2681. Power of Heroes

Problem

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

Solution (Java)

class Solution {
    public int sumOfPower(int[] nums) {
        long mod = (long) 1e9 + 7, pre = 0, res = 0;
        Arrays.sort(nums);
        for (long x : nums) {
            res = (res + (x * x % mod) * x % mod + (x * x % mod) * pre % mod) % mod;
            pre = (pre * 2 + x) % mod;
        }
        return (int) res;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).