#1931Hard~60 min
Painting a Grid With Three Different Colors
Time O(2^(3 * m) * logn) · Space O(2^(2 * m)) · Official statement on LeetCode
Solutions
// Time: O(m * 2^m + 3^m + 2^(3 * m) * logn) = O(2^(3 * m) * logn)
// Space: O(2^(2 * m))
// better complexity for small m, super large n
class Solution {
public:
int colorTheGrid(int m, int n) {
if (m > n) {
swap(m, n);
}
const int basis = pow(3, m - 1);
vector<int> masks;
backtracking(-1, -1, basis, &masks); // Time: O(2^m), Space: O(2^m)
assert(size(masks) == 3 * pow(2, m - 1));
unordered_map<int, int> lookup;
for (const auto& mask : masks) { // Time: O(m * 2^m)
lookup[mask] = normalize(basis, mask);
}
unordered_map<int, int> normalized_mask_cnt;
for (const auto& mask : masks) {
normalized_mask_cnt[lookup[mask]] = (normalized_mask_cnt[lookup[mask]] + 1) % MOD;
}
assert(size(normalized_mask_cnt) == 3 * pow(2, m - 1) / 3 / (m >= 2 ? 2 : 1)); // divided by 3 * 2 is since the first two colors are normalized to speed up performance
unordered_map<int, vector<int>> adj;
for (const auto& [mask, _] : normalized_mask_cnt) { // O(3^m) leaves which are all in depth m => Time: O(3^m), Space: O(3^m)
backtracking(mask, -1, basis, &adj[mask]);
}
unordered_map<int, unordered_map<int, int>> normalized_adj;
for (const auto& [mask1, mask2s] : adj) {
for (const auto& mask2 : mask2s) {
normalized_adj[mask1][lookup[mask2]] = (normalized_adj[mask1][lookup[mask2]] + 1) % MOD;
}
}
// divided by 3 * 2 is since the first two colors in upper row are normalized to speed up performance
assert(accumulate(cbegin(normalized_adj), cend(normalized_adj), 0,
[](const auto& total, const auto& kvp) {
return total + size(kvp.second);
}) <= 2 * pow(3, m) / 3 / 2);
// since first two colors in lower row which has at most 3 choices could be also normalized, lower bound is upper bound divided by at most 3
assert(accumulate(cbegin(normalized_adj), cend(normalized_adj), 0,
[](const auto& total, const auto& kvp) {
return total + size(kvp.second);
}) >= 2 * pow(3, m) / 3 / 2 / 3);
vector<vector<int>> matrix;
vector<vector<int>> counts(1);
for (const auto& [mask1, cnt] : normalized_mask_cnt) {
matrix.emplace_back();
for (const auto& [mask2, cnt] : normalized_mask_cnt) {
matrix.back().emplace_back(normalized_adj[mask1][mask2]);
}
counts[0].emplace_back(cnt);
}
const auto& result = matrixMult(counts, matrixExpo(matrix, n - 1)); // Time: O((2^m)^3 * logn), Space: O((2^m)^2)
return accumulate(cbegin(result[0]), cend(result[0]), 0,
[](const auto& total, const auto& x) {
return (total + x) % MOD;
}); // Time: O(2^m)
}
private:
void backtracking(int mask1, int mask2, int basis, vector<int> *result) { // Time: O(2^m), Space: O(2^m)
if (!basis) {
result->emplace_back(mask2);
return;
}
for (int i = 0; i < 3; ++i) {
if ((mask1 == -1 || mask1 / basis % 3 != i) && (mask2 == -1 || mask2 / (basis * 3) % 3 != i)) {
backtracking(mask1, mask2 != -1 ? mask2 + i * basis : i * basis, basis / 3, result);
}
}
}
vector<vector<int>> matrixExpo(const vector<vector<int>>& A, int pow) {
vector<vector<int>> result(A.size(), vector<int>(A.size()));
vector<vector<int>> A_exp(A);
for (int i = 0; i < A.size(); ++i) {
result[i][i] = 1;
}
while (pow) {
if (pow % 2 == 1) {
result = matrixMult(result, A_exp);
}
A_exp = matrixMult(A_exp, A_exp);
pow /= 2;
}
return result;
}
vector<vector<int>> matrixMult(const vector<vector<int>>& A, const vector<vector<int>>& B) {
vector<vector<int>> result(A.size(), vector<int>(B[0].size()));
