(m + 1)]
    xor_values = []
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1]
            xor_values.append(prefix[i][j])
    xor_values.sort(reverse=True)
    return xor_values[k - 1]

# Example usage
if __name__ == '__main__':
    matrix = [[5,2],[1,6]]
    k = 3
    print(kthLargestValue(matrix, k))  # Output: 4

Line Notes

prefix = [[0] * (n + 1) for _ in range(m + 1)]Create prefix XOR matrix with extra row and column for boundary

for i in range(1, m + 1):Iterate rows starting from 1 due to prefix indexing

for j in range(1, n + 1):Iterate columns starting from 1

prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1]Compute prefix XOR using inclusion-exclusion

xor_values.append(prefix[i][j])Collect XOR coordinate values

xor_values.sort(reverse=True)Sort descending to find kth largest

return xor_values[k - 1]Return kth largest XOR coordinate value

import java.util.*;
public class Solution {
    public int kthLargestValue(int[][] matrix, int k) {
        int m = matrix.length, n = matrix[0].length;
        int[][] prefix = new int[m + 1][n + 1];
        List<Integer> xorValues = new ArrayList<>();
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1];
                xorValues.add(prefix[i][j]);
            }
        }
        Collections.sort(xorValues, Collections.reverseOrder());
        return xorValues.get(k - 1);
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        int[][] matrix = {{5,2},{1,6}};
        int k = 3;
        System.out.println(sol.kthLargestValue(matrix, k)); // Output: 4
    }
}

Line Notes

int[][] prefix = new int[m + 1][n + 1];Prefix XOR matrix with extra row and column for boundaries

for (int i = 1; i <= m; i++) {Iterate rows starting at 1 for prefix indexing

for (int j = 1; j <= n; j++) {Iterate columns starting at 1

prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1];Calculate prefix XOR using inclusion-exclusion

xorValues.add(prefix[i][j]);Collect XOR coordinate values

Collections.sort(xorValues, Collections.reverseOrder());Sort descending for kth largest

return xorValues.get(k - 1);Return kth largest XOR value

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int kthLargestValue(vector<vector<int>>& matrix, int k) {
    int m = matrix.size(), n = matrix[0].size();
    vector<vector<int>> prefix(m + 1, vector<int>(n + 1, 0));
    vector<int> xorValues;
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1];
            xorValues.push_back(prefix[i][j]);
        }
    }
    sort(xorValues.rbegin(), xorValues.rend());
    return xorValues[k - 1];
}

int main() {
    vector<vector<int>> matrix = {{5,2},{1,6}};
    int k = 3;
    cout << kthLargestValue(matrix, k) << endl; // Output: 4
    return 0;
}

Line Notes

vector<vector<int>> prefix(m + 1, vector<int>(n + 1, 0));Prefix XOR matrix with boundary padding

for (int i = 1; i <= m; i++) {Iterate rows starting at 1

for (int j = 1; j <= n; j++) {Iterate columns starting at 1

prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1];Compute prefix XOR with inclusion-exclusion

xorValues.push_back(prefix[i][j]);Collect XOR coordinate values

sort(xorValues.rbegin(), xorValues.rend());Sort descending for kth largest

return xorValues[k - 1];Return kth largest XOR value

function kthLargestValue(matrix, k) {
    const m = matrix.length, n = matrix[0].length;
    const prefix = Array.from({length: m + 1}, () => Array(n + 1).fill(0));
    const xorValues = [];
    for (let i = 1; i <= m; i++) {
        for (let j = 1; j <= n; j++) {
            prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1];
            xorValues.push(prefix[i][j]);
        }
    }
    xorValues.sort((a, b) => b - a);
    return xorValues[k - 1];
}

// Example usage
const matrix = [[5,2],[1,6]];
const k = 3;
console.log(kthLargestValue(matrix, k)); // Output: 4

