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Problem solution · Python

Maximize Grid Happiness

Maximize Grid Happiness: a Python solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
68 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

Dynamic programming

For Maximize Grid Happiness, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 68 lines of Python from the credited upstream file 1659.py.
  • The implementation visibly relies on hash lookup, cached states.
  • No explicit loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximize Grid Happiness · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def getMaxGridHappiness(      self,      m: int,      n: int,      introvertsCount: int,      extrovertsCount: int,  ) -> int:    def getPlacementCost(        i: int,        j: int,        inMask: int,        exMask: int,        diff: int,    ) -> int:      """Calculates the cost based on left and up neighbors.       The `diff` parameter represents the happiness change due to the current      placed person in (i, j). We add `diff` each time we encounter a neighbor      (left or up) who is already placed.       1. If the neighbor is an introvert, we subtract 30 from cost.      2. If the neighbor is an extrovert, we add 20 to from cost.      """      cost = 0      if i > 0:        if (1 << (n - 1)) & inMask:          cost += diff - 30        if (1 << (n - 1)) & exMask:          cost += diff + 20      if j > 0:        if 1 & inMask:          cost += diff - 30        if 1 & exMask:          cost += diff + 20      return cost     @functools.lru_cache(None)    def dp(        pos: int, inMask: int, exMask: int, inCount: int, exCount: int    ) -> int:      # `inMask` is the placement of introvert people in the last n cells.      # e.g. if we have m = 2, n = 3, i = 1, j = 1, then inMask = 0b101 means      #      # ? 1 0      # 1 x ? (x := current position)      i, j = divmod(pos, n)      if i == m:        return 0       shiftedInMask = (inMask << 1) & ((1 << n) - 1)      shiftedExMask = (exMask << 1) & ((1 << n) - 1)       skip = dp(pos + 1, shiftedInMask, shiftedExMask, inCount, exCount)      placeIntrovert = (          120 + getPlacementCost(i, j, inMask, exMask, -30) +          dp(pos + 1, shiftedInMask + 1, shiftedExMask, inCount - 1, exCount)          if inCount > 0          else -math.inf)      placeExtrovert = (          40 + getPlacementCost(i, j, inMask, exMask, 20) +          dp(pos + 1, shiftedInMask, shiftedExMask + 1, inCount, exCount - 1)          if exCount > 0          else -math.inf)      return max(skip, placeIntrovert, placeExtrovert)     return dp(0, 0, 0, introvertsCount, extrovertsCount)

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