How do you solve Count Islands With Total Value Divisible by K on LeetCode?

Count Islands With Total Value Divisible by K (LeetCode #3619) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Islands are defined by 4-directional connectivity.

What is the time complexity of Count Islands With Total Value Divisible by K?

The optimal solution for Count Islands With Total Value Divisible by K runs in O(n) time and uses O(n) space. The time complexity is O(n) as we visit each cell once. The space complexity is O(n) due to the visited matrix.

What is the brute force solution for Count Islands With Total Value Divisible by K?

The brute force approach for Count Islands With Total Value Divisible by K is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore each cell in the grid and perform a DFS to find all connected land cells, summing their values. After finding each island, we check if the sum is divisible by k. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Islands With Total Value Divisible by K?

Count Islands With Total Value Divisible by K uses the following concepts: Array, Depth-First Search, Breadth-First Search, Union-Find, Matrix. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Islands With Total Value Divisible by K?

Explain your thought process clearly. Discuss edge cases and constraints upfront. Practice DFS/BFS patterns before the interview.

What are common mistakes when solving Count Islands With Total Value Divisible by K?

Forgetting to mark cells as visited. Not handling edge cases like empty grids.

#3619

Count Islands With Total Value Divisible by K

Medium

ArrayDepth-First SearchBreadth-First SearchUnion-FindMatrixHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using DFS or BFS, we can efficiently find all islands while tracking their total values. We can mark cells as visited to avoid counting them multiple times.

⚙️

Algorithm

3 steps

1Step 1: Initialize a visited matrix to keep track of explored cells.
2Step 2: For each unvisited land cell, perform DFS/BFS to calculate the total value of the island and mark cells as visited.
3Step 3: Check if the total value is divisible by k and update the count accordingly.

solution.py16 lines

1def countIslands(grid, k):
2    def dfs(x, y):
3        if x < 0 or x >= len(grid) or y < 0 or y >= len(grid[0]) or grid[x][y] <= 0:
4            return 0
5        total = grid[x][y]
6        grid[x][y] = -1
7        total += dfs(x+1, y) + dfs(x-1, y) + dfs(x, y+1) + dfs(x, y-1)
8        return total
9    count = 0
10    for i in range(len(grid)):
11        for j in range(len(grid[0])):
12            if grid[i][j] > 0:
13                total_value = dfs(i, j)
14                if total_value % k == 0:
15                    count += 1
16    return count

ℹ

Complexity note: The time complexity is O(n) as we visit each cell once. The space complexity is O(n) due to the visited matrix.

1Islands are defined by 4-directional connectivity.
2The sum of island values must be checked for divisibility by k.

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