1906. Minimum Absolute Difference Queries

Problem Link

Description

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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 not 0 because a[i] and a[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 if x >= 0.
• -x if x < 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

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Python3

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