How do you solve Power of Heroes on LeetCode?

Power of Heroes (LeetCode #2681) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting the array allows for efficient calculation of contributions.

What is the brute force solution for Power of Heroes?

The brute force approach for Power of Heroes is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible non-empty groups of heroes and calculate their power. This approach is straightforward but inefficient due to the exponential number of groups. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Power of Heroes?

Power of Heroes uses the following concepts: Array, Math, Dynamic Programming, Sorting, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Prefix Sum.

What are the interview tips for Power of Heroes?

Explain your thought process clearly while coding. Discuss edge cases and how your solution handles them. Practice similar problems to recognize patterns.

What are common mistakes when solving Power of Heroes?

Forgetting to take modulo during calculations. Not considering the contribution of each hero correctly.

#2681

Power of Heroes

Hard

ArrayMathDynamic ProgrammingSortingPrefix SumDynamic ProgrammingPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

By sorting the array, we can efficiently calculate the contribution of each hero to the total power based on their position in the sorted order. This allows us to avoid generating all subsets explicitly.

⚙️

Algorithm

4 steps

1Step 1: Sort the array in non-decreasing order.
2Step 2: Initialize a total power variable and a prefix sum array to keep track of contributions.
3Step 3: For each hero in the sorted array, calculate how many times they will be the max and min in the groups they belong to.
4Step 4: Use the prefix sum to calculate the contributions of each hero efficiently.

solution.py16 lines

1# Full working Python code
2def sum_of_powers(nums):
3    MOD = 10**9 + 7
4    nums.sort()
5    total_power = 0
6    n = len(nums)
7    prefix_sum = 0
8
9    for i in range(n):
10        max_contrib = (nums[i] ** 2) * (pow(2, i, MOD) - 1)
11        min_contrib = nums[i] * prefix_sum * pow(2, n - i - 1, MOD)
12        total_power = (total_power + max_contrib - min_contrib) % MOD
13        prefix_sum = (prefix_sum + nums[i] * pow(2, i, MOD)) % MOD
14
15    return total_power
16

ℹ

Complexity note: The time complexity is O(n log n) due to sorting the array, and O(n) for calculating contributions, making it efficient for large inputs.

1Sorting the array allows for efficient calculation of contributions.
2Using prefix sums helps in avoiding recomputation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.