Line Notes

const prefix = Array.from({length: m + 1}, () => Array(n + 1).fill(0));Initialize prefix XOR matrix with boundary padding

for (let i = 1; i <= m; i++) {Iterate rows starting at 1

for (let j = 1; j <= n; j++) {Iterate columns starting at 1

prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1];Compute prefix XOR with inclusion-exclusion

xorValues.push(prefix[i][j]);Collect XOR coordinate values

xorValues.sort((a, b) => b - a);Sort descending for kth largest

return xorValues[k - 1];Return kth largest XOR value

⏱

Complexity

TimeO(m * n log(m * n)) due to sorting all XOR values

SpaceO(m * n) for prefix matrix and XOR values list

Prefix XOR computation is O(m*n), sorting all values dominates with O(m*n log(m*n)).

💡 For a 1000x1000 matrix, prefix XOR is fast, but sorting 1 million values is still costly but feasible.

Interview Verdict: Accepted

This approach is efficient enough for large inputs and is a common interview solution.

🧠

Prefix XOR Matrix + Min-Heap of Size k

💡 This approach optimizes space and time by avoiding sorting all XOR values, instead maintaining a min-heap of size k to track the kth largest value dynamically.

Intuition

Use prefix XOR to compute all XOR coordinate values, but instead of sorting all, keep a min-heap of size k to store the top k values seen so far.

Algorithm

1. Create prefix XOR matrix as before.
2. Initialize a min-heap to store up to k XOR values.
3. For each prefix XOR value, if heap size < k, push it; else if current value > heap top, pop top and push current.
4. After processing all values, the heap top is the kth largest XOR coordinate value.

💡 Maintaining a min-heap of size k ensures efficient top-k tracking without sorting all values.

</>

Code

import heapq

def kthLargestValue(matrix, k):
    m, n = len(matrix), len(matrix[0])
    prefix = [[0] * (n + 1) for _ in range(m + 1)]
    min_heap = []
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1]
            val = prefix[i][j]
            if len(min_heap) < k:
                heapq.heappush(min_heap, val)
            elif val > min_heap[0]:
                heapq.heapreplace(min_heap, val)
    return min_heap[0]

# Example usage
if __name__ == '__main__':
    matrix = [[5,2],[1,6]]
    k = 3
    print(kthLargestValue(matrix, k))  # Output: 4

Line Notes

import heapqImport heapq module for min-heap operations

min_heap = []Initialize empty min-heap to track top k values

if len(min_heap) < k:If heap not full, push current value

heapq.heappush(min_heap, val)Add value to min-heap

elif val > min_heap[0]:If current value larger than smallest in heap, replace

heapq.heapreplace(min_heap, val)Pop smallest and push current value to maintain size k

return min_heap[0]Heap root is kth largest value after processing all

import java.util.*;
public class Solution {
    public int kthLargestValue(int[][] matrix, int k) {
        int m = matrix.length, n = matrix[0].length;
        int[][] prefix = new int[m + 1][n + 1];
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1];
                int val = prefix[i][j];
                if (minHeap.size() < k) {
                    minHeap.offer(val);
                } else if (val > minHeap.peek()) {
                    minHeap.poll();
                    minHeap.offer(val);
                }
            }
        }
        return minHeap.peek();
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        int[][] matrix = {{5,2},{1,6}};
        int k = 3;
        System.out.println(sol.kthLargestValue(matrix, k)); // Output: 4
    }
}

