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2926. Maximum Balanced Subsequence Sum

Description

You are given a 0-indexed integer array nums.

A subsequence of nums having length k and consisting of indices i0 < i1 < ... < ik-1 is balanced if the following holds:

  • nums[ij] - nums[ij-1] >= ij - ij-1, for every j in the range [1, k - 1].

A subsequence of nums having length 1 is considered balanced.

Return an integer denoting the maximum possible sum of elements in a balanced subsequence of nums.

A subsequence of an array is a new non-empty array that is formed from the original array by deleting some (possibly none) of the elements without disturbing the relative positions of the remaining elements.

 

Example 1:

Input: nums = [3,3,5,6]
Output: 14
Explanation: In this example, the subsequence [3,5,6] consisting of indices 0, 2, and 3 can be selected.
nums[2] - nums[0] >= 2 - 0.
nums[3] - nums[2] >= 3 - 2.
Hence, it is a balanced subsequence, and its sum is the maximum among the balanced subsequences of nums.
The subsequence consisting of indices 1, 2, and 3 is also valid.
It can be shown that it is not possible to get a balanced subsequence with a sum greater than 14.

Example 2:

Input: nums = [5,-1,-3,8]
Output: 13
Explanation: In this example, the subsequence [5,8] consisting of indices 0 and 3 can be selected.
nums[3] - nums[0] >= 3 - 0.
Hence, it is a balanced subsequence, and its sum is the maximum among the balanced subsequences of nums.
It can be shown that it is not possible to get a balanced subsequence with a sum greater than 13.

Example 3:

Input: nums = [-2,-1]
Output: -1
Explanation: In this example, the subsequence [-1] can be selected.
It is a balanced subsequence, and its sum is the maximum among the balanced subsequences of nums.

 

Constraints:

  • 1 <= nums.length <= 105
  • -109 <= nums[i] <= 109

Solutions

Solution 1: Dynamic Programming + Binary Indexed Tree

According to the problem description, we can transform the inequality $nums[i] - nums[j] \ge i - j$ into $nums[i] - i \ge nums[j] - j$. Therefore, we consider defining a new array $arr$, where $arr[i] = nums[i] - i$. A balanced subsequence satisfies that for any $j < i$, $arr[j] \le arr[i]$. The problem is transformed into selecting an increasing subsequence in $arr$ such that the corresponding sum in $nums$ is maximized.

Suppose $i$ is the index of the last element in the subsequence, then we consider the index $j$ of the second to last element in the subsequence. If $arr[j] \le arr[i]$, we can consider whether to add $j$ to the subsequence.

Therefore, we define $f[i]$ as the maximum sum of $nums$ when the index of the last element in the subsequence is $i$. The answer is $\max_{i=0}^{n-1} f[i]$.

The state transition equation is:

$$ f[i] = \max(\max_{j=0}^{i-1} f[j], 0) + nums[i] $$

where $j$ satisfies $arr[j] \le arr[i]$.

We can use a Binary Indexed Tree to maintain the maximum value of the prefix, i.e., for each $arr[i]$, we maintain the maximum value of $f[i]$ in the prefix $arr[0..i]$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

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class BinaryIndexedTree:
    def __init__(self, n: int):
        self.n = n
        self.c = [-inf] * (n + 1)

    def update(self, x: int, v: int):
        while x <= self.n:
            self.c[x] = max(self.c[x], v)
            x += x & -x

    def query(self, x: int) -> int:
        mx = -inf
        while x:
            mx = max(mx, self.c[x])
            x -= x & -x
        return mx


class Solution:
    def maxBalancedSubsequenceSum(self, nums: List[int]) -> int:
        arr = [x - i for i, x in enumerate(nums)]
        s = sorted(set(arr))
        tree = BinaryIndexedTree(len(s))
        for i, x in enumerate(nums):
            j = bisect_left(s, x - i) + 1
            v = max(tree.query(j), 0) + x
            tree.update(j, v)
        return tree.query(len(s))

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class BinaryIndexedTree {
    private int n;
    private long[] c;
    private final long inf = 1L << 60;

    public BinaryIndexedTree(int n) {
        this.n = n;
        c = new long[n + 1];
        Arrays.fill(c, -inf);
    }

    public void update(int x, long v) {
        while (x <= n) {
            c[x] = Math.max(c[x], v);
            x += x & -x;
        }
    }

    public long query(int x) {
        long mx = -inf;
        while (x > 0) {
            mx = Math.max(mx, c[x]);
            x -= x & -x;
        }
        return mx;
    }
}

class Solution {
    public long maxBalancedSubsequenceSum(int[] nums) {
        int n = nums.length;
        int[] arr = new int[n];
        for (int i = 0; i < n; ++i) {
            arr[i] = nums[i] - i;
        }
        Arrays.sort(arr);
        int m = 0;
        for (int i = 0; i < n; ++i) {
            if (i == 0 || arr[i] != arr[i - 1]) {
                arr[m++] = arr[i];
            }
        }
        BinaryIndexedTree tree = new BinaryIndexedTree(m);
        for (int i = 0; i < n; ++i) {
            int j = search(arr, nums[i]