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Problem solution · Java

Maximize Grid Happiness

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

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
67 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 Maximize Grid Happiness, 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

  • 67 lines of Java from the credited upstream file 1659.java.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 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 walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximize Grid Happiness · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int getMaxGridHappiness(int m, int n, int introvertsCount, int extrovertsCount) {    final int twoToThePowerOfN = (int) Math.pow(2, n);    int[][][][][] mem = new int[m * n][twoToThePowerOfN][twoToThePowerOfN][introvertsCount + 1]                               [extrovertsCount + 1];    return getMaxGridHappiness(m, n, 0, 0, 0, introvertsCount, extrovertsCount, mem);  }   // 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.  private int getPlacementCost(int n, int i, int j, int inMask, int exMask, int diff) {    int cost = 0;    if (i > 0) {      if (((1 << (n - 1)) & inMask) > 0)        cost += diff - 30;      if (((1 << (n - 1)) & exMask) > 0)        cost += diff + 20;    }    if (j > 0) {      if ((1 & inMask) > 0)        cost += diff - 30;      if ((1 & exMask) > 0)        cost += diff + 20;    }    return cost;  }   private int getMaxGridHappiness(int m, int n, int pos, int inMask, int exMask, int inCount,                                  int exCount, int[][][][][] mem) {    // `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)    final int i = pos / n;    final int j = pos % n;    if (i == m)      return 0;    if (mem[pos][inMask][exMask][inCount][exCount] > 0)      return mem[pos][inMask][exMask][inCount][exCount];     final int shiftedInMask = (inMask << 1) & ((1 << n) - 1);    final int shiftedExMask = (exMask << 1) & ((1 << n) - 1);     final int skip =        getMaxGridHappiness(m, n, pos + 1, shiftedInMask, shiftedExMask, inCount, exCount, mem);    final int placeIntrovert =        inCount > 0 ? 120 + getPlacementCost(n, i, j, inMask, exMask, -30) +                          getMaxGridHappiness(m, n, pos + 1, shiftedInMask | 1, shiftedExMask,                                              inCount - 1, exCount, mem)                    : Integer.MIN_VALUE;    final int placeExtrovert =        exCount > 0 ? 40 + getPlacementCost(n, i, j, inMask, exMask, 20) +                          getMaxGridHappiness(m, n, pos + 1, shiftedInMask, shiftedExMask | 1,                                              inCount, exCount - 1, mem)                    : Integer.MIN_VALUE;    return mem[pos][inMask][exMask][inCount][exCount] =               Math.max(skip, Math.max(placeIntrovert, placeExtrovert));  }}

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