Welcome to Subscribe On Youtube

3974. Maximum Total Sum of K Selected Elements

Description

You are given an integer array nums and two integers k and mul.

Select exactly k elements from nums. Process these elements one by one in any order you choose.

For each selected element, independently choose one of the following:

  • Add the element's value to the total sum, or
  • Multiply the element by the current value of mul and add the result to the total sum.

After processing each selected element, mul decreases by 1, regardless of which option was chosen. The current value of mul may become 0 or negative.

Return an integer denoting the maximum possible total sum.

 

Example 1:

Input: nums = [6,1,2,9], k = 3, mul = 2

Output: 26

Explanation:

One optimal way:

  • One optimal selection is nums[3] = 9, nums[0] = 6, and nums[2] = 2.
  • Process nums[3] = 9 first: choose multiplication, so it contributes 9 * 2 = 18. Now, mul becomes 1.
  • Process nums[0] = 6 next: choose multiplication, so it contributes 6 * 1 = 6. Now, mul becomes 0.
  • Process nums[2] = 2 last: choose addition, so it contributes 2.
  • The total sum is 18 + 6 + 2 = 26.

Example 2:

Input: nums = [3,7,5,2], k = 2, mul = 4

Output: 43

Explanation:

One optimal way:

  • One optimal selection is nums[1] = 7 and nums[2] = 5.
  • Process nums[1] = 7 first: choose multiplication, so it contributes 7 * 4 = 28. Now, mul becomes 3.
  • Process nums[2] = 5 next: choose multiplication, so it contributes 5 * 3 = 15.
  • The total sum is 28 + 15 = 43.

Example 3:

Input: nums = [4,4], k = 1, mul = 1

Output: 4

Explanation:

One optimal way:

  • One optimal selection is nums[0] = 4.
  • Process nums[0] = 4: choose multiplication, so it contributes 4 * 1 = 4.
  • The total sum is 4.

 

Constraints:

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

Solutions

Solution 1: Greedy + Sorting

We can sort the array $\textit{nums}$ and then select the $k$ largest elements from the sorted array. For the $i$-th element, we can choose to multiply it by $\max(1, \textit{mul})$ and add it to the total sum, and then $\textit{mul}$ decreases by $1$. Finally, we return the total sum.

The time complexity is $O(n \log n)$, and the space complexity is $O(\log n)$. Where $n$ is the length of the array $\textit{nums}$.

  • class Solution {
    public long maxSum(int[] nums, int k, int mul) {
        Arrays.sort(nums);
        int n = nums.length;
        long ans = 0;

        for (int i = n - 1; i >= n - k; i--) {
            int m = Math.max(1, mul);
            ans += (long) nums[i] * m;
            mul--;
        }

        return ans;
    }
}

  • class Solution {
public:
    long long maxSum(vector<int>& nums, int k, int mul) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        long long ans = 0;

        for (int i = n - 1; i >= n - k; --i) {
            int m = max(1, mul);
            ans += 1LL * nums[i] * m;
            mul--;
        }

        return ans;
    }
};

  • class Solution:
    def maxSum(self, nums: list[int], k: int, mul: int) -> int:
        nums.sort()
        n = len(nums)
        ans = 0
        for i in range(n - 1, n - 1 - k, -1):
            ans += nums[i] * max(1, mul)
            mul -= 1
        return ans


  • func maxSum(nums []int, k int, mul int) int64 {
    sort.Ints(nums)
    n := len(nums)
    var ans int64 = 0

    for i := n - 1; i >= n-k; i-- {
        m := max(1, mul)
        ans += int64(nums[i]) * int64(m)
        mul--
    }

    return ans
}

  • function maxSum(nums: number[], k: number, mul: number): number {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    let ans = 0;

    for (let i = n - 1; i >= n - k; i--) {
        const m = Math.max(1, mul);
        ans += nums[i] * m;
        mul--;
    }

    return ans;
}

All Problems

All Solutions