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2681. Power of Heroes

Description

You are given a 0-indexed integer array nums representing the strength of some heroes. The power of a group of heroes is defined as follows:

  • Let i0, i1, ... ,ik be the indices of the heroes in a group. Then, the power of this group is max(nums[i0], nums[i1], ... ,nums[ik])2 * min(nums[i0], nums[i1], ... ,nums[ik]).

Return the sum of the power of all non-empty groups of heroes possible. Since the sum could be very large, return it modulo 109 + 7.

 

Example 1:

Input: nums = [2,1,4]
Output: 141
Explanation: 
1st group: [2] has power = 22 * 2 = 8.
2nd group: [1] has power = 12 * 1 = 1. 
3rd group: [4] has power = 42 * 4 = 64. 
4th group: [2,1] has power = 22 * 1 = 4. 
5th group: [2,4] has power = 42 * 2 = 32. 
6th group: [1,4] has power = 42 * 1 = 16. 
​​​​​​​7th group: [2,1,4] has power = 42​​​​​​​ * 1 = 16. 
The sum of powers of all groups is 8 + 1 + 64 + 4 + 32 + 16 + 16 = 141.

Example 2:

Input: nums = [1,1,1]
Output: 7
Explanation: A total of 7 groups are possible, and the power of each group will be 1. Therefore, the sum of the powers of all groups is 7.

 

Constraints:

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

Solutions

Solution 1: Sorting + Mathematics

We notice that the problem involves the maximum and minimum values of a subsequence, and the order of elements in the array does not affect the final result. Therefore, we can sort the array first.

Next, we enumerate each element as the minimum value of the subsequence. Let's denote each element of the array as $a_1, a_2, \cdots, a_n$. The contribution of the subsequence with $a_i$ as the minimum value is:

$$ a_i \times (a_{i}^{2} + a_{i+1}^2 + 2 \times a_{i+2}^2 + 4 \times a_{i+3}^2 + \cdots + 2^{n-i-1} \times a_n^2) $$

We notice that each $a_i$ will be multiplied by $a_i^2$, which we can directly add to the answer. For the remaining part, we can maintain it with a variable $p$, initially set to $0$.

Then, we enumerate $a_i$ from right to left. Each time, we add $a_i \times p$ to the answer, and then set $p = p \times 2 + a_i^2$.

After enumerating all $a_i$, return the answer.

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

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class Solution:
    def sumOfPower(self, nums: List[int]) -> int:
        mod = 10**9 + 7
        nums.sort()
        ans = 0
        p = 0
        for x in nums[::-1]:
            ans = (ans + (x * x % mod) * x) % mod
            ans = (ans + x * p) % mod
            p = (p * 2 + x * x) % mod
        return ans

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class Solution {
    public int sumOfPower(int[] nums) {
        final int mod = (int) 1e9 + 7;
        Arrays.sort(nums);
        long ans = 0, p = 0;
        for (int i = nums.length - 1; i >= 0; --i) {
            long x = nums[i];
            ans = (ans + (x * x % mod) * x) % mod;
            ans = (ans + x * p % mod) % mod;
            p = (p * 2 + x * x % mod) % mod;
        }
        return (int) ans;
    }
}

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class Solution {
public:
    int sumOfPower(vector<int>& nums) {
        const int mod = 1e9 + 7;
        sort(nums.rbegin(), nums.rend());
        long long ans = 0, p = 0;
        for (long long x : nums) {
            ans = (ans + (x * x % mod) * x) % mod;
            ans = (ans + x * p % mod) % mod;
            p = (p * 2 + x * x % mod) % mod;
        }
        return ans;
    }
};

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func sumOfPower(nums []int) (ans int) {
    const mod = 1e9 + 7
    sort.Ints(nums)
    p := 0
    for i := len(nums) - 1; i >= 0; i-- {
        x := nums[i]
        ans = (ans + (x*x%mod)*x) % mod
        ans = (ans + x*p%mod) % mod
        p = (p*2 + x*x%mod) % mod
    }
    return
}

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function sumOfPower(nums: number[]): number {
    const mod = 10 ** 9 + 7;
    nums.sort((a, b) => a - b);
    let ans = 0;
    let p = 0;
    for (let i = nums.length - 1; i >= 0; --i) {
        const x = BigInt(nums[i]);
        ans = (ans + Number((x * x * x) % BigInt(mod))) % mod;
        ans = (ans + Number((x * BigInt(p)) % BigInt(mod))) % mod;
        p = Number((BigInt(p) * 2n + x * x) % BigInt(mod));
    }
    return ans;
}

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