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Minimum Absolute Difference Queries
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
Program:
class Solution:
def minDifference(self, nums: List[int], queries: List[List[int]]) -> List[int]:
max_el, answer, counter = max(nums), [], [[0]*(max(nums)+1)]
set_of_nums = sorted(list(set(nums)))
for n in nums:
t = counter[-1][:]
t[n] += 1
counter.append(t)
def f(start, finish):
arr = [num for num in set_of_nums if counter[finish+1][num]-counter[start][num]>0]
if len(arr)==1:
return -1
return min([b-a for a,b in zip(arr, arr[1:])])
for q in queries:
answer.append(f(q[0],q[1]))
return answer
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