【Likou】2681. The Power of Heroes

The following are other people's solution ideas, and my own code

topic

Title meaning

gives you a subscript from $0-$ based integer array$nums$ , which represents the hero's ability value. If we select some heroes,the poweris defined as:
$i_0，i_1，...i_k$Indicates the subscript of this group of heroes in the array. Then the strength of this group of heroes is $max(nums[i_0],nums[i_1]...nums[i_k])2\ast min(nums[i_0],nums[i_1]...nums[i_k])$ . Please return the sum of the power
all possiblenon-emptyhero groups. Since the answer may be very large, please compare the result to1 0 9 + 7 10^9 + 7$10^9+7$ to take the remainder.

Example 1

Input : $nums=[2,1,4]$
Output:$141$
Explanation:
No.$Group\; 1$ :$The\; power\; of\; [2]$ is$2^2\ast 2=8$ .
Second2Group$2$$\left[1\right]$ has a power of$1^2\ast 1=1$ .
3rd$3$ groups:$The\; power\; of\; [4]$ is$4^2\ast 4=64$ .
4th$4$ groups:$[2,1]$ has a power of$2^2\ast 1=4$ .
5th$5$ groups:$[2,4]$ has a power of$4^2\ast 2=32$ .
6th$6$ groups:$[1,4]$ has a power of$4^2\ast 1=16$ .
$7$ groups:$[2,1,4]$ has a power of$4^2\ast 1=16$ .
The sum of the power of all hero groups is$8+1+64+4+32+16+16=141$ 。

Example 2

Input : $nums=[1,1,1]$
Output:$7$
Explanation: There are$7$ hero groups, the power of each group is$1$ . So the sum of the power of all hero groups is$7$ 。

hint

1. $1<=nums.length<=10^5$
2. $1<=nums[i]<=10^9$

other people's ideas

contribution method

Algorithm summary

Since the order of the elements does not affect the answer, sort first.

设有 $a,b,c,d,e$ five numbers, in ascending order.

If put $d$ as the maximum value:

• If only choose $d$ is a single number, then the power is$d^3$。
• Choose $a$ is the minimum value, due to the middle$b$ and$c$ is optional, there are a total of$2^2$ schemes, so the sum of power is$d^2\ast a\ast 2^2$。
• choose $b$ is the minimum value due to the middle$c$ is optional, there are a total of$2^1$ scheme, so the sum of forces is$d^2\ast b\ast 2^1$。
• choose $c$ is the minimum value, only$2^0=1$ scheme, so the sum of forces is$d^2\cdot c\cdot 2^0$。

Therefore, when ddWhen$d$$The\; contribution\; of\; d$ and its left elements to the answer is
$$d^3+d^2\ast (a\ast 2^2+b\ast 2^1+c\ast 2^0)$$

令 $s=a\ast 2^2+b\ast 2^1+c\ast 2^0$，上式的
$$d^3+d^2\ast s=d^2\ast (d+s)$$

Go ahead and put $e$ as the maximum value, observe$How\; s$ changes, that is,$a,b,c,Contribution\; of\; d$ as minimum:
$$\begin{gathered} a\ast 2^3+b\ast 2^3+c\ast 2^1+d\ast 2^0 \\ =2\ast (a\ast 2^2+b\ast 2^1+c\ast 2^0)+d\ast 2^0 \\ =2\ast s+d \end{gathered}$$

This means that we don't need to enumerate the minimum value, we only need to enumerate the maximum value, we can put $s$ is recursively calculated.

the complexity

• Time Complexity: $O(nlogn)$ , of which$n$ 为 $The\; length\; of\; nums$ . The bottleneck is sorting.
• Space complexity: $O\left(1\right)$ . Ignoring stack space for sorting, only a few extra variables are used.

My code

Java

class Solution {
    
    
    public int sumOfPower(int[] nums) {
    
    
        Arrays.sort(nums);

        long mod = (long) 1e9 + 7;
        long ans = 0, s = 0;
        for (long i : nums) {
    
    
            ans = (ans + i * i % mod * (i + s)) % mod;
            s = (2 * s + i) % mod;
        }

        return (int) ans;
    }
}

Submission result: passed

• Execution time: $14ms$
• Memory Consumption: $51.72MB$