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| 1 | +package com.github.dkoval.leetcode.challenge; |
| 2 | + |
| 3 | +import java.util.ArrayList; |
| 4 | +import java.util.function.BiFunction; |
| 5 | + |
| 6 | +/** |
| 7 | + * <a href="https://leetcode.com/problems/trionic-array-ii/">Trionic Array II (Hard)</a> |
| 8 | + * <p> |
| 9 | + * You are given an integer array nums of length n. |
| 10 | + * <p> |
| 11 | + * A trionic subarray is a contiguous subarray nums[l...r] (with 0 <= l < r < n) for which there exist indices l < p < q < r such that: |
| 12 | + * <ul> |
| 13 | + * <li>nums[l...p] is strictly increasing,</li> |
| 14 | + * <li>nums[p...q] is strictly decreasing,</li> |
| 15 | + * <li>nums[q...r] is strictly increasing.</li> |
| 16 | + * </ul> |
| 17 | + * Return the maximum sum of any trionic subarray in nums. |
| 18 | + * <p> |
| 19 | + * Constraints: |
| 20 | + * <ul> |
| 21 | + * <li>4 <= n = nums.length <= 10^5</li> |
| 22 | + * <li>-10^9 <= nums[i] <= 10^9</li> |
| 23 | + * <li>It is guaranteed that at least one trionic subarray exists.</li> |
| 24 | + * </ul> |
| 25 | + */ |
| 26 | +public interface TrionicArray2 { |
| 27 | + |
| 28 | + long maxSumTrionic(int[] nums); |
| 29 | + |
| 30 | + class TrionicArray2Rev1 implements TrionicArray2 { |
| 31 | + |
| 32 | + @Override |
| 33 | + public long maxSumTrionic(int[] nums) { |
| 34 | + final var n = nums.length; |
| 35 | + |
| 36 | + // step 1: find decreasing chunks |
| 37 | + final var decreasing = new ArrayList<Interval>(); |
| 38 | + var start = -1; |
| 39 | + var sum = 0L; |
| 40 | + for (var i = 0; i < n - 1; i++) { |
| 41 | + if (nums[i] > nums[i + 1]) { |
| 42 | + if (start < 0) { |
| 43 | + start = i; |
| 44 | + } |
| 45 | + sum += nums[i]; |
| 46 | + } else { |
| 47 | + if (start >= 0) { |
| 48 | + sum += nums[i]; |
| 49 | + decreasing.add(new Interval(start, i, sum)); |
| 50 | + start = -1; |
| 51 | + sum = 0; |
| 52 | + } |
| 53 | + } |
| 54 | + } |
| 55 | + |
| 56 | + // step 2: find increasing chunks to the left and to the right of a decreasing one |
| 57 | + var best = Long.MIN_VALUE; |
| 58 | + for (var interval : decreasing) { |
| 59 | + if (interval.start == 0 || interval.end == n - 1) { |
| 60 | + continue; |
| 61 | + } |
| 62 | + |
| 63 | + var leftSum = maxSumOfIncreasing(nums, interval.start, -1, (a, b) -> a > b); |
| 64 | + var rightSum = maxSumOfIncreasing(nums, interval.end, 1, (a, b) -> a < b); |
| 65 | + best = Math.max(best, interval.sum + leftSum + rightSum); |
| 66 | + } |
| 67 | + return best; |
| 68 | + } |
| 69 | + |
| 70 | + private long maxSumOfIncreasing(int[] nums, int start, int step, BiFunction<Integer, Integer, Boolean> comparator) { |
| 71 | + var i = start; |
| 72 | + var total = 0L; |
| 73 | + var best = Long.MIN_VALUE; |
| 74 | + while (i > 0 && i < nums.length - 1 && comparator.apply(nums[i], nums[i + step])) { |
| 75 | + i += step; |
| 76 | + total += nums[i]; |
| 77 | + best = Math.max(best, total); |
| 78 | + } |
| 79 | + return best; |
| 80 | + } |
| 81 | + |
| 82 | + record Interval(int start, int end, long sum) { |
| 83 | + } |
| 84 | + } |
| 85 | +} |
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