Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
You are given two integers m and n. Consider an m x n grid where each cell is initially white. You can paint each cell red, green, or blue. All cells must be painted.
Return the number of ways to color the grid with no two adjacent cells having the same color. Since the answer can be very large, return it modulo 109 + 7.
Example 1:
Input: m = 1, n = 1 Output: 3 Explanation: The three possible colorings are shown in the image above.
Example 2:
Input: m = 1, n = 2 Output: 6 Explanation: The six possible colorings are shown in the image above.
Example 3:
Input: m = 5, n = 5 Output: 580986
Constraints:
1 <= m <= 51 <= n <= 1000Problem summary: You are given two integers m and n. Consider an m x n grid where each cell is initially white. You can paint each cell red, green, or blue. All cells must be painted. Return the number of ways to color the grid with no two adjacent cells having the same color. Since the answer can be very large, return it modulo 109 + 7.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming
1 1
1 2
5 5
number-of-ways-to-paint-n-3-grid)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1931: Painting a Grid With Three Different Colors
class Solution {
private int m;
public int colorTheGrid(int m, int n) {
this.m = m;
final int mod = (int) 1e9 + 7;
int mx = (int) Math.pow(3, m);
Set<Integer> valid = new HashSet<>();
int[] f = new int[mx];
for (int i = 0; i < mx; ++i) {
if (f1(i)) {
valid.add(i);
f[i] = 1;
}
}
Map<Integer, List<Integer>> d = new HashMap<>();
for (int i : valid) {
for (int j : valid) {
if (f2(i, j)) {
d.computeIfAbsent(i, k -> new ArrayList<>()).add(j);
}
}
}
for (int k = 1; k < n; ++k) {
int[] g = new int[mx];
for (int i : valid) {
for (int j : d.getOrDefault(i, List.of())) {
g[i] = (g[i] + f[j]) % mod;
}
}
f = g;
}
int ans = 0;
for (int x : f) {
ans = (ans + x) % mod;
}
return ans;
}
private boolean f1(int x) {
int last = -1;
for (int i = 0; i < m; ++i) {
if (x % 3 == last) {
return false;
}
last = x % 3;
x /= 3;
}
return true;
}
private boolean f2(int x, int y) {
for (int i = 0; i < m; ++i) {
if (x % 3 == y % 3) {
return false;
}
x /= 3;
y /= 3;
}
return true;
}
}
// Accepted solution for LeetCode #1931: Painting a Grid With Three Different Colors
func colorTheGrid(m int, n int) (ans int) {
f1 := func(x int) bool {
last := -1
for i := 0; i < m; i++ {
if x%3 == last {
return false
}
last = x % 3
x /= 3
}
return true
}
f2 := func(x, y int) bool {
for i := 0; i < m; i++ {
if x%3 == y%3 {
return false
}
x /= 3
y /= 3
}
return true
}
mx := int(math.Pow(3, float64(m)))
valid := map[int]bool{}
f := make([]int, mx)
for i := 0; i < mx; i++ {
if f1(i) {
valid[i] = true
f[i] = 1
}
}
d := map[int][]int{}
for i := range valid {
for j := range valid {
if f2(i, j) {
d[i] = append(d[i], j)
}
}
}
const mod int = 1e9 + 7
for k := 1; k < n; k++ {
g := make([]int, mx)
for i := range valid {
for _, j := range d[i] {
g[i] = (g[i] + f[j]) % mod
}
}
f = g
}
for _, x := range f {
ans = (ans + x) % mod
}
return
}
# Accepted solution for LeetCode #1931: Painting a Grid With Three Different Colors
class Solution:
def colorTheGrid(self, m: int, n: int) -> int:
def f1(x: int) -> bool:
last = -1
for _ in range(m):
if x % 3 == last:
return False
last = x % 3
x //= 3
return True
def f2(x: int, y: int) -> bool:
for _ in range(m):
if x % 3 == y % 3:
return False
x, y = x // 3, y // 3
return True
mod = 10**9 + 7
mx = 3**m
valid = {i for i in range(mx) if f1(i)}
d = defaultdict(list)
for x in valid:
for y in valid:
if f2(x, y):
d[x].append(y)
f = [int(i in valid) for i in range(mx)]
for _ in range(n - 1):
g = [0] * mx
for i in valid:
for j in d[i]:
g[i] = (g[i] + f[j]) % mod
f = g
return sum(f) % mod
// Accepted solution for LeetCode #1931: Painting a Grid With Three Different Colors
/**
* [1931] Painting a Grid With Three Different Colors
*
* You are given two integers m and n. Consider an m x n grid where each cell is initially white. You can paint each cell red, green, or blue. All cells must be painted.
