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3592. Find X Sum Of All K Long Subarrays II

Array Hash Table Sliding Window Heap (Priority Queue)

Problem - Find X Sum Of All K Long Subarrays II

Hard

You are given an array nums of n integers and two integers k and x.

The x-sum of an array is calculated by the following procedure:

  • Count the occurrences of all elements in the array.
  • Keep only the occurrences of the top x most frequent elements. If two elements have the same number of occurrences, the element with the bigger value is considered more frequent.
  • Calculate the sum of the resulting array.

Note that if an array has less than x distinct elements, its x-sum is the sum of the array.

Return an integer array answer of length n - k + 1 where answer[i] is the x-sum of the subarray nums[i..i + k - 1].

 

Example 1:

Input: nums = [1,1,2,2,3,4,2,3], k = 6, x = 2

Output: [6,10,12]

Explanation:

  • For subarray [1, 1, 2, 2, 3, 4], only elements 1 and 2 will be kept in the resulting array. Hence, answer[0] = 1 + 1 + 2 + 2.
  • For subarray [1, 2, 2, 3, 4, 2], only elements 2 and 4 will be kept in the resulting array. Hence, answer[1] = 2 + 2 + 2 + 4. Note that 4 is kept in the array since it is bigger than 3 and 1 which occur the same number of times.
  • For subarray [2, 2, 3, 4, 2, 3], only elements 2 and 3 are kept in the resulting array. Hence, answer[2] = 2 + 2 + 2 + 3 + 3.

Example 2:

Input: nums = [3,8,7,8,7,5], k = 2, x = 2

Output: [11,15,15,15,12]

Explanation:

Since k == x, answer[i] is equal to the sum of the subarray nums[i..i + k - 1].

 

Constraints:

  • nums.length == n
  • 1 <= n <= 105
  • 1 <= nums[i] <= 109
  • 1 <= x <= k <= nums.length

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class Solution:
    def findXSum(self, nums: List[int], k: int, x: int) -> List[int]:
        left = SortedList()
        right = SortedList()
        count = Counter()
        sum_val = 0
        n = len(nums)
        result = [0] * (n - k + 1)

        def add(val: int):
            if count[val] == 0:
                return
            temp = (count[val], val)
            if left and temp > left[0]:
                nonlocal sum_val
                sum_val += temp[0] * temp[1]
                left.add(temp)
            else:
                right.add(temp)

        def remove(val: int):
            if count[val] == 0:
                return
            temp = (count[val], val)
            if temp in left:
                nonlocal sum_val
                sum_val -= temp[0] * temp[1]
                left.remove(temp)
            else:
                right.remove(temp)

        for i, val in enumerate(nums):
            remove(val)
            count[val] += 1
            add(val)
            j = i - k + 1
            if j < 0:
                continue
            while right and len(left) < x:
                temp = right.pop()
                left.add(temp)
                sum_val += temp[0] * temp[1]

            while len(left) > x:
                temp = left.pop(0)
                sum_val -= temp[0] * temp[1]
                right.add(temp)
            result[j] = sum_val

            remove(nums[j])
            count[nums[j]] -= 1
            add(nums[j])

        return result

Submission Stats:

  • Runtime: 2986 ms (76.19%)
  • Memory: 43.7 MB (95.24%)