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Problem solution · Python

Minimum Number of Visited Cells in a Grid

Minimum Number of Visited Cells in a Grid: a Python solution using disjoint set union. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Disjoint set union
Source
Kamyu LeetCode Solutions
Length
102 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Disjoint set union

For Minimum Number of Visited Cells in a Grid, the implementation maintains connected components and merges them as relationships are processed.

  1. Give each element a component representative.
  2. Merge representatives when a connection is accepted.
  3. Answer connectivity or component queries from the compressed representatives.

Code notes

  • 102 lines of Python from the credited upstream file minimum-number-of-visited-cells-in-a-grid.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Account for every find and union operation; with path compression and ranked merging, the amortized cost is nearly constant per operation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Number of Visited Cells in a Grid · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(m * n * alpha(m + n)) = O(m + n)# Space: O(m * n) # bfs, union findclass UnionFind(object):  # Time: O(n * alpha(n)), Space: O(n)    def __init__(self, n):        self.set = range(n)        self.rank = [0]*n        self.right = range(n)  # added     def find_set(self, x):        stk = []        while self.set[x] != x:  # path compression            stk.append(x)            x = self.set[x]        while stk:            self.set[stk.pop()] = x        return x     def union_set(self, x, y):        x, y = self.find_set(x), self.find_set(y)        if x == y:            return False        if self.rank[x] > self.rank[y]:  # union by rank            x, y = y, x        self.set[x] = self.set[y]        if self.rank[x] == self.rank[y]:            self.rank[y] += 1        self.right[y] = max(self.right[x], self.right[y])  # added        return True     def right_set(self, x):  # added        return self.right[self.find_set(x)]  class Solution(object):    def minimumVisitedCells(self, grid):        """        :type grid: List[List[int]]        :rtype: int        """        m, n = len(grid), len(grid[0])        uf1 = [UnionFind(n+1) for _ in xrange(m)]        uf2 = [UnionFind(m+1) for _ in xrange(n)]        d, i, j = 1, 0, 0        q = [(i, j)]        uf1[i].union_set(j, j+1)        uf2[j].union_set(i, i+1)        while q:            new_q = []            for i, j in q:                if (i, j) == (m-1, n-1):                    return d                while uf1[i].right_set(j) <= min(j+grid[i][j], n-1):                    k = uf1[i].right_set(j)                    new_q.append((i, k))                    uf2[k].union_set(i, i+1)                    uf1[i].union_set(k, k+1)                while uf2[j].right_set(i) <= min(i+grid[i][j], m-1):                    k = uf2[j].right_set(i)                    new_q.append((k, j))                    uf1[k].union_set(j, j+1)                    uf2[j].union_set(k, k+1)            q = new_q            d += 1        return -1  # Time:  O(m * n * log(m + n))# Space: O(m * n)from sortedcontainers import SortedList  # bfs, sorted listclass Solution2_TLE(object):    def minimumVisitedCells(self, grid):        """        :type grid: List[List[int]]        :rtype: int        """        m, n = len(grid), len(grid[0])        sl1 = [SortedList(xrange(n)) for _ in xrange(m)]        sl2 = [SortedList(xrange(m)) for _ in xrange(n)]        d, i, j = 1, 0, 0        q = [(i, j)]        while q:            new_q = []            for i, j in q:                if (i, j) == (m-1, n-1):                    return d                for k in list(sl1[i].irange(j+1, min(j+grid[i][j], n-1))):                    new_q.append((i, k))                    sl2[k].remove(i)                    sl1[i].remove(k)                for k in list(sl2[j].irange(i+1, min(i+grid[i][j], m-1))):                    new_q.append((k, j))                    sl1[k].remove(j)                    sl2[j].remove(k)            q = new_q            d += 1        return -1

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