for (int i = 0; i < A.size(); ++i) {
for (int j = 0; j < B[0].size(); ++j) {
int64_t entry = 0;
for (int k = 0; k < B.size(); ++k) {
entry = (static_cast<int64_t>(A[i][k]) * B[k][j] % MOD + entry) % MOD;
}
result[i][j] = static_cast<int>(entry);
}
}
return result;
}
int normalize(int basis, int mask) {
unordered_map<int, int> norm;
int result = 0;
for (; basis; basis /= 3) {
int x = mask / basis % 3;
if (!norm.count(x)) {
norm[x] = size(norm);
}
result += norm[x] * basis;
}
return result;
}
static const int MOD = 1e9 + 7;
};
// Time: O(n * 3^m)
// Space: O(3^m)
// better complexity for small m, large n
class Solution2 {
public:
int colorTheGrid(int m, int n) {
static const int MOD = 1e9 + 7;
if (m > n) {
swap(m, n);
}
const int basis = pow(3, m - 1);
const auto& masks = find_masks(m, basis); // alternative of backtracking, Time: O(2^m), Space: O(2^m)
assert(size(masks) == 3 * pow(2, m - 1));
unordered_map<int, int> lookup;
for (const auto& mask : masks) { // Time: O(m * 2^m)
lookup[mask] = normalize(basis, mask);
}
unordered_map<int, int> dp;
for (const auto& mask : masks) { // normalize colors to speed up performance
++dp[lookup[mask]];
}
const auto& adj = find_adj(m, basis, dp); // alternative of backtracking, Time: O(3^m), Space: O(3^m)
// proof:
// 'o' uses the same color with its bottom-left one,
// 'x' uses the remaining color different from its left one and bottom-left one,
// k is the cnt of 'o',
// [3, 1(o), 1(x), 1(o), ..., 1(o), 1(x)] => nCr(m-1, k) * 3 * 2 * 2^k for k in xrange(m) = 3 * 2 * (2+1)^(m-1) = 2*3^m combinations
// [2, 2, 1, 2, ..., 2, 1]
// another proof:
// given previous pair of colors, each pair of '?' has 3 choices of colors
// [3, ?, ?, ..., ?] => 3 * 2 * 3^(m-1) = 2*3^m combinations
// | | |
// 3 3 3
// | | |
// [2, ?, ?, ..., ?]
unordered_map<int, unordered_map<int, int>> normalized_adj;
for (const auto& [mask1, mask2s] : adj) {
for (const auto& mask2 : mask2s) {
normalized_adj[lookup[mask1]][lookup[mask2]] = (normalized_adj[lookup[mask1]][lookup[mask2]] + 1) % MOD;
}
}
// divided by 3 * 2 is since the first two colors in upper row are normalized to speed up performance
assert(accumulate(cbegin(normalized_adj), cend(normalized_adj), 0,
[](const auto& total, const auto& kvp) {
return total + size(kvp.second);
}) <= 2 * pow(3, m) / 3 / 2);
// since first two colors in lower row which has at most 3 choices could be also normalized, lower bound is upper bound divided by at most 3
assert(accumulate(cbegin(normalized_adj), cend(normalized_adj), 0,
[](const auto& total, const auto& kvp) {
return total + size(kvp.second);
}) >= 2 * pow(3, m) / 3 / 2 / 3);
for (int i = 0; i < n - 1; ++i) { // Time: O(n * 3^m), Space: O(2^m)
assert(size(dp) == 3 * pow(2, m - 1) / 3 / (m >= 2 ? 2 : 1)); // divided by 3 * 2 is since the first two colors are normalized to speed up performance
unordered_map<int, int> new_dp;
for (const auto [mask, v] : dp) {
for (const auto& [new_mask, cnt] : normalized_adj[mask]) {
new_dp[lookup[new_mask]] = (new_dp[lookup[new_mask]] + (v * int64_t(cnt)) % MOD) % MOD;
}
}
dp = move(new_dp);
}
return accumulate(cbegin(dp), cend(dp), 0,
[](const auto& total, const auto& kvp) {
return (total + kvp.second) % MOD;
}); // Time: O(2^m)
}
private:
vector<int> find_masks(int m, int basis) { // Time: 3 + 3*2 + 3*2*2 + ... + 3*2^(m-1) = 3 * (2^m - 1) = O(2^m), Space: O(2^m)
vector<int> masks = {0};
for (int c = 0; c < m; ++c) {
