Leetcode 2841: Maximum Sum of Almost Unique Subarray
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 in nums. A subarray is almost unique if it contains at least m distinct elements. If no such subarray exists, return 0.
Problem
Approach
Steps
Complexity
Input: The input consists of an array nums and two integers m and k.
Example: nums = [1, 3, 5, 7, 1, 8], m = 2, k = 3
Constraints:
• 1 <= nums.length <= 2 * 10^4
• 1 <= m <= k <= nums.length
• 1 <= nums[i] <= 10^9
Output: Return the maximum sum of any almost unique subarray of length k, or return 0 if no such subarray exists.
Example: 16
Constraints:
• The output should be the maximum sum of a valid subarray or 0 if no valid subarray is found.
Goal: The goal is to find the maximum sum of subarrays of length k with at least m distinct elements.
Steps:
• Initialize a sliding window of size k and a map to track the frequency of elements in the window.
• Iterate through the array, adjusting the window to include k elements while maintaining the condition of having at least m distinct elements.
• Track the sum of the elements in the window and update the maximum sum when the window contains at least m distinct elements.
Goal: The constraints on the input are as follows:
Steps:
• The array nums has at least one element.
• 1 <= m <= k <= nums.length
• nums contains integer elements within the range 1 <= nums[i] <= 10^9.
Assumptions:
• The array contains only positive integers.
• The solution should be efficient enough to handle arrays up to the length of 2 * 10^4.
• Input: nums = [1, 3, 5, 7, 1, 8], m = 2, k = 3
• Explanation: We can form the following subarrays of length 3: [1, 3, 5], [3, 5, 7], [5, 7, 1], [7, 1, 8]. The maximum sum of the valid subarray with at least 2 distinct elements is 16, from the subarray [7, 1, 8].
• Input: nums = [9, 9, 1, 5, 6, 6], m = 2, k = 3
• Explanation: The subarray [9, 9, 1] has a sum of 19 and contains at least 2 distinct elements, so it is valid. The maximum sum is 20, as no other valid subarray has a greater sum.
• Input: nums = [1, 1, 1, 1], m = 2, k = 2
• Explanation: No subarrays of length 2 contain at least 2 distinct elements, so the result is 0.
Approach: The approach involves using a sliding window to efficiently find subarrays of size k and check if they contain at least m distinct elements.
Observations:
• We need to efficiently find subarrays of length k with at least m distinct elements.
• A sliding window approach with a hashmap can help track distinct elements and their frequencies.
• By iterating through the array and maintaining a window of size k, we can ensure that the solution works in linear time.
Steps:
• Use a sliding window to examine subarrays of length k.
• Track the number of distinct elements in the current window using a hashmap.
• If the number of distinct elements is at least m, compute the sum of the subarray and update the maximum sum if necessary.
Empty Inputs:
• The array will not be empty, as nums.length >= 1.
Large Inputs:
• The algorithm should efficiently handle arrays with up to 20,000 elements.
Special Values:
• All elements in nums could be the same, leading to no valid subarray if m > 1.
Constraints:
• Ensure the sliding window is implemented efficiently to handle the largest input sizes.
long long maxSum(vector<int>& nums, int size, int k) {
unordered_map<int,int> m;
long long sum=0;
long long maxi=0;
int i=0;
int j=0;
while(j<nums.size())
{
m[nums[j]]++;
sum+=nums[j];
if(j-i+1>k)
{
m[nums[i]]--;
sum-=nums[i];
if(m[nums[i]]==0)
{
m.erase(nums[i]);
}
i++;
}
if(j-i+1==k)
{
if(m.size() >= size)
{
maxi=max(sum , maxi);
// cout<<maxi<<endl;
}
}
j++;
}
return maxi;
}
1 : Initialization
long long maxSum(vector<int>& nums, int size, int k) {
Function definition to find the maximum sum of subarray with at least 'size' distinct elements and subarray length 'k'.
2 : Initialization
unordered_map<int,int> m;
Create a map to track the frequency of elements within the sliding window.
3 : Initialization
long long sum=0;
Variable to store the sum of elements within the current sliding window.
4 : Initialization
long long maxi=0;
Variable to store the maximum sum found during the iteration.
5 : Initialization
int i=0;
Left pointer of the sliding window, used to maintain the window size.
6 : Initialization
int j=0;
Right pointer of the sliding window, used to iterate through the array.
7 : Loop
while(j<nums.size())
Start the sliding window loop over the array.
8 : Window Expansion
m[nums[j]]++;
Increase the count of the current element in the map.
9 : Window Expansion
sum+=nums[j];
Add the current element to the sum of the sliding window.
10 : Check Window Size
if(j-i+1>k)
Check if the window size exceeds 'k'. If true, adjust the window by moving the left pointer.
11 : Window Shrinking
m[nums[i]]--;
Decrease the count of the element at the left pointer.
12 : Window Shrinking
sum-=nums[i];
Subtract the element at the left pointer from the sum.
13 : Window Shrinking
if(m[nums[i]]==0)
If the element at the left pointer has no more occurrences in the map, remove it.
14 : Window Shrinking
m.erase(nums[i]);
Erase the element from the map if its count reaches zero.
15 : Update Left Pointer
i++;
Move the left pointer to shrink the window.
16 : Window Validation
if(j-i+1==k)
Check if the current window size matches 'k'.
17 : Max Sum Check
if(m.size() >= size)
If the window contains at least 'size' distinct elements, check if the sum is a new maximum.
18 : Max Sum Update
maxi=max(sum , maxi);
Update the maximum sum if the current sum is larger.
19 : Window Expansion
j++;
Move the right pointer to expand the window.
20 : Return Result
return maxi;
Return the maximum sum found during the loop.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear, O(n), because we only pass through the array once, adjusting the sliding window.
Best Case: O(k)
Worst Case: O(k)
Description: The space complexity is O(k) due to the storage required for the sliding window and the hashmap of size k.
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