Formatted question description: https://leetcode.ca/all/1906.html
1906. Minimum Absolute Difference Queries
Level
Medium
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]
is2  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] = [l_i, r_i]
. For each query i
, compute the minimum absolute difference of the subarray nums[l_i...r_i]
containing the elements of nums
between the 0based indices l_i
and r_i
(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 13 = 2. 
queries[1] = [1,2]: The subarray is [3,4] and the minimum absolute difference is 34 = 1. 
queries[2] = [2,3]: The subarray is [4,8] and the minimum absolute difference is 48 = 4. 
queries[3] = [0,3]: The subarray is [1,3,4,8] and the minimum absolute difference is 34 = 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 45 = 1. 
queries[2] = [0,5]: The subarray is [4,5,2,2,7,10] and the minimum absolute difference is 45 = 1. 
queries[3] = [3,5]: The subarray is [2,7,10] and the minimum absolute difference is 710 = 3.
Constraints:
2 <= nums.length <= 10^5
1 <= nums[i] <= 100
1 <= queries.length <= 2 * 10^4
0 <= l_i < r_i < nums.length
Solution
Since the number range of nums
is limited, store the number of occurrences for each number from 1 to 100 for each prefix of nums
. Then for each query, the number of occurrences for each number from 1 to 100 in the query range can be calculated efficiently, and the minimum absolute difference can be calculated by looping over the numbers from 1 to 100 and calculating the differences between each paif of adjacent numbers that both occur in the range.
class Solution {
public int[] minDifference(int[] nums, int[][] queries) {
Map<Integer, int[]> map = new HashMap<Integer, int[]>();
int length = nums.length;
int[] counts = new int[101];
for (int i = 0; i < length; i++) {
int num = nums[i];
counts[num]++;
int[] copy = new int[101];
System.arraycopy(counts, 0, copy, 0, 101);
map.put(i, copy);
}
int queriesCount = queries.length;
int[] ans = new int[queriesCount];
for (int i = 0; i < queriesCount; i++) {
int[] query = queries[i];
int[] rangeCounts = getRangeCounts(map, query[0], query[1]);
int curMin = Integer.MAX_VALUE;
int prev = 1;
for (int j = 1; j <= 100; j++) {
if (rangeCounts[j] > 0) {
if (prev > 0) {
int diff = j  prev;
curMin = Math.min(curMin, diff);
}
prev = j;
}
}
ans[i] = curMin == Integer.MAX_VALUE ? 1 : curMin;
}
return ans;
}
public int[] getRangeCounts(Map<Integer, int[]> map, int start, int end) {
int[] endCounts = map.get(end);
if (start == 0)
return endCounts;
int[] prevCounts = map.get(start  1);
int[] rangeCounts = new int[101];
for (int i = 0; i <= 100; i++)
rangeCounts[i] = endCounts[i]  prevCounts[i];
return rangeCounts;
}
}