vector<int> new_masks;
for (const auto& mask : masks) {
vector<bool> used(3);
if (c > 0) {
used[mask / basis] = true; // get left grid
}
for (int x = 0; x < 3; ++x) {
if (used[x]) {
continue;
}
new_masks.emplace_back((x * basis) + (mask / 3)); // encoding mask
}
}
masks = move(new_masks);
}
return masks;
}
unordered_map<int, vector<int>> find_adj(int m, int basis, const unordered_map<int, int>& dp) {
// Time: 3*2^(m-1) * (1 + 2 + 2 * (3/2) + 2 * (3/2)^2 + ... + 2 * (3/2)^(m-2)) =
// 3*2^(m-1) * (1+2*((3/2)^(m-1)-1)/((3/2)-1)) =
// 3*2^(m-1) * (1+4*((3/2)^(m-1)-1)) =
// 3*2^(m-1) * (4*(3/2)^(m-1)-3) =
// 4*3^m-9*2^(m-1) =
// O(3^m),
// Space: O(3^m)
unordered_map<int, vector<int>> adj;
for (const auto& [mask, _] : dp) { // O(2^m)
adj[mask].emplace_back(mask);
}
for (int c = 0; c < m; ++c) {
assert(accumulate(cbegin(adj), cend(adj), 0,
[](const auto& total, const auto& kvp) {
return total + size(kvp.second);
}) == (c ? pow(3, c) * pow(2, m - (c - 1)) : 3 * pow(2, m - 1)) / 3 / (m >= 2 ? 2 : 1)); // divided by 3 * 2 is since the first two colors are normalized to speed up performance
unordered_map<int, vector<int>> new_adj;
for (const auto& [mask1, mask2s] : adj) {
for (const auto& mask : mask2s) {
vector<bool> used(3);
used[mask % 3] = true; // get up grid
if (c > 0) {
used[mask / basis] = true; // get left grid
}
for (int x = 0; x < 3; ++x) {
if (used[x]) {
continue;
}
new_adj[mask1].emplace_back((x * basis) + (mask / 3)); // encoding mask
}
}
}
adj = move(new_adj);
}
return adj;
}
int normalize(int basis, int mask) {
unordered_map<int, int> norm;
int result = 0;
for (; basis; basis /= 3) {
int x = mask / basis % 3;
if (!norm.count(x)) {
norm[x] = size(norm);
}
result += norm[x] * basis;
}
return result;
}
};
// Time: (m * n grids) * (O(3*3*2^(m-2)) possible states per grid) = O(n * m * 2^m)
// Space: O(3*3*2^(m-2)) = O(2^m)
// better complexity for large m, large n
class Solution3 {
public:
int colorTheGrid(int m, int n) {
static const int MOD = 1e9 + 7;
if (m > n) {
swap(m, n);
}
const int basis = pow(3, m - 1);
int b = basis;
unordered_map<int, unordered_map<int, int>> lookup;
unordered_map<int, int> dp = {{0, 1}};
for (int idx = 0; idx < m * n; ++idx) {
int r = idx / m;
int c = idx % m;
// sliding window with size m doesn't cross rows:
// [3, 2, ..., 2] => 3*2^(m-1) combinations
assert(r != 0 || c != 0 || size(dp) == 1);
assert(r != 0 || c == 0 || size(dp) == 3 * pow(2, c - 1) / 3 / (c >= 2 ? 2 : 1)); // divided by 3 * 2 is since the first two colors are normalized to speed up performance
assert(r == 0 || c != 0 || size(dp) == 3 * pow(2, m - 1) / 3 / (m >= 2 ? 2 : 1)); // divided by 3 * 2 is since the first two colors are normalized to speed up performance
// sliding window with size m crosses rows:
// [*, ..., *, *, 3, 2, ..., 2] => 3*3 * 2^(m-2) combinations
// [2, ..., 2, 3, *, *, ..., *]
assert(r == 0 || c == 0 || size(dp) == (m == 1 ? 1 : m == 2 ? 2 : (3 * 3 * pow(2, m - 2) / 3 / 2))); // divided by 3 * 2 for m >= 3 is since the first two colors of window are normalized to speed up performance
unordered_map<int, int> new_dp;
for (const auto [mask, v] : dp) {
vector<bool> used(3);
if (r > 0) {
used[mask % 3] = true; // get up grid
}
if (c > 0) {
used[mask / basis] = true; // get left grid
}
for (int x = 0; x < 3; ++x) {
if (used[x]) {
continue;
}
const auto new_mask = normalize(basis / b, ((x * basis) + (mask / 3)) / b, &lookup) * b; // encoding mask
new_dp[new_mask] = (new_dp[new_mask] + v) % MOD;