Line Notes

PriorityQueue<Integer> minHeap = new PriorityQueue<>();Min-heap to track top k XOR values

if (minHeap.size() < k) {Add values until heap size reaches k

minHeap.offer(val);Add current larger value

else if (val > minHeap.peek()) {Replace smallest if current value is larger

minHeap.poll();Remove smallest value from heap

return minHeap.peek();Heap root is kth largest XOR value

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

int kthLargestValue(vector<vector<int>>& matrix, int k) {
    int m = matrix.size(), n = matrix[0].size();
    vector<vector<int>> prefix(m + 1, vector<int>(n + 1, 0));
    priority_queue<int, vector<int>, greater<int>> minHeap;
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1];
            int val = prefix[i][j];
            if ((int)minHeap.size() < k) {
                minHeap.push(val);
            } else if (val > minHeap.top()) {
                minHeap.pop();
                minHeap.push(val);
            }
        }
    }
    return minHeap.top();
}

int main() {
    vector<vector<int>> matrix = {{5,2},{1,6}};
    int k = 3;
    cout << kthLargestValue(matrix, k) << endl; // Output: 4
    return 0;
}

Line Notes

priority_queue<int, vector<int>, greater<int>> minHeap;Min-heap to maintain top k XOR values

if ((int)minHeap.size() < k) {Push values until heap size reaches k

minHeap.push(val);Add current larger value

else if (val > minHeap.top()) {Replace smallest if current value is larger

minHeap.pop();Remove smallest value

return minHeap.top();Heap root is kth largest XOR value

class MinHeap {
    constructor() {
        this.heap = [];
    }
    size() {
        return this.heap.length;
    }
    peek() {
        return this.heap[0];
    }
    push(val) {
        this.heap.push(val);
        this.bubbleUp();
    }
    pop() {
        const top = this.heap[0];
        const end = this.heap.pop();
        if (this.heap.length > 0) {
            this.heap[0] = end;
            this.bubbleDown();
        }
        return top;
    }
    bubbleUp() {
        let idx = this.heap.length - 1;
        const element = this.heap[idx];
        while (idx > 0) {
            let parentIdx = Math.floor((idx - 1) / 2);
            let parent = this.heap[parentIdx];
            if (element >= parent) break;
            this.heap[parentIdx] = element;
            this.heap[idx] = parent;
            idx = parentIdx;
        }
    }
    bubbleDown() {
        let idx = 0;
        const length = this.heap.length;
        const element = this.heap[0];
        while (true) {
            let leftIdx = 2 * idx + 1;
            let rightIdx = 2 * idx + 2;
            let swap = null;
            if (leftIdx < length) {
                if (this.heap[leftIdx] < element) swap = leftIdx;
            }
            if (rightIdx < length) {
                if ((swap === null && this.heap[rightIdx] < element) || (swap !== null && this.heap[rightIdx] < this.heap[leftIdx])) {
                    swap = rightIdx;
                }
            }
            if (swap === null) break;
            this.heap[idx] = this.heap[swap];
            this.heap[swap] = element;
            idx = swap;
        }
    }
}

function kthLargestValue(matrix, k) {
    const m = matrix.length, n = matrix[0].length;
    const prefix = Array.from({length: m + 1}, () => Array(n + 1).fill(0));
    const minHeap = new MinHeap();
    for (let i = 1; i <= m; i++) {
        for (let j = 1; j <= n; j++) {
            prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1];
            const val = prefix[i][j];
            if (minHeap.size() < k) {
                minHeap.push(val);
            } else if (val > minHeap.peek()) {
                minHeap.pop();
                minHeap.push(val);
            }
        }
    }
    return minHeap.peek();
}

// Example usage
const matrix = [[5,2],[1,6]];
const k = 3;
console.log(kthLargestValue(matrix, k)); // Output: 4

Line Notes

class MinHeap {Custom min-heap implementation for top k tracking

minHeap.push(val);Add value to min-heap when size < k

minHeap.peek()Get smallest value in heap to compare

minHeap.pop();Remove smallest value when current is larger

prefix[i][j] = prefix[i - 1][j] ^ prefix[i][j - 1] ^ prefix[i - 1][j - 1] ^ matrix[i - 1][j - 1];Compute prefix XOR with inclusion-exclusion

return minHeap.peek();Return kth largest XOR value after processing all

⏱

Complexity

TimeO(m * n * log k) due to heap operations for each coordinate

SpaceO(m * n) for prefix matrix plus O(k) for heap

Prefix XOR is O(m*n), each heap operation is O(log k), total O(m*n log k).

