How do you solve Find the Grid of Region Average on LeetCode?

Find the Grid of Region Average (LeetCode #3030) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n * α(m * n)) time complexity and O(m * n) space complexity. Key insight: Regions can overlap, and a pixel can belong to multiple regions.

What is the brute force solution for Find the Grid of Region Average?

The brute force approach for Find the Grid of Region Average is Brute Force, with O(m * n * 9) = O(m * n) time complexity and O(1) space complexity. The brute force approach involves checking every possible 3x3 subgrid in the image to determine if it forms a valid region based on the threshold. If valid, we calculate the average intensity for that region and update the result grid accordingly. The optimal Optimal Solution approach improves this to O(m * n * α(m * n)).

What algorithm or data structure is used to solve Find the Grid of Region Average?

Find the Grid of Region Average uses the following concepts: Array, Matrix. The recommended patterns to study are: Union-Find, Graph Traversal.

What are the interview tips for Find the Grid of Region Average?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases and how they might affect your solution. Explain your thought process clearly to the interviewer as you code.

What are common mistakes when solving Find the Grid of Region Average?

Not considering edge cases where pixels are at the borders of the grid. Failing to correctly calculate the average when a pixel belongs to multiple regions.

#3030

Find the Grid of Region Average

Medium

ArrayMatrixUnion-FindGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m * n * 9) = O(m * n)	O(m * n * α(m * n))
Space	O(1)	O(m * n)

💡

Intuition

Time O(m * n * α(m * n))Space O(m * n)

The optimal approach uses a more efficient method to identify regions by leveraging a union-find structure or a similar grouping technique. This allows us to efficiently calculate the average intensity of overlapping regions without redundant checks.

⚙️

Algorithm

3 steps

1Step 1: Create a union-find structure to group pixels into regions based on the threshold condition.
2Step 2: For each pixel, check its neighbors and union them if they meet the threshold condition.
3Step 3: After forming regions, calculate the average intensity for each region and update the result grid accordingly.

solution.py59 lines

1# Full working Python code
2class UnionFind:
3    def __init__(self, size):
4        self.parent = list(range(size))
5        self.rank = [1] * size
6        self.sum = [0] * size
7        self.count = [0] * size
8
9    def find(self, x):
10        if self.parent[x] != x:
11            self.parent[x] = self.find(self.parent[x])
12        return self.parent[x]
13
14    def union(self, x, y, image):
15        rootX = self.find(x)
16        rootY = self.find(y)
17        if rootX != rootY:
18            if self.rank[rootX] > self.rank[rootY]:
19                self.parent[rootY] = rootX
20                self.sum[rootX] += self.sum[rootY]
21                self.count[rootX] += self.count[rootY]
22            elif self.rank[rootX] < self.rank[rootY]:
23                self.parent[rootX] = rootY
24                self.sum[rootY] += self.sum[rootX]
25                self.count[rootY] += self.count[rootX]
26            else:
27                self.parent[rootY] = rootX
28                self.sum[rootX] += self.sum[rootY]
29                self.count[rootX] += self.count[rootY]
30                self.rank[rootX] += 1
31
32    def add_pixel(self, index, value):
33        self.sum[index] = value
34        self.count[index] = 1
35
36
37def find_region_average(image, threshold):
38    m, n = len(image), len(image[0])
39    uf = UnionFind(m * n)
40    for i in range(m):
41        for j in range(n):
42            index = i * n + j
43            uf.add_pixel(index, image[i][j])
44            for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
45                ni, nj = i + dx, j + dy
46                if 0 <= ni < m and 0 <= nj < n:
47                    if abs(image[i][j] - image[ni][nj]) <= threshold:
48                        uf.union(index, ni * n + nj, image)
49    result = [[0] * n for _ in range(m)]
50    for i in range(m):
51        for j in range(n):
52            index = i * n + j
53            root = uf.find(index)
54            if uf.count[root] > 0:
55                result[i][j] = uf.sum[root] // uf.count[root]
56            else:
57                result[i][j] = image[i][j]
58    return result
59

ℹ

Complexity note: The time complexity is O(m * n * α(m * n)) due to the union-find operations, where α is the inverse Ackermann function, which grows very slowly. The space complexity is O(m * n) for storing the union-find structure.

1Regions can overlap, and a pixel can belong to multiple regions.
2Efficiently grouping pixels using union-find can significantly reduce redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.