}
}
if (b > 1) {
b /= 3;
}
dp = move(new_dp);
}
return accumulate(cbegin(dp), cend(dp), 0,
[](const auto& total, const auto& kvp) {
return (total + kvp.second) % MOD;
}); // Time: O(2^m)
}
int normalize(int basis, int mask, unordered_map<int, unordered_map<int, int>> *lookup) { // compute and cache, at most O(3*2^(m-3)) time and space
if (!(*lookup)[basis].count(mask)) {
unordered_map<int, int> norm;
int result = 0;
for (int b = basis; b; b /= 3) {
int x = mask / b % 3;
if (!norm.count(x)) {
norm[x] = size(norm);
}
result += norm[x] * b;
}
(*lookup)[basis][mask] = result;
}
return (*lookup)[basis][mask];
}
};
Beginner Explanation
What is Painting a Grid With Three Different Colors?
Painting a Grid With Three Different Colors (LeetCode #1931) is a Hard 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 and dfs backtracking.
- Only then translate the idea into code.
Why this problem matters
Hard problems force you to combine patterns and prove complexity carefully — interview gold. Official solution notes mention: DP, Backtracking, Matrix Exponentiation, State Compression.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Painting a Grid With Three Different Colors
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 and dfs backtracking.
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(2^(3 * m) * logn)) and space (O(2^(2 * m))) 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(2^(3 * m) * logn) time and O(2^(2 * m)) space.
Pattern focus: dynamic programming and dfs backtracking
Use the pattern as a checklist:
- dynamic programming — confirm the invariant holds after each step
- dfs backtracking — 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(2^(3 * m) * logn) |
| Space | O(2^(2 * m)) |
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 Painting a Grid With Three Different Colors
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for dynamic programming and dfs backtracking — 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 and dfs backtracking:
- 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 Painting a Grid With Three Different Colors 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
Conceptual diagram for Painting a Grid With Three Different Colors: show input structure (dynamic programming), highlight the moving parts of the dynamic programming and dfs backtracking approach, and annotate each step with the maintained invariant and complexity.
Study checklist
- Read the official problem statement on LeetCode
- Solve on paper / whiteboard first
- Implement the dynamic programming and dfs backtracking 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
Painting a Grid With Three Different Colors (#1931) — Hard. Pattern: dynamic programming and dfs backtracking. Complexity: O(2^(3 * m) * logn) time / O(2^(2 * m)) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Painting a Grid With Three Different Colors?+
The reference solutions aim for O(2^(3 * m) * logn) time and O(2^(2 * m)) space. Always re-derive complexity from the code you write in the interview.
What pattern does Painting a Grid With Three Different Colors use?+
It primarily maps to dynamic programming and dfs backtracking, within the broader topic of dynamic programming.
Is Painting a Grid With Three Different Colors good for interviews?+
Yes — as a Hard 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/painting-a-grid-with-three-different-colors/