Description#
There is a battle and n heroes are trying to defeat m monsters. You are given two 1-indexed arrays of positive integers heroes and monsters of length n and m, respectively. heroes[i] is the power of ith hero, and monsters[i] is the power of ith monster.
The ith hero can defeat the jth monster if monsters[j] <= heroes[i].
You are also given a 1-indexed array coins of length m consisting of positive integers. coins[i] is the number of coins that each hero earns after defeating the ith monster.
Return an array ans of length n where ans[i] is the maximum number of coins that the ith hero can collect from this battle.
Notes
- The health of a hero doesn't get reduced after defeating a monster.
- Multiple heroes can defeat a monster, but each monster can be defeated by a given hero only once.
Example 1:
Input: heroes = [1,4,2], monsters = [1,1,5,2,3], coins = [2,3,4,5,6]
Output: [5,16,10]
Explanation: For each hero, we list the index of all the monsters he can defeat:
1st hero: [1,2] since the power of this hero is 1 and monsters[1], monsters[2] <= 1. So this hero collects coins[1] + coins[2] = 5 coins.
2nd hero: [1,2,4,5] since the power of this hero is 4 and monsters[1], monsters[2], monsters[4], monsters[5] <= 4. So this hero collects coins[1] + coins[2] + coins[4] + coins[5] = 16 coins.
3rd hero: [1,2,4] since the power of this hero is 2 and monsters[1], monsters[2], monsters[4] <= 2. So this hero collects coins[1] + coins[2] + coins[4] = 10 coins.
So the answer would be [5,16,10].
Example 2:
Input: heroes = [5], monsters = [2,3,1,2], coins = [10,6,5,2]
Output: [23]
Explanation: This hero can defeat all the monsters since monsters[i] <= 5. So he collects all of the coins: coins[1] + coins[2] + coins[3] + coins[4] = 23, and the answer would be [23].
Example 3:
Input: heroes = [4,4], monsters = [5,7,8], coins = [1,1,1]
Output: [0,0]
Explanation: In this example, no hero can defeat a monster. So the answer would be [0,0],
Constraints:
1 <= n == heroes.length <= 1051 <= m == monsters.length <= 105coins.length == m1 <= heroes[i], monsters[i], coins[i] <= 109
Solutions#
Solution 1: Sorting + Prefix Sum + Binary Search#
We can sort the monsters and coins in ascending order of the monsters’ combat power, and then use prefix sum to calculate the total number of coins each hero can get by defeating the first $i$ monsters.
Next, for each hero, we can use binary search to find the strongest monster he can defeat, and then use prefix sum to calculate the total number of coins he can get.
The time complexity is $O((m + n) \times \log n)$, and the space complexity is $O(m)$. Here, $m$ and $n$ are the number of monsters and heroes, respectively.
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| class Solution:
def maximumCoins(
self, heroes: List[int], monsters: List[int], coins: List[int]
) -> List[int]:
m = len(monsters)
idx = sorted(range(m), key=lambda i: monsters[i])
s = list(accumulate((coins[i] for i in idx), initial=0))
ans = []
for h in heroes:
i = bisect_right(idx, h, key=lambda i: monsters[i])
ans.append(s[i])
return ans
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| class Solution {
public long[] maximumCoins(int[] heroes, int[] monsters, int[] coins) {
int m = monsters.length;
Integer[] idx = new Integer[m];
for (int i = 0; i < m; ++i) {
idx[i] = i;
}
Arrays.sort(idx, Comparator.comparingInt(j -> monsters[j]));
long[] s = new long[m + 1];
for (int i = 0; i < m; ++i) {
s[i + 1] = s[i] + coins[idx[i]];
}
int n = heroes.length;
long[] ans = new long[n];
for (int k = 0; k < n; ++k) {
int i = search(monsters, idx, heroes[k]);
ans[k] = s[i];
}
return ans;
}
private int search(int[] nums, Integer[] idx, int x) {
int l = 0, r = idx.length;
while (l < r) {
int mid = (l + r) >> 1;
if (nums[idx[mid]] > x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
}
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| class Solution {
public:
vector<long long> maximumCoins(vector<int>& heroes, vector<int>& monsters, vector<int>& coins) {
int m = monsters.size();
vector<int> idx(m);
iota(idx.begin(), idx.end(), 0);
sort(idx.begin(), idx.end(), [&](int i, int j) {
return monsters[i] < monsters[j];
});
long long s[m + 1];
s[0] = 0;
for (int i = 1; i <= m; ++i) {
s[i] = s[i - 1] + coins[idx[i - 1]];
}
vector<long long> ans;
auto search = [&](int x) {
int l = 0, r = m;
while (l < r) {
int mid = (l + r) >> 1;
if (monsters[idx[mid]] > x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
};
for (int h : heroes) {
ans.push_back(s[search(h)]);
}
return ans;
}
};
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| func maximumCoins(heroes []int, monsters []int, coins []int) (ans []int64) {
m := len(monsters)
idx := make([]int, m)
for i := range idx {
idx[i] = i
}
sort.Slice(idx, func(i, j int) bool { return monsters[idx[i]] < monsters[idx[j]] })
s := make([]int64, m+1)
for i, j := range idx {
s[i+1] = s[i] + int64(coins[j])
}
for _, h := range heroes {
i := sort.Search(m, func(i int) bool { return monsters[idx[i]] > h })
ans = append(ans, s[i])
}
return
}
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| function maximumCoins(heroes: number[], monsters: number[], coins: number[]): number[] {
const m = monsters.length;
const idx: number[] = Array.from({ length: m }, (_, i) => i);
idx.sort((i, j) => monsters[i] - monsters[j]);
const s: number[] = Array(m + 1).fill(0);
for (let i = 0; i < m; ++i) {
s[i + 1] = s[i] + coins[idx[i]];
}
const search = (x: number): number => {
let l = 0;
let r = m;
while (l < r) {
const mid = (l + r) >> 1;
if (monsters[idx[mid]] > x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
};
return heroes.map(h => s[search(h)]);
}
|
