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Problem solution · Java

Maximum Total Beauty of the Gardens

Maximum Total Beauty of the Gardens: a Java solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Binary search
Source
walkccc LeetCode Solutions
Length
50 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

Binary search

For Maximum Total Beauty of the Gardens, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 50 lines of Java from the credited upstream file 2234.java.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

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 Total Beauty of the Gardens · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public long maximumBeauty(int[] flowers, long newFlowers, int target, int full, int partial) {    final int n = flowers.length;     // If a garden is already complete, clamp it to the target.    for (int i = 0; i < n; ++i)      flowers[i] = Math.min(flowers[i], target);    Arrays.sort(flowers);     // All gardens are complete, so nothing we can do.    if (flowers[0] == target)      return (long) n * full;     // Having many new flowers maximizes the beauty value.    if (newFlowers >= (long) n * target - Arrays.stream(flowers).asLongStream().sum())      return Math.max((long) n * full, (n - 1L) * full + (target - 1L) * partial);     long ans = 0;    long leftFlowers = newFlowers;    // cost[i] := the cost to make flowers[0..i] the same    long[] cost = new long[flowers.length];     for (int i = 1; i < flowers.length; ++i)      // Plant (flowers[i] - flowers[i - 1]) flowers for flowers[0..i - 1].      cost[i] = cost[i - 1] + i * (flowers[i] - flowers[i - 1]);     int i = flowers.length - 1; // flowers' index (flowers[i + 1..n) are complete)    while (flowers[i] == target)      --i;     for (; leftFlowers >= 0; --i) {      // To maximize the minimum number of incomplete flowers, we find the first      // index j that we can't make flowers[0..j] equal to flowers[j], then we      // know we can make flowers[0..j - 1] equal to flowers[j - 1]. In the      // meantime, evenly increase each of them to seek a bigger minimum value.      final int j = firstGreater(cost, i, leftFlowers);      final long minIncomplete = flowers[j - 1] + (leftFlowers - cost[j - 1]) / j;      ans = Math.max(ans, (n - 1L - i) * full + (long) minIncomplete * partial);      leftFlowers -= Math.max(0, target - flowers[i]);    }     return ans;  }   private int firstGreater(long[] A, int maxIndex, long target) {    final int i = Arrays.binarySearch(A, 0, maxIndex + 1, target + 1);    return i < 0 ? -i - 1 : i;  }}

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