Description
You are given an integer array nums
of length n
and a 2D array queries
where queries[i] = [li, ri]
.
Each queries[i]
represents the following action on nums
:
- Decrement the value at each index in the range
[li, ri]
innums
by at most 1. - The amount by which the value is decremented can be chosen independently for each index.
A Zero Array is an array with all its elements equal to 0.
Return the maximum number of elements that can be removed from queries
, such that nums
can still be converted to a zero array using the remaining queries. If it is not possible to convert nums
to a zero array, return -1.
Β
Example 1:
Input: nums = [2,0,2], queries = [[0,2],[0,2],[1,1]]
Output: 1
Explanation:
After removing queries[2]
, nums
can still be converted to a zero array.
- Using
queries[0]
, decrementnums[0]
andnums[2]
by 1 andnums[1]
by 0. - Using
queries[1]
, decrementnums[0]
andnums[2]
by 1 andnums[1]
by 0.
Example 2:
Input: nums = [1,1,1,1], queries = [[1,3],[0,2],[1,3],[1,2]]
Output: 2
Explanation:
We can remove queries[2]
and queries[3]
.
Example 3:
Input: nums = [1,2,3,4], queries = [[0,3]]
Output: -1
Explanation:
nums
cannot be converted to a zero array even after using all the queries.
Β
Constraints:
1 <= nums.length <= 105
0 <= nums[i] <= 105
1 <= queries.length <= 105
queries[i].length == 2
0 <= li <= ri < nums.length
Solution
Python3
class Solution:
def maxRemoval(self, nums: List[int], queries: List[List[int]]) -> int:
N = len(nums)
queries.sort(key = lambda x : x[0])
pq = []
j = 0
operations = 0
delta = [0] * (N + 1)
for i, num in enumerate(nums):
operations += delta[i]
while j < len(queries) and i == queries[j][0]:
heappush(pq, -queries[j][1])
j += 1
while operations < num and pq and i <= -pq[0]:
operations += 1
delta[-heappop(pq) + 1] -= 1
if operations < num:
return -1
return len(pq)