2681. Power of Heroes LeetCode Solution

Last updated: March 5, 2025

In this guide, you will get 2681. Power of Heroes LeetCode Solution with the best time and space complexity. The solution to Power of Heroes problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Power of Heroes solution in C++
Power of Heroes solution in Java
Power of Heroes solution in Python
Additional Resources

2681. Power of Heroes LeetCode Solution image

Problem Statement of Power of Heroes

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

Complexity Analysis

Time Complexity: O(\texttt{sort})
Space Complexity: O(\texttt{sort})

2681. Power of Heroes LeetCode Solution in C++

class Solution {
 public:
  int sumOfPower(vector<int>& nums) {
    constexpr int kMod = 1'000'000'007;
    long ans = 0;
    long sum = 0;

    ranges::sort(nums);

    for (const int num : nums) {
      ans = (ans + (num + sum) * num % kMod * num % kMod) % kMod;
      sum = (sum * 2 + num) % kMod;
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

2681. Power of Heroes LeetCode Solution in Java

class Solution {
  public int sumOfPower(int[] nums) {
    final int kMod = 1_000_000_007;
    long ans = 0;
    long sum = 0;

    Arrays.sort(nums);

    for (final int num : nums) {
      ans = (ans + (num + sum) * num % kMod * num % kMod) % kMod;
      sum = (sum * 2 + num) % kMod;
    }

    return (int) ans;
  }
}
// code provided by PROGIEZ

2681. Power of Heroes LeetCode Solution in Python

class Solution:
  def sumOfPower(self, nums: list[int]) -> int:
    kMod = 1_000_000_007
    ans = 0
    summ = 0

    for num in sorted(nums):
      ans += (num + summ) * num**2
      ans %= kMod
      summ = (summ * 2 + num) % kMod

    return ans
 # code by PROGIEZ

Additional Resources

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