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Problem solution · Python

Largest Local Values in a Matrix II

Largest Local Values in a Matrix II: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Direct simulation
Source
Kamyu LeetCode Solutions
Length
66 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

Direct simulation

For Largest Local Values in a Matrix II, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 66 lines of Python from the credited upstream file largest-local-values-in-a-matrix-ii.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 codeLargest Local Values in a Matrix II · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n * m * logn * logm)# Space: O(n * m * logn * logm) # 2d sparse tableclass Solution(object):    def countLocalMaximums(self, matrix):        """        :type matrix: List[List[int]]        :rtype: int        """        # RMQ - 2D Sparse Table        # Time:  ctor:  O(N * M * logN * logM) * O(fn)        #        query: O(fn)        # Space: O(N * M * logN * logM)        class SparseTable2D(object):            def __init__(self, matrix, fn):                self.fn = fn                n = len(matrix)                m = len(matrix[0])                logn = n.bit_length()-1  # log2_floor(n)                logm = m.bit_length()-1  # log2_floor(m)                self.bit_length = [0]                for i in xrange(max(logn, logm)+1):                    self.bit_length.extend(i+1 for _ in xrange(min(1<<i, (max(n, m)+1)-len(self.bit_length))))                self.st = [[[[0]*m for _ in xrange(n)] for _ in xrange(logm+1)] for _ in xrange(logn+1)]                for r in xrange(n):                    for c in xrange(m):                        self.st[0][0][r][c] = matrix[r][c]                for j in xrange(1, logm+1):                    for r in xrange(n):                        for c in xrange((m-(1<<j))+1):                            self.st[0][j][r][c] = fn(self.st[0][j-1][r][c], self.st[0][j-1][r][c+(1<<(j-1))])                for i in xrange(1, logn+1):                    for j in xrange(logm+1):                        for r in xrange((n-(1<<i))+1):                            for c in xrange((m-(1<<j))+1):                                self.st[i][j][r][c] = fn(self.st[i-1][j][r][c], self.st[i-1][j][r+(1<<(i-1))][c])             def query(self, r1, c1, r2, c2):  # Time: O(fn)                i = self.bit_length[r2-r1+1]-1  # log2_floor(r2-r1+1)                j = self.bit_length[c2-c1+1]-1  # log2_floor(c2-c1+1)                return self.fn(                    self.fn(self.st[i][j][r1][c1], self.st[i][j][r1][c2-(1<<j)+1]),                    self.fn(self.st[i][j][r2-(1<<i)+1][c1], self.st[i][j][r2-(1<<i)+1][c2-(1<<j)+1])                )         n, m = len(matrix), len(matrix[0])        st = SparseTable2D(matrix, max)        result = 0        for r in xrange(n):            row = matrix[r]            for c in xrange(m):                x = row[c]                if x == 0:                    continue                r1, r2 = max(0, r-x), min(n-1, r+x)                c1, c2 = max(0, c-x), min(m-1, c+x)                tl = r-x >= 0 and c-x >= 0                tr = r-x >= 0 and c+x <= m-1                bl = r+x <= n-1 and c-x >= 0                br = r+x <= n-1 and c+x <= m-1                if max(st.query(r1, c1+(1 if tl or bl else 0), r2, c2-(1 if tr or br else 0)),                       st.query(r1+(1 if tl or tr else 0), c1, r2-(1 if bl or br else 0), c2)) <= x:                    result += 1        return result

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