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Problem solution · Java

Maximum AND Sum of Array

Maximum AND Sum of Array: a Java 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
21 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 Maximum AND Sum of Array, 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

  • 21 lines of Java from the credited upstream file 2172.java.
  • The implementation visibly relies on sequence storage, cached states.
  • 2 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 codeMaximum AND Sum of Array · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int maximumANDSum(int[] nums, int numSlots) {    final int n = 2 * numSlots;    final int nSelected = 1 << n;    // dp[i] := the maximum value, where i is the bitmask of the selected    // numbers    int[] dp = new int[nSelected];    int[] arr = Arrays.copyOf(nums, n);     for (int mask = 1; mask < nSelected; ++mask) {      final int selected = Integer.bitCount(mask);      final int slot = (selected + 1) / 2; // (1, 2) -> 1, (3, 4) -> 2      for (int i = 0; i < n; ++i)        if ((mask >> i & 1) == 1) // Assign `arr[i]` to the `slot`-th slot.          dp[mask] = Math.max(dp[mask], dp[mask ^ 1 << i] + (slot & arr[i]));    }     return dp[nSelected - 1];  }}

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