Maximum Coins Heroes Can Collect

class Solution {
public:
    vector<long long> collectMaxCoins(vector<int>& heroes, vector<int>& monsters, vector<int>& coins) {
        vector<long long> result(heroes.size());
        vector<pair<int, int>> pairedMonsters(monsters.size());
    
        for (int i = 0; i < monsters.size(); i++) {
            pairedMonsters[i] = make_pair(monsters[i], coins[i]);
        }
    
        // Sorting monsters and coins together by monster strength
        sort(pairedMonsters.begin(), pairedMonsters.end());
    
        vector<long long> accumulatedCoins(coins.size());
        long long cumulativeSum = 0;
        for (int i = 0; i < pairedMonsters.size(); i++) {
            cumulativeSum += pairedMonsters[i].second;
            accumulatedCoins[i] = cumulativeSum;
        }
    
        for (int i = 0; i < heroes.size(); i++) {
            result[i] = calculateMaximumCoins(pairedMonsters, heroes[i], accumulatedCoins);
        }
    
        return result;
    }
    
private:
    long long calculateMaximumCoins(const vector<pair<int, int>>& pairedMonsters,
                                    int heroStrength, const vector<long long>& accumulatedCoins) {
        int left = 0;
        int right = pairedMonsters.size() - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (pairedMonsters[mid].first > heroStrength)
                right = mid - 1;
            else
                left = mid + 1;
        }
            
        if (left == 0 && pairedMonsters[left].first > heroStrength) {
            return 0; // no monsters can be defeated
        }
    
        return accumulatedCoins[right];
    }
};

This solution outlines an approach to finding out the maximum number of coins that heroes can collect by defeating monsters, as determined by their respective strengths. The solution employs the C++ programming language, utilizing standard library features such as vectors and pairs.

Implementation Details:

• Hero and Monster Modeling:

  • Two sets of vectors are used—one to represent the heroes' strengths and another to represent monsters with associated coins, modeled as pairs.

• Sorting Monsters:

  • Create pairs from monsters and their respective coin values. These pairs are sorted based on the monster's strength, ensuring that heroes can potentially maximize coin collection by defeating weaker monsters first.

• Coin Accumulation Calculation:

  • Compute a prefix sum for the coins that reflects the total coins achievable from defeating a sequence of the weakest monsters up to each position in the sorted list.

• Main Calculation Loop:

  • For each hero, determine how many monsters they can defeat based on their strength using a binary search mechanism. The binary search in the calculateMaximumCoins function finds the upper bound of monster's strength a hero can defeat and uses this position to index into the precomputed prefix sum array to find the maximum coins collectible.

• Binary Search Mechanism:

  • Adjust the search space based on the comparison between hero's strength and the monster's strength at the midpoint. This iteratively narrows down the section of the array, which directly corresponds to the range of monsters a hero can potentially defeat.

• Edge Case Handling:

  • Consider scenarios where no monsters can be defeated if all are stronger than the hero by checking the strength of the weakest monster against the hero's strength.

Output Details:

• The result is a vector where each entry corresponds to the maximum coins that can be collected by each hero based on their strength against the sorted list of monsters.

This method provides an efficient and logical approach to allocating resources (heroes) against challenges (monsters) while maximizing rewards (coins). The use of sorting, paired data structuring, and binary search ensures that the solution remains computationally feasible even for larger numbers of heroes or monsters.