Hard
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:
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
class Solution {
private val mod = 1_000_000_007L
fun sumOfPower(nums: IntArray): Int {
nums.sort()
var sumOfMin = 0L
var res = 0L
for (num in nums) {
val max = (num.toLong() * num.toLong()) % mod
val mySumOfMin = (sumOfMin + num) % mod
res = (res + max * mySumOfMin) % mod
sumOfMin = (sumOfMin + mySumOfMin) % mod
}
return res.toInt()
}
}