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 subarraynums[li...ri] containing the elements of nums between the 0-based indices li and ri (inclusive).
Return an arrayanswhereans[i]is the answer to theithquery.
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.
functionminDifference(nums:number[],queries:number[][]):number[]{letm=nums.length,n=queries.length;letmax=100;// let max = Math.max(...nums);letpre:number[][]=[];pre.push(newArray(max+1).fill(0));for(leti=0;i<m;++i){letnum=nums[i];pre.push(pre[i].slice());pre[i+1][num]+=1;}letans=[];for(let[left,right]ofqueries){letlast=-1;letmin=Infinity;for(letj=1;j<max+1;++j){if(pre[left][j]<pre[right+1][j]){if(last!=-1){min=Math.min(min,j-last);}last=j;}}ans.push(min==Infinity?-1:min);}returnans;}