Description#
You are given an integer array nums and two positive integers m and k.
Return the maximum sum out of all almost unique subarrays of length k of nums. If no such subarray exists, return 0.
A subarray of nums is almost unique if it contains at least m distinct elements.
A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums = [2,6,7,3,1,7], m = 3, k = 4
Output: 18
Explanation: There are 3 almost unique subarrays of size k = 4. These subarrays are [2, 6, 7, 3], [6, 7, 3, 1], and [7, 3, 1, 7]. Among these subarrays, the one with the maximum sum is [2, 6, 7, 3] which has a sum of 18.
Example 2:
Input: nums = [5,9,9,2,4,5,4], m = 1, k = 3
Output: 23
Explanation: There are 5 almost unique subarrays of size k. These subarrays are [5, 9, 9], [9, 9, 2], [9, 2, 4], [2, 4, 5], and [4, 5, 4]. Among these subarrays, the one with the maximum sum is [5, 9, 9] which has a sum of 23.
Example 3:
Input: nums = [1,2,1,2,1,2,1], m = 3, k = 3
Output: 0
Explanation: There are no subarrays of size k = 3 that contain at least m = 3 distinct elements in the given array [1,2,1,2,1,2,1]. Therefore, no almost unique subarrays exist, and the maximum sum is 0.
Constraints:
1 <= nums.length <= 2 * 1041 <= m <= k <= nums.length1 <= nums[i] <= 109
Solutions#
Solution 1: Sliding Window + Hash Table#
We can traverse the array $nums$, maintain a window of size $k$, use a hash table $cnt$ to count the occurrence of each element in the window, and use a variable $s$ to sum all elements in the window. If the number of different elements in $cnt$ is greater than or equal to $m$, then we update the answer $ans = \max(ans, s)$.
After the traversal ends, return the answer.
The time complexity is $O(n)$, and the space complexity is $O(k)$. Here, $n$ is the length of the array.
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| class Solution:
def maxSum(self, nums: List[int], m: int, k: int) -> int:
cnt = Counter(nums[:k])
s = sum(nums[:k])
ans = 0
if len(cnt) >= m:
ans = s
for i in range(k, len(nums)):
cnt[nums[i]] += 1
cnt[nums[i - k]] -= 1
s += nums[i] - nums[i - k]
if cnt[nums[i - k]] == 0:
cnt.pop(nums[i - k])
if len(cnt) >= m:
ans = max(ans, s)
return ans
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| class Solution {
public long maxSum(List<Integer> nums, int m, int k) {
Map<Integer, Integer> cnt = new HashMap<>();
int n = nums.size();
long s = 0;
for (int i = 0; i < k; ++i) {
cnt.merge(nums.get(i), 1, Integer::sum);
s += nums.get(i);
}
long ans = 0;
if (cnt.size() >= m) {
ans = s;
}
for (int i = k; i < n; ++i) {
cnt.merge(nums.get(i), 1, Integer::sum);
if (cnt.merge(nums.get(i - k), -1, Integer::sum) == 0) {
cnt.remove(nums.get(i - k));
}
s += nums.get(i) - nums.get(i - k);
if (cnt.size() >= m) {
ans = Math.max(ans, s);
}
}
return ans;
}
}
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| class Solution {
public:
long long maxSum(vector<int>& nums, int m, int k) {
unordered_map<int, int> cnt;
long long s = 0;
int n = nums.size();
for (int i = 0; i < k; ++i) {
cnt[nums[i]]++;
s += nums[i];
}
long long ans = cnt.size() >= m ? s : 0;
for (int i = k; i < n; ++i) {
cnt[nums[i]]++;
if (--cnt[nums[i - k]] == 0) {
cnt.erase(nums[i - k]);
}
s += nums[i] - nums[i - k];
if (cnt.size() >= m) {
ans = max(ans, s);
}
}
return ans;
}
};
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| func maxSum(nums []int, m int, k int) int64 {
cnt := map[int]int{}
var s int64
for _, x := range nums[:k] {
cnt[x]++
s += int64(x)
}
var ans int64
if len(cnt) >= m {
ans = s
}
for i := k; i < len(nums); i++ {
cnt[nums[i]]++
cnt[nums[i-k]]--
if cnt[nums[i-k]] == 0 {
delete(cnt, nums[i-k])
}
s += int64(nums[i]) - int64(nums[i-k])
if len(cnt) >= m {
ans = max(ans, s)
}
}
return ans
}
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| function maxSum(nums: number[], m: number, k: number): number {
const n = nums.length;
const cnt: Map<number, number> = new Map();
let s = 0;
for (let i = 0; i < k; ++i) {
cnt.set(nums[i], (cnt.get(nums[i]) || 0) + 1);
s += nums[i];
}
let ans = cnt.size >= m ? s : 0;
for (let i = k; i < n; ++i) {
cnt.set(nums[i], (cnt.get(nums[i]) || 0) + 1);
cnt.set(nums[i - k], cnt.get(nums[i - k])! - 1);
if (cnt.get(nums[i - k]) === 0) {
cnt.delete(nums[i - k]);
}
s += nums[i] - nums[i - k];
if (cnt.size >= m) {
ans = Math.max(ans, s);
}
}
return ans;
}
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