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Problem solution · Java

Maximum Strength of K Disjoint Subarrays

Maximum Strength of K Disjoint Subarrays: 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
31 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 Maximum Strength of K Disjoint Subarrays, 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

  • 31 lines of Java from the credited upstream file 3077.java.
  • 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 walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximum Strength of K Disjoint Subarrays · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public long maximumStrength(int[] nums, int k) {    Long[][][] mem = new Long[nums.length][k + 1][2];    return maximumStrength(nums, 0, k, /*fresh=*/true, mem);  }   private static final long MIN = Long.MIN_VALUE / 2;   // Returns the maximum strength of nums[i..n) with k operations left, where  // `fresh` means we're starting a new subarray.  private long maximumStrength(int[] nums, int i, int k, boolean fresh, Long[][][] mem) {    if (nums.length - i < k)      return MIN;    if (k == 0)      return 0;    if (i == nums.length)      return k == 0 ? 0 : MIN;    if (mem[i][k][fresh ? 1 : 0] != null)      return mem[i][k][fresh ? 1 : 0];    // If it's not fresh, we can't skip the current number and consider it as a    // fresh start, since the case where it's fresh is already covered by    // `includeAndFreshStart`.    final long skip = fresh ? maximumStrength(nums, i + 1, k, true, mem) : MIN;    final long gain = (k % 2 == 0 ? -1 : 1) * 1L * nums[i] * k;    final long includeAndContinue = maximumStrength(nums, i + 1, k, false, mem) + gain;    final long includeAndFreshStart = maximumStrength(nums, i + 1, k - 1, true, mem) + gain;    return mem[i][k][fresh ? 1 : 0] =               Math.max(skip, Math.max(includeAndContinue, includeAndFreshStart));  }}

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