Leetcode #711: Number of Distinct Islands II
In this guide, we solve Leetcode #711 Number of Distinct Islands II in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an m x n binary matrix grid. An island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Depth-First Search, Breadth-First Search, Union Find, Hash Table, Hash Function
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: grid = [[1,1,0,0,0],[1,0,0,0,0],[0,0,0,0,1],[0,0,0,1,1]]
Output: 1
Explanation: The two islands are considered the same because if we make a 180 degrees clockwise rotation on the first island, then two islands will have the same shapes.
Python Solution
class Solution:
def numDistinctIslands2(self, grid: List[List[int]]) -> int:
def dfs(i, j, shape):
shape.append([i, j])
grid[i][j] = 0
for a, b in [[1, 0], [-1, 0], [0, 1], [0, -1]]:
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n and grid[x][y] == 1:
dfs(x, y, shape)
def normalize(shape):
shapes = [[] for _ in range(8)]
for i, j in shape:
shapes[0].append([i, j])
shapes[1].append([i, -j])
shapes[2].append([-i, j])
shapes[3].append([-i, -j])
shapes[4].append([j, i])
shapes[5].append([j, -i])
shapes[6].append([-j, i])
shapes[7].append([-j, -i])
for e in shapes:
e.sort()
for i in range(len(e) - 1, -1, -1):
e[i][0] -= e[0][0]
e[i][1] -= e[0][1]
shapes.sort()
return tuple(tuple(e) for e in shapes[0])
m, n = len(grid), len(grid[0])
s = set()
for i in range(m):
for j in range(n):
if grid[i][j]:
shape = []
dfs(i, j, shape)
s.add(normalize(shape))
return len(s)
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.