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:
k, andReturn 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] <= 105class Solution {
public:
long long maximumSubarraySum(vector<int>& nums, int k) {
long long int res = 0, currSum=0;
unordered_set<int> s;
for(int lt=0, rt=0; lt<=nums.size()-k; rt++){
if (s.find(nums[rt])!=s.end()) // found
s.erase(nums[lt]), currSum-=nums[lt], lt++, rt--;
else s.insert(nums[rt]), currSum+=nums[rt];
if(s.size()==k)
res=max(res, currSum), currSum-=nums[lt], s.erase(nums[lt]), lt++;
}
return res;
}
};class Solution {
public long maximumSubarraySum(int[] nums, int k) {
Set<Integer> window = new HashSet<>();
int n = nums.length;
long sum = 0, ans = 0;
for (int i = 0; i < n; i++) {
System.out.println(i + "\t" + sum + "\t" + window);
if (!window.contains(nums[i])) {
if (window.size() < k) {
window.add(nums[i]);
sum += nums[i];
} else {
window.add(nums[i]);
window.remove(nums[i - k]);
sum += nums[i] - nums[i - k];
ans = Math.max(sum, ans);
}
} else {
while (!window.isEmpty() && window.contains(nums[i])) {
sum -= nums[i - window.size()];
window.remove(nums[i - window.size()]);
}
}
}
return ans;
}
}