You are given an integer array nums and an integer k. Find the maximum subarray sum of all the subarrays of nums that meet the following conditions:
- The length of the subarray is
k, and - All the elements of the subarray are distinct.
Return the maximum subarray sum of all the subarrays that meet the conditions. If no subarray meets the conditions, return 0.
A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums = [1,5,4,2,9,9,9], k = 3 Output: 15 Explanation: The subarrays of nums with length 3 are: - [1,5,4] which meets the requirements and has a sum of 10. - [5,4,2] which meets the requirements and has a sum of 11. - [4,2,9] which meets the requirements and has a sum of 15. - [2,9,9] which does not meet the requirements because the element 9 is repeated. - [9,9,9] which does not meet the requirements because the element 9 is repeated. We return 15 because it is the maximum subarray sum of all the subarrays that meet the conditions
Example 2:
Input: nums = [4,4,4], k = 3 Output: 0 Explanation: The subarrays of nums with length 3 are: - [4,4,4] which does not meet the requirements because the element 4 is repeated. We return 0 because no subarrays meet the conditions.
Constraints:
1 <= k <= nums.length <= 1051 <= nums[i] <= 105
Companies: Amazon
Related Topics:
Array, Hash Table, Sliding Window
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- Longest Nice Subarray (Medium)
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// OJ: https://leetcode.com/problems/maximum-sum-of-distinct-subarrays-with-length-k
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(K)
class Solution {
public:
long long maximumSubarraySum(vector<int>& A, int k) {
unordered_map<int, int> cnt;
long long sum = 0, N = A.size(), ans = 0;
for (int i = 0; i < N; ++i) {
sum += A[i];
cnt[A[i]]++;
if (i - k >= 0) {
sum -= A[i - k];
if (--cnt[A[i - k]] == 0) cnt.erase(A[i - k]);
}
if (cnt.size() == k) ans = max(ans, sum);
}
return ans;
}
};