3478. Choose K Elements With Maximum Sum
Description
You are given two integer arrays, nums1 and nums2, both of length n, along with a positive integer k.
For each index i from 0 to n - 1, perform the following:
- Find all indices
jwherenums1[j]is less thannums1[i]. - Choose at most
kvalues ofnums2[j]at these indices to maximize the total sum.
Return an array answer of size n, where answer[i] represents the result for the corresponding index i.
Example 1:
Input: nums1 = [4,2,1,5,3], nums2 = [10,20,30,40,50], k = 2
Output: [80,30,0,80,50]
Explanation:
- For
i = 0: Select the 2 largest values fromnums2at indices[1, 2, 4]wherenums1[j] < nums1[0], resulting in50 + 30 = 80. - For
i = 1: Select the 2 largest values fromnums2at index[2]wherenums1[j] < nums1[1], resulting in 30. - For
i = 2: No indices satisfynums1[j] < nums1[2], resulting in 0. - For
i = 3: Select the 2 largest values fromnums2at indices[0, 1, 2, 4]wherenums1[j] < nums1[3], resulting in50 + 30 = 80. - For
i = 4: Select the 2 largest values fromnums2at indices[1, 2]wherenums1[j] < nums1[4], resulting in30 + 20 = 50.
Example 2:
Input: nums1 = [2,2,2,2], nums2 = [3,1,2,3], k = 1
Output: [0,0,0,0]
Explanation:
Since all elements in nums1 are equal, no indices satisfy the condition nums1[j] < nums1[i] for any i, resulting in 0 for all positions.
Constraints:
n == nums1.length == nums2.length1 <= n <= 1051 <= nums1[i], nums2[i] <= 1061 <= k <= n
Solutions
Solution 1: Sorting + Priority Queue (Min-Heap)
We can convert the array $\textit{nums1}$ into an array $\textit{arr}$, where each element is a tuple $(x, i)$, representing the value $x$ at index $i$ in $\textit{nums1}$. Then, we sort the array $\textit{arr}$ in ascending order by $x$.
We use a min-heap $\textit{pq}$ to maintain the elements from the array $\textit{nums2}$. Initially, $\textit{pq}$ is empty. We use a variable $\textit{s}$ to record the sum of the elements in $\textit{pq}$. Additionally, we use a pointer $j$ to maintain the current position in the array $\textit{arr}$ that needs to be added to $\textit{pq}$.
We traverse the array $\textit{arr}$. For the $h$-th element $(x, i)$, we add all elements $\textit{nums2}[\textit{arr}[j][1]]$ to $\textit{pq}$ that satisfy $j < h$ and $\textit{arr}[j][0] < x$, and add these elements to $\textit{s}$. If the size of $\textit{pq}$ exceeds $k$, we pop the smallest element from $\textit{pq}$ and subtract it from $\textit{s}$. Then, we update the value of $\textit{ans}[i]$ to $\textit{s}$.
After traversing, we return the answer array $\textit{ans}$.
The time complexity is $O(n \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array.
Python3
class Solution:
def findMaxSum(self, nums1: List[int], nums2: List[int], k: int) -> List[int]:
arr = [(x, i) for i, x in enumerate(nums1)]
arr.sort()
pq = []
s = j = 0
n = len(arr)
ans = [0] * n
for h, (x, i) in enumerate(arr):
while j < h and arr[j][0] < x:
y = nums2[arr[j][1]]
heappush(pq, y)
s += y
if len(pq) > k:
s -= heappop(pq)
j += 1
ans[i] = s
return ans
Java
class Solution {
public long[] findMaxSum(int[] nums1, int[] nums2, int k) {
int n = nums1.length;
int[][] arr = new int[n][0];
for (int i = 0; i < n; ++i) {
arr[i] = new int[] {nums1[i], i};
}
Arrays.sort(arr, (a, b) -> a[0] - b[0]);
PriorityQueue<Integer> pq = new PriorityQueue<>();
long s = 0;
long[] ans = new long[n];
int j = 0;
for (int h = 0; h < n; ++h) {
int x = arr[h][0], i = arr[h][1];
while (j < h && arr[j][0] < x) {
int y = nums2[arr[j][1]];
pq.offer(y);
s += y;
if (pq.size() > k) {
s -= pq.poll();
}
++j;
}
ans[i] = s;
}
return ans;
}
}
C++
class Solution {
public:
vector<long long> findMaxSum(vector<int>& nums1, vector<int>& nums2, int k) {
int n = nums1.size();
vector<pair<int, int>> arr(n);
for (int i = 0; i < n; ++i) {
arr[i] = {nums1[i], i};
}
ranges::sort(arr);
priority_queue<int, vector<int>, greater<int>> pq;
long long s = 0;
int j = 0;
vector<long long> ans(n);
for (int h = 0; h < n; ++h) {
auto [x, i] = arr[h];
while (j < h && arr[j].first < x) {
int y = nums2[arr[j].second];
pq.push(y);
s += y;
if (pq.size() > k) {
s -= pq.top();
pq.pop();
}
++j;
}
ans[i] = s;
}
return ans;
}
};
Go
func findMaxSum(nums1 []int, nums2 []int, k int) []int64 {
n := len(nums1)
arr := make([][2]int, n)
for i, x := range nums1 {
arr[i] = [2]int{x, i}
}
ans := make([]int64, n)
sort.Slice(arr, func(i, j int) bool { return arr[i][0] < arr[j][0] })
pq := hp{}
var s int64
j := 0
for h, e := range arr {
x, i := e[0], e[1]
for j < h && arr[j][0] < x {
y := nums2[arr[j][1]]
heap.Push(&pq, y)
s += int64(y)
if pq.Len() > k {
s -= int64(heap.Pop(&pq).(int))
}
j++
}
ans[i] = s
}
return ans
}
type hp struct{ sort.IntSlice }
func (h hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }
func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
a := h.IntSlice
v := a[len(a)-1]
h.IntSlice = a[:len(a)-1]
return v
}
TypeScript
function findMaxSum(nums1: number[], nums2: number[], k: number): number[] {
const n = nums1.length;
const arr = nums1.map((x, i) => [x, i]).sort((a, b) => a[0] - b[0]);
const pq = new MinPriorityQueue();
let [s, j] = [0, 0];
const ans: number[] = Array(k).fill(0);
for (let h = 0; h < n; ++h) {
const [x, i] = arr[h];
while (j < h && arr[j][0] < x) {
const y = nums2[arr[j++][1]];
pq.enqueue(y);
s += y;
if (pq.size() > k) {
s -= pq.dequeue();
}
}
ans[i] = s;
}
return ans;
}