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1911. Maximum Alternating Subsequence Sum

SourceBiweekly Contest 55 Q3DifficultyMediumRating1785

Description

The alternating sum of a 0-indexed array is defined as the sum of the elements at even indices minus the sum of the elements at odd indices.

  • For example, the alternating sum of [4,2,5,3] is (4 + 5) - (2 + 3) = 4.

Given an array nums, return the maximum alternating sum of any subsequence of nums (after reindexing the elements of the subsequence).

A subsequence of an array is a new array generated from the original array by deleting some elements (possibly none) without changing the remaining elements' relative order. For example, [2,7,4] is a subsequence of [4,2,3,7,2,1,4] (the underlined elements), while [2,4,2] is not.

 

Example 1:

Input: nums = [4,2,5,3]
Output: 7
Explanation: It is optimal to choose the subsequence [4,2,5] with alternating sum (4 + 5) - 2 = 7.

Example 2:

Input: nums = [5,6,7,8]
Output: 8
Explanation: It is optimal to choose the subsequence [8] with alternating sum 8.

Example 3:

Input: nums = [6,2,1,2,4,5]
Output: 10
Explanation: It is optimal to choose the subsequence [6,1,5] with alternating sum (6 + 5) - 1 = 10.

 

Constraints:

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

Solutions

Solution 1: Dynamic Programming

Thinking

Enumerating subsequences is exponential. With \(n\le 10^5\) we need a linear DP on the parity of the last chosen index.

Let \(f[i]\) be the best alternating sum of the first \(i\) elements whose last pick sits on an odd slot (subtracted), and \(g[i]\) the best whose last pick sits on an even slot (added). Element \(x\) is either appended to the opposite parity or skipped.

The recurrences are \(f[i]=\max(g[i-1]-x,f[i-1])\) and \(g[i]=\max(f[i-1]+x,g[i-1])\); the answer is the larger of \(f[n]\) and \(g[n]\).

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class Solution:
    def maxAlternatingSum(self, nums: List[int]) -> int:
        n = len(nums)
        f = [0] * (n + 1)
        g = [0] * (n + 1)
        for i, x in enumerate(nums, 1):
            f[i] = max(g[i - 1] - x, f[i - 1])
            g[i] = max(f[i - 1] + x, g[i - 1])
        return max(f[n], g[n])

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class Solution {
    public long maxAlternatingSum(int[] nums) {
        int n = nums.length;
        long[] f = new long[n + 1];
        long[] g = new long[n + 1];
        for (int i = 1; i <= n; ++i) {
            f[i] = Math.max(g[i - 1] - nums[i - 1], f[i - 1]);
            g[i] = Math.max(f[i - 1] + nums[i - 1], g[i - 1]);
        }
        return Math.max(f[n], g[n]);
    }
}

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class Solution {
public:
    long long maxAlternatingSum(vector<int>& nums) {
        int n = nums.size();
        vector<long long> f(n + 1), g(n + 1);
        for (int i = 1; i <= n; ++i) {
            f[i] = max(g[i - 1] - nums[i - 1], f[i - 1]);
            g[i] = max(f[i - 1] + nums[i - 1], g[i - 1]);
        }
        return max(f[n], g[n]);
    }
};

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func maxAlternatingSum(nums []int) int64 {
    n := len(nums)
    f := make([]int, n+1)
    g := make([]int, n+1)
    for i, x := range nums {
        i++
        f[i] = max(g[i-1]-x, f[i-1])
        g[i] = max(f[i-1]+x, g[i-1])
    }
    return int64(max(f[n], g[n]))
}

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function maxAlternatingSum(nums: number[]): number {
    const n = nums.length;
    const f: number[] = new Array(n + 1).fill(0);
    const g = f.slice();
    for (let i = 1; i <= n; ++i) {
        f[i] = Math.max(g[i - 1] + nums[i - 1], f[i - 1]);
        g[i] = Math.max(f[i - 1] - nums[i - 1], g[i - 1]);
    }
    return Math.max(f[n], g[n]);
}

Solution 2: Dynamic Programming (Space Optimization)

Thinking

Only the previous \(f\) and \(g\) are read, so two scalars updated left to right replace the arrays and drop extra space to \(O(1)\).

\(f[i]\) and \(g[i]\) depend only on the previous index, so two variables are enough and the space complexity is \(O(1)\).

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class Solution:
    def maxAlternatingSum(self, nums: List[int]) -> int:
        f = g = 0
        for x in nums:
            f, g = max(g - x, f), max(f + x, g)
        return max(f, g)

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class Solution {
    public long maxAlternatingSum(int[] nums) {
        long f = 0, g = 0;
        for (int x : nums) {
            long ff = Math.max(g - x, f);
            long gg = Math.max(f + x, g);
            f = ff;
            g = gg;
        }
        return Math.max(f, g);
    }
}

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class Solution {
public:
    long long maxAlternatingSum(vector<int>& nums) {
        long long f = 0, g = 0;
        for (int& x : nums) {
            long ff = max(g - x, f), gg = max(f + x, g);
            f = ff, g = gg;
        }
        return max(f, g);
    }
};

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func maxAlternatingSum(nums []int) int64 {
    var f, g int
    for _, x := range nums {
        f, g = max(g-x, f), max(f+x, g)
    }
    return int64(max(f, g))
}

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function maxAlternatingSum(nums: number[]): number {
    let [f, g] = [0, 0];
    for (const x of nums) {
        [f, g] = [Math.max(g - x, f), Math.max(f + x, g)];
    }
    return g;
}

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