Minimum XOR Path in a Grid
Time O(m * n * r) · Space O(n * r) · Official statement on LeetCode
Solutions
// Time: O(m * n * r), r = max(x for row in grid for x in row)
// Space: O(n * r / 64)
// dp, bitset
class Solution {
public:
int minCost(vector<vector<int>>& grid) {
static const int MAX_LEN = 1024;
vector<bitset<MAX_LEN>> dp(size(grid[0]));
dp[0].set(0);
for (int i = 0; i < size(grid); ++i) {
vector<bitset<MAX_LEN>> new_dp(size(grid[0]));
for (int j = 0; j < size(grid[0]); ++j) {
for (int k = 0; k < MAX_LEN; ++k) {
if (dp[j][k] || (j - 1 >= 0 && new_dp[j - 1][k])) {
new_dp[j].set(k ^ grid[i][j]);
}
}
}
dp = move(new_dp);
}
for (int i = 0; i < MAX_LEN; ++i) {
if (dp.back()[i]) {
return i;
}
}
return -1;
}
};
// Time: O(m * n * r), r = max(x for row in grid for x in row)
// Space: O(n * r)
// dp
class Solution2 {
public:
int minCost(vector<vector<int>>& grid) {
uint32_t mx = 0;
for (const auto& row : grid) {
mx = max(mx, static_cast<uint32_t>(ranges::max(row)));
}
const auto& l = 1 << bit_width(mx);
vector<vector<bool>> dp(size(grid[0]), vector<bool>(l));
dp[0][0] = true;
for (int i = 0; i < size(grid); ++i) {
vector<vector<bool>> new_dp(size(grid[0]), vector<bool>(l));
for (int j = 0; j < size(grid[0]); ++j) {
for (int k = 0; k < l; ++k) {
if (dp[j][k] || (j - 1 >= 0 && new_dp[j - 1][k])) {
new_dp[j][k ^ grid[i][j]] = true;
}
}
}
dp = move(new_dp);
}
for (int i = 0; i < l; ++i) {
if (dp.back()[i]) {
return i;
}
}
return -1;
}
};
// Time: O(m * n * r), r = max(x for row in grid for x in row)
// Space: O(n * r)
// dp
class Solution_TLE {
public:
int minCost(vector<vector<int>>& grid) {
vector<unordered_set<int>> dp(size(grid[0]));
dp[0].emplace(0);
for (int i = 0; i < size(grid); ++i) {
vector<unordered_set<int>> new_dp(size(grid[0]));
for (int j = 0; j < size(grid[0]); ++j) {
for (const auto& k : dp[j]) {
new_dp[j].emplace(k ^ grid[i][j]);
}
if (j - 1 < 0) {
continue;
}
for (const auto& k : new_dp[j - 1]) {
new_dp[j].emplace(k ^ grid[i][j]);
}
}
dp = move(new_dp);
}
return ranges::min(dp.back());
}
};
Beginner Explanation
What is Minimum XOR Path in a Grid?
Minimum XOR Path in a Grid (LeetCode #3882) is a Medium problem that primarily trains dynamic programming.
How to think about it
- Restate the goal in your own words before coding.
- Work a tiny example by hand so the invariant becomes obvious.
- Identify the pattern — this problem aligns with dynamic programming.
- Only then translate the idea into code.
Why this problem matters
It sits in the sweet spot of interview difficulty: multiple valid approaches, clear trade-offs. Official solution notes mention: DP.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Minimum XOR Path in a Grid
Opening (30–60 seconds)
- Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
- State a brute force so the interviewer knows you can solve it naively.
- Propose the optimal direction tied to dynamic programming.
Core solution narrative
- Define the state you track (pointers, DP cell, set membership, stack top, etc.).
- Explain the transition when you process the next element.
- Call out time (O(m * n * r)) and space (O(n * r)) before coding.
- Code cleanly; narrate variable names.
What interviewers listen for
- Correctness on edge cases
- Complexity honesty
- Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)
Follow-up questions they may ask
- Can you solve it with less memory?
- What if the input stream is infinite / doesn't fit in RAM?
- How would tests look for adversarial inputs?
Optimized Approach
Optimized solution notes
The reference solutions on AlgoForge target O(m * n * r) time and O(n * r) space.
Pattern focus: dynamic programming
Use the pattern as a checklist:
- dynamic programming — confirm the invariant holds after each step
Multiple methods appear in the source solutions — compare them and explain when each is preferable.
Implementation tips
- Prefer readable names over micro-optimizations in interviews.
- Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
- After AC-level logic, re-scan for off-by-one and null checks.
Complexity Analysis
Complexity
| Measure | Bound |
|---|---|
| Time | O(m * n * r) |
| Space | O(n * r) |
How to justify this in an interview
- Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
- Space: include hash maps, recursion stack, and output allocation when the problem asks for it.
If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.
Common Mistakes
Common mistakes on Minimum XOR Path in a Grid
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for dynamic programming — updating state too early or too late.
- Mutating input unexpectedly when the problem forbids it.
- Off-by-one in windows, ranges, or binary search bounds.
- Ignoring overflow / precision for integer arithmetic problems.
- Overengineering — jumping to an advanced structure when a simpler approach works.
Alternative Approaches
Alternatives
The source file includes more than one method. Compare:
- Primary optimized path — best complexity for typical interviews.
- Secondary approach — often brute force, sorting-based, or space-optimized variant.
Practice articulating when you would pick each (constraints, readability, follow-ups).
Edge Cases
Edge cases checklist
- Minimum input size
- Maximum input size / time limits
- Duplicates and already-sorted input
- Negative numbers / zeros (if applicable)
- Disconnected structures (graphs/trees)
- Single path vs branching recursion depth
Pattern Recognition
Spotting this pattern
Signal phrases that point to dynamic programming:
- Sorted input or ability to sort without changing the answer class
- Need for contiguous subarray / substring → consider sliding window
- Need for O(1) membership → hash set/map
- Optimal substructure + overlapping subproblems → DP
- Connectivity / components → graph DFS/BFS or Union-Find
Primary topics: dynamic programming.
Follow-up Interview Questions
Follow-ups
- How does the solution change if the input is a stream?
- Can you solve it in-place?
- What if duplicates must be handled differently?
- How would you parallelize the approach?
- Design tests that would break a buggy implementation.
Practice Recommendations
What to practice next
- Re-solve Minimum XOR Path in a Grid in a second language (cpp, python).
- Drill 3–5 more problems tagged dynamic programming.
- Teach the solution out loud in under 5 minutes.
- Add this problem to your revision calendar in 3 days and 14 days.
Visualization
Study checklist
- Read the official problem statement on LeetCode
- Solve on paper / whiteboard first
- Implement the dynamic programming approach
- Verify edge cases from the checklist
- State time and space complexity aloud
- Compare with the AlgoForge reference solution
- Schedule a revision session
Revision notes
Minimum XOR Path in a Grid (#3882) — Medium. Pattern: dynamic programming. Complexity: O(m * n * r) time / O(n * r) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Minimum XOR Path in a Grid?+
The reference solutions aim for O(m * n * r) time and O(n * r) space. Always re-derive complexity from the code you write in the interview.
What pattern does Minimum XOR Path in a Grid use?+
It primarily maps to dynamic programming, within the broader topic of dynamic programming.
Is Minimum XOR Path in a Grid good for interviews?+
Yes — as a Medium problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.
Where can I read the official statement?+
Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/minimum-xor-path-in-a-grid/