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 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] = [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:

  • 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 <= 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;
    }
}

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