💡 For k=10 and a 1000x1000 matrix, this is about 1,000,000 * log 10 ≈ 2,000,000 operations, efficient for interviews.

Interview Verdict: Accepted

This is the optimal approach balancing time and space, preferred in interviews for large inputs.

📊

All Approaches - One-Glance Tradeoffs

💡 In interviews, code the prefix XOR + min-heap approach for best balance of clarity and efficiency.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(m^2 * n^2)	O(m * n)	No	N/A	Mention only - never code due to inefficiency
2. Prefix XOR + Sorting	O(m * n log(m * n))	O(m * n)	No	N/A	Good for explaining prefix XOR, acceptable for coding
3. Prefix XOR + Min-Heap	O(m * n log k)	O(m * n + k)	No	N/A	Optimal approach to code in interviews

💼

Interview Strategy

💡 Use this guide to understand the problem deeply, practice coding each approach, and prepare to explain tradeoffs clearly in interviews.

How to Present

Step 1: Clarify problem inputs, outputs, and constraints.Step 2: Describe brute force approach to show understanding.Step 3: Introduce prefix XOR optimization to reduce computation.Step 4: Explain min-heap usage to efficiently find kth largest.Step 5: Code the optimal solution and test with examples.

Time Allocation

Clarify: 3min → Approach: 5min → Code: 10min → Test: 5min. Total ~23min

What the Interviewer Tests

Interviewer tests your ability to optimize nested loops using prefix computations and your knowledge of heaps for top-k selection.

Common Follow-ups

• What if k is very small compared to matrix size? → Min-heap is efficient here.
• Can you do this without sorting all values? → Yes, use min-heap to maintain top k.

💡 Follow-ups often focus on improving efficiency and handling edge cases with data structures.

🔍

Pattern Recognition

When to Use

1) Problem involves 2D prefix computations, 2) XOR or similar associative operation, 3) Need kth largest or smallest value, 4) Efficient top-k data structure needed.

Signature Phrases

'XOR coordinate value''kth largest''prefix XOR matrix''min-heap of size k'

NOT This Pattern When

Problems that only require prefix sums without top-k selection or XOR operations.

Similar Problems

Kth Largest Element in an Array - classic top-k problem using heapsMaximum XOR of Two Numbers in an Array - related XOR prefix techniqueFind K Pairs with Smallest Sums - heap-based top-k selection

Practice

1. You are given an n x n matrix where each row and each column is sorted in ascending order. You need to find the kth smallest element in the matrix efficiently. Which of the following approaches guarantees the optimal time complexity for this problem?

easy

A. Use a brute force approach by flattening the matrix into a list and sorting it.

B. Perform a binary search on the value range between the smallest and largest elements, counting how many elements are less than or equal to mid.

C. Use a dynamic programming approach to build a table of smallest elements up to k.

D. Use a greedy approach by repeatedly picking the smallest element from the first row and first column.

Solution

1. Step 1: Understand problem constraints

   The matrix is sorted both row-wise and column-wise, allowing binary search on the value range.

2. Step 2: Identify optimal approach

   Binary search on the value range uses the sorted property to count elements ≤ mid efficiently, achieving O(n log(max-min)) time.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Binary search on values leverages matrix sorting [OK]

Hint: Binary search on values exploits matrix sorting [OK]

Common Mistakes:

• Thinking brute force or greedy is optimal
• Confusing DP with this problem

2. You need to hire exactly K workers from a pool, each with a quality and a minimum wage expectation. You must pay every hired worker at least their minimum wage, and all hired workers must be paid in the same ratio to their quality. Which approach guarantees finding the minimum total cost to hire exactly K workers under these constraints?

easy

A. Use a brute force approach checking all subsets of size K and calculating total cost for each.

B. Use dynamic programming to select workers minimizing total wage while respecting quality constraints.

C. Sort workers by quality and greedily pick the first K workers with the lowest quality.

D. Sort workers by their wage-to-quality ratio, then use a max-heap or balanced tree to maintain the K smallest qualities while iterating through workers in ratio order.

