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]is|2 - 3| = 1. Note that it is not0becausea[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 0-based indices l_i and r_i (inclusive).
Return an array ans where ans[i] is the answer to the i-th query.
A subarray is a contiguous sequence of elements in an array.
The value of |x| is defined as:
xifx >= 0.-xifx < 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 <= 10^51 <= nums[i] <= 1001 <= queries.length <= 2 * 10^40 <= 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;
}
}