Question

1911. Maximum Alternating Subsequence Sum

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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

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])

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]);
  }
}

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]);
  }
};

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]))
}

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

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)

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);
  }
}

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);
  }
};

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))
}

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;
}