Solution

1. Step 1: Understand the fairness constraint

   All hired workers must be paid at the same wage-to-quality ratio, so the maximum ratio among chosen workers determines the pay rate.

2. Step 2: Use ratio sorting and a data structure to maintain K workers

   Sorting by ratio ensures we consider possible pay rates in ascending order. Maintaining the K smallest qualities with a max-heap or balanced tree minimizes total quality sum, thus minimizing total cost = ratio * sum of qualities.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Sorting by ratio + max-heap for qualities is the known optimal approach [OK]

Hint: Sort by ratio and maintain K smallest qualities [OK]

Common Mistakes:

• Greedy by quality only ignores wage constraints
• DP approach is not efficient or straightforward here
• Brute force is correct but not efficient

3. What is the time complexity of the add(val) operation in the optimal min-heap approach for maintaining the kth largest element in a stream, assuming the heap size is k and the stream length is n?

medium

A. O(n ∙ log n) per add call, because the stream grows with n

B. O(1) per add call, as heap root access is constant time

C. O(k) per add call, because we may need to scan the heap to find the position

D. O(log k) per add call, since heap operations depend only on heap size k

Solution

1. Step 1: Identify heap size and operations

   The heap size is fixed at k, independent of stream length n. Each add involves either a push or a replace operation on the heap.

2. Step 2: Analyze complexity of heap operations

   Heap push or replace operations take O(log k) time because the heap has size k. Accessing the root is O(1), but add involves heapify steps.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Heap operations depend on k, not n, so O(log k) per add [OK]

Hint: Heap operations scale with heap size k, not stream length n [OK]

Common Mistakes:

• Confusing n with k in complexity
• Assuming linear scan of heap
• Ignoring heapify cost

4. What is the time complexity of the optimal binary search on value range approach to find the kth smallest element in an n x n sorted matrix, where the matrix elements range between min and max values?

medium

A. O(n^2 log n)

B. O(n log(max - min))

C. O(n^2 log(max - min))

D. O(n log n)

Solution

1. Step 1: Identify binary search iterations

   Binary search on value range performs O(log(max - min)) iterations.

2. Step 2: Analyze countLessEqual complexity

   Counting elements ≤ mid takes O(n) time by traversing from bottom-left to top-right.

3. Step 3: Combine complexities

   Total time is O(n * log(max - min)). However, since n is matrix dimension, counting is O(n), but matrix size is n², so counting is O(n) per iteration, total O(n log(max-min)).

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Counting is O(n), binary search O(log(max-min)) [OK]

Hint: Counting ≤ mid is O(n), not O(n²) [OK]

Common Mistakes:

• Confusing n with n² in counting
• Assuming sorting complexity applies

5. What is the time complexity of the optimal bucket sort approach for finding the top k frequent words from a list of n words, assuming the maximum frequency is m and average word length is constant?

medium

A. O(n log n) due to sorting all unique words

B. O(n m) because sorting words within buckets depends on max frequency m

C. O(n) because frequency counting and bucket placement are linear, and sorting within buckets is bounded

D. O(k log n) because extracting top k from a heap dominates

Solution

1. Step 1: Analyze frequency counting

   Counting frequencies is O(n) since each word is processed once.

2. Step 2: Analyze bucket placement and sorting

   Words are placed into buckets indexed by frequency; sorting within each bucket is lexicographic but total words sum to n, so total sorting cost is O(n).

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Overall time is linear in n with lex sort bounded by total words [OK]

Hint: Bucket sort + lex sort within buckets is O(n) total [OK]

Common Mistakes:

• Assuming full sort is needed (O(n log n))
• Confusing max frequency m as multiplicative factor
• Thinking heap extraction dominates