Class Solution
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public final class Solution
2681 - Power of Heroes.
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:Let <code>i<sub>0</sub></code>, <code>i<sub>1</sub></code>, ... ,<code>i<sub>k</sub></code> be the indices of the heroes in a group. Then, the power of this group is <code>max(numsi<sub>0</sub>, numsi<sub>1</sub>, ... ,numsi<sub>k</sub>)<sup>2</sup> * min(numsi<sub>0</sub>, numsi<sub>1</sub>, ... ,numsi<sub>k</sub>)</code>.
Return the sum of the power of all non-empty groups of heroes possible. Since the sum could be very large, return it modulo <code>10<sup>9</sup> + 7</code>.
Example 1:
Input: nums = 2,1,4
Output: 141
Explanation:
1<sup>st</sup> group: 2 has power = 2<sup>2</sup> \* 2 = 8.
2<sup>nd</sup> group: 1 has power = 1<sup>2</sup> \* 1 = 1.
3<sup>rd</sup> group: 4 has power = 4<sup>2</sup> \* 4 = 64.
4<sup>th</sup> group: 2,1 has power = 2<sup>2</sup> \* 1 = 4.
5<sup>th</sup> group: 2,4 has power = 4<sup>2</sup> \* 2 = 32.
6<sup>th</sup> group: 1,4 has power = 4<sup>2</sup> \* 1 = 16.
7<sup>th</sup> group: 2,1,4 has power = 4<sup>2</sup> \* 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:
<code>1 <= nums.length <= 10<sup>5</sup></code>
<code>1 <= numsi<= 10<sup>9</sup></code>
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Constructor Summary
Constructors Constructor Description Solution()
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Method Summary
Modifier and Type Method Description final Integer
sumOfPower(IntArray nums)
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Method Detail
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sumOfPower
final Integer sumOfPower(IntArray nums)
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