1906. Minimum Absolute Difference Queries
Description
The minimum absolute difference of an array a
is defined as the minimum value of |a[i] - a[j]|
, where 0 <= i < j < a.length
and a[i] != a[j]
. If all elements of a
are the same, the minimum absolute difference is -1
.
- For example, the minimum absolute difference of the array
[5,2,3,7,2]
is|2 - 3| = 1
. Note that it is not0
becausea[i]
anda[j]
must be different.
You are given an integer array nums
and the array queries
where queries[i] = [li, ri]
. For each query i
, compute the minimum absolute difference of the subarray nums[li...ri]
containing the elements of nums
between the 0-based indices li
and ri
(inclusive).
Return an array ans
where ans[i]
is the answer to the ith
query.
A subarray is a contiguous sequence of elements in an array.
The value of |x|
is defined as:
x
ifx >= 0
.-x
ifx < 0
.
Example 1:
Input: nums = [1,3,4,8], queries = [[0,1],[1,2],[2,3],[0,3]] Output: [2,1,4,1] Explanation: The queries are processed as follows: - queries[0] = [0,1]: The subarray is [1,3] and the minimum absolute difference is |1-3| = 2. - queries[1] = [1,2]: The subarray is [3,4] and the minimum absolute difference is |3-4| = 1. - queries[2] = [2,3]: The subarray is [4,8] and the minimum absolute difference is |4-8| = 4. - queries[3] = [0,3]: The subarray is [1,3,4,8] and the minimum absolute difference is |3-4| = 1.
Example 2:
Input: nums = [4,5,2,2,7,10], queries = [[2,3],[0,2],[0,5],[3,5]] Output: [-1,1,1,3] Explanation: The queries are processed as follows: - queries[0] = [2,3]: The subarray is [2,2] and the minimum absolute difference is -1 because all the elements are the same. - queries[1] = [0,2]: The subarray is [4,5,2] and the minimum absolute difference is |4-5| = 1. - queries[2] = [0,5]: The subarray is [4,5,2,2,7,10] and the minimum absolute difference is |4-5| = 1. - queries[3] = [3,5]: The subarray is [2,7,10] and the minimum absolute difference is |7-10| = 3.
Constraints:
2 <= nums.length <= 105
1 <= nums[i] <= 100
1 <= queries.length <= 2 * 104
0 <= li < ri < nums.length
Solution
minimum-absolute-difference-queries.py
class Solution:
def minDifference(self, nums: List[int], queries: List[List[int]]) -> List[int]:
dp = [[0] * (max(nums) + 1)]
for i, x in enumerate(nums):
t = dp[-1][:]
t[x] += 1
dp.append(t)
res = []
for start, end in queries:
A = [i for i, (a,b) in enumerate(zip(dp[start], dp[end + 1])) if a != b]
res.append(min([b - a for a,b in zip(A, A[1:])] or [-1]))
return res