LeetCode #2681 — HARD

Power of Heroes

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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, ... ,i_k be the indices of the heroes in a group. Then, the power of this group is max(nums[i₀], nums[i₁], ... ,nums[i_k])² * min(nums[i₀], nums[i₁], ... ,nums[i_k]).

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: 
1^st group: [2] has power = 2² * 2 = 8.
2^nd group: [1] has power = 1² * 1 = 1. 
3^rd group: [4] has power = 4² * 4 = 64. 
4^th group: [2,1] has power = 2² * 1 = 4. 
5^th group: [2,4] has power = 4² * 2 = 32. 
6^th group: [1,4] has power = 4² * 1 = 16. 
7^th 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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Dynamic Programming

Example 1

[2,1,4]

Example 2

[1,1,1]

Step 02

Core Insight

What unlocks the optimal approach

Try something with sorting the array.
For a pair of array elements nums[i] and nums[j] (i < j), the power would be nums[i]*nums[j]^2 regardless of how many elements in between are included.
The number of subsets with the above as power will correspond to 2^(j-i-1).
Try collecting the terms for nums[0], nums[1], …, nums[j-1] when computing the power of heroes ending at index j to get the power in a single pass.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2681: Power of Heroes
class Solution {
    public int sumOfPower(int[] nums) {
        final int mod = (int) 1e9 + 7;
        Arrays.sort(nums);
        long ans = 0, p = 0;
        for (int i = nums.length - 1; i >= 0; --i) {
            long x = nums[i];
            ans = (ans + (x * x % mod) * x) % mod;
            ans = (ans + x * p % mod) % mod;
            p = (p * 2 + x * x % mod) % mod;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #2681: Power of Heroes
func sumOfPower(nums []int) (ans int) {
	const mod = 1e9 + 7
	sort.Ints(nums)
	p := 0
	for i := len(nums) - 1; i >= 0; i-- {
		x := nums[i]
		ans = (ans + (x*x%mod)*x) % mod
		ans = (ans + x*p%mod) % mod
		p = (p*2 + x*x%mod) % mod
	}
	return
}

# Accepted solution for LeetCode #2681: Power of Heroes
class Solution:
    def sumOfPower(self, nums: List[int]) -> int:
        mod = 10**9 + 7
        nums.sort()
        ans = 0
        p = 0
        for x in nums[::-1]:
            ans = (ans + (x * x % mod) * x) % mod
            ans = (ans + x * p) % mod
            p = (p * 2 + x * x) % mod
        return ans

// Accepted solution for LeetCode #2681: Power of Heroes
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2681: Power of Heroes
// class Solution {
//     public int sumOfPower(int[] nums) {
//         final int mod = (int) 1e9 + 7;
//         Arrays.sort(nums);
//         long ans = 0, p = 0;
//         for (int i = nums.length - 1; i >= 0; --i) {
//             long x = nums[i];
//             ans = (ans + (x * x % mod) * x) % mod;
//             ans = (ans + x * p % mod) % mod;
//             p = (p * 2 + x * x % mod) % mod;
//         }
//         return (int) ans;
//     }
// }

// Accepted solution for LeetCode #2681: Power of Heroes
function sumOfPower(nums: number[]): number {
    const mod = 10 ** 9 + 7;
    nums.sort((a, b) => a - b);
    let ans = 0;
    let p = 0;
    for (let i = nums.length - 1; i >= 0; --i) {
        const x = BigInt(nums[i]);
        ans = (ans + Number((x * x * x) % BigInt(mod))) % mod;
        ans = (ans + Number((x * BigInt(p)) % BigInt(mod))) % mod;
        p = Number((BigInt(p) * 2n + x * x) % BigInt(mod));
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.