* Return the number of ways to color the grid with no two adjacent cells having the same color. Since the answer can be very large, return it modulo 10^9 + 7.
*
* Example 1:
* <img alt="" src="https://assets.leetcode.com/uploads/2021/06/22/colorthegrid.png" style="width: 200px; height: 50px;" />
* Input: m = 1, n = 1
* Output: 3
* Explanation: The three possible colorings are shown in the image above.
*
* Example 2:
* <img alt="" src="https://assets.leetcode.com/uploads/2021/06/22/copy-of-colorthegrid.png" style="width: 321px; height: 121px;" />
* Input: m = 1, n = 2
* Output: 6
* Explanation: The six possible colorings are shown in the image above.
*
* Example 3:
*
* Input: m = 5, n = 5
* Output: 580986
*
*
* Constraints:
*
* 1 <= m <= 5
* 1 <= n <= 1000
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/painting-a-grid-with-three-different-colors/
// discuss: https://leetcode.com/problems/painting-a-grid-with-three-different-colors/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
impl Solution {
pub fn color_the_grid(m: i32, n: i32) -> i32 {
0
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
#[ignore]
fn test_1931_example_1() {
let m = 1;
let n = 1;
let result = 3;
assert_eq!(Solution::color_the_grid(m, n), result);
}
#[test]
#[ignore]
fn test_1931_example_2() {
let m = 1;
let n = 2;
let result = 6;
assert_eq!(Solution::color_the_grid(m, n), result);
}
#[test]
#[ignore]
fn test_1931_example_3() {
let m = 5;
let n = 5;
let result = 580986;
assert_eq!(Solution::color_the_grid(m, n), result);
}
}
// Accepted solution for LeetCode #1931: Painting a Grid With Three Different Colors
function colorTheGrid(m: number, n: number): number {
const f1 = (x: number): boolean => {
let last = -1;
for (let i = 0; i < m; ++i) {
if (x % 3 === last) {
return false;
}
last = x % 3;
x = Math.floor(x / 3);
}
return true;
};
const f2 = (x: number, y: number): boolean => {
for (let i = 0; i < m; ++i) {
if (x % 3 === y % 3) {
return false;
}
x = Math.floor(x / 3);
y = Math.floor(y / 3);
}
return true;
};
const mx = 3 ** m;
const valid = new Set<number>();
const f: number[] = Array(mx).fill(0);
for (let i = 0; i < mx; ++i) {
if (f1(i)) {
valid.add(i);
f[i] = 1;
}
}
const d: Map<number, number[]> = new Map();
for (const i of valid) {
for (const j of valid) {
if (f2(i, j)) {
d.set(i, (d.get(i) || []).concat(j));
}
}
}
const mod = 10 ** 9 + 7;
for (let k = 1; k < n; ++k) {
const g: number[] = Array(mx).fill(0);
for (const i of valid) {
for (const j of d.get(i) || []) {
g[i] = (g[i] + f[j]) % mod;
}
}
f.splice(0, f.length, ...g);
}
let ans = 0;
for (const x of f) {
ans = (ans + x) % mod;
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.