Leetcode #1659: Maximize Grid Happiness
In this guide, we solve Leetcode #1659 Maximize Grid Happiness 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 four integers, m, n, introvertsCount, and extrovertsCount. You have an m x n grid, and there are two types of people: introverts and extroverts.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Memoization, Dynamic Programming, Bitmask
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: m = 2, n = 3, introvertsCount = 1, extrovertsCount = 2
Output: 240
Explanation: Assume the grid is 1-indexed with coordinates (row, column).
We can put the introvert in cell (1,1) and put the extroverts in cells (1,3) and (2,3).
- Introvert at (1,1) happiness: 120 (starting happiness) - (0 * 30) (0 neighbors) = 120
- Extrovert at (1,3) happiness: 40 (starting happiness) + (1 * 20) (1 neighbor) = 60
- Extrovert at (2,3) happiness: 40 (starting happiness) + (1 * 20) (1 neighbor) = 60
The grid happiness is 120 + 60 + 60 = 240.
The above figure shows the grid in this example with each person's happiness. The introvert stays in the light green cell while the extroverts live on the light purple cells.
Python Solution
class Solution:
def getMaxGridHappiness(
self, m: int, n: int, introvertsCount: int, extrovertsCount: int
) -> int:
def dfs(i: int, pre: int, ic: int, ec: int) -> int:
if i == m or (ic == 0 and ec == 0):
return 0
ans = 0
for cur in range(mx):
if ix[cur] <= ic and ex[cur] <= ec:
a = f[cur] + g[pre][cur]
b = dfs(i + 1, cur, ic - ix[cur], ec - ex[cur])
ans = max(ans, a + b)
return ans
mx = pow(3, n)
f = [0] * mx
g = [[0] * mx for _ in range(mx)]
h = [[0, 0, 0], [0, -60, -10], [0, -10, 40]]
bits = [[0] * n for _ in range(mx)]
ix = [0] * mx
ex = [0] * mx
for i in range(mx):
mask = i
for j in range(n):
mask, x = divmod(mask, 3)
bits[i][j] = x
if x == 1:
ix[i] += 1
f[i] += 120
elif x == 2:
ex[i] += 1
f[i] += 40
if j:
f[i] += h[x][bits[i][j - 1]]
for i in range(mx):
for j in range(mx):
for k in range(n):
g[i][j] += h[bits[i][k]][bits[j][k]]
return dfs(0, 0, introvertsCount, extrovertsCount)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.