g1401_1500.s1425_constrained_subsequence_sum.Solution Maven / Gradle / Ivy

Go to download

Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation

Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.38

package g1401_1500.s1425_constrained_subsequence_sum;

// #Hard #Array #Dynamic_Programming #Heap_Priority_Queue #Sliding_Window #Queue #Monotonic_Queue
// #2022_03_28_Time_69_ms_(30.52%)_Space_129.5_MB_(19.86%)

import java.util.LinkedList;

/**
 * 1425 - Constrained Subsequence Sum\.
 *
 * Hard
 *
 * Given an integer array `nums` and an integer `k`, return the maximum sum of a **non-empty** subsequence of that array such that for every two **consecutive** integers in the subsequence, `nums[i]` and `nums[j]`, where `i < j`, the condition `j - i <= k` is satisfied.
 *
 * A _subsequence_ of an array is obtained by deleting some number of elements (can be zero) from the array, leaving the remaining elements in their original order.
 *
 * **Example 1:**
 *
 * **Input:** nums = [10,2,-10,5,20], k = 2
 *
 * **Output:** 37
 *
 * **Explanation:** The subsequence is [10, 2, 5, 20].
 *
 * **Example 2:**
 *
 * **Input:** nums = [-1,-2,-3], k = 1
 *
 * **Output:** -1
 *
 * **Explanation:** The subsequence must be non-empty, so we choose the largest number.
 *
 * **Example 3:**
 *
 * **Input:** nums = [10,-2,-10,-5,20], k = 2
 *
 * **Output:** 23
 *
 * **Explanation:** The subsequence is [10, -2, -5, 20].
 *
 * **Constraints:**
 *
 * *   1 <= k <= nums.length <= 10⁵
 * *   -10⁴ <= nums[i] <= 10⁴
**/
public class Solution {
    public int constrainedSubsetSum(int[] nums, int k) {
        int n = nums.length;
        int res = Integer.MIN_VALUE;
        LinkedList mono = new LinkedList<>();

        for (int i = 0; i < n; i++) {
            int take = nums[i];

            while (!mono.isEmpty() && i - mono.getFirst()[0] > k) {
                mono.removeFirst();
            }
            if (!mono.isEmpty()) {
                int mx = Math.max(0, mono.getFirst()[1]);
                take += mx;
            }
            while (!mono.isEmpty() && take > mono.getLast()[1]) {
                mono.removeLast();
            }

            mono.add(new int[] {i, take});
            res = Math.max(res, take);
        }
        return res;
    }
}