1906. Minimum Absolute Difference Queries

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] = [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 arrayans 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

 

Solutions

Solution: Hash Table

• Time complexity: O(n+q∗100∗log(n))
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} nums
 * @param {number[][]} queries
 * @return {number[]}
 */
const minDifference = function (nums, queries) {
  const MAX_NUM = 100;
  const numIndices = Array.from({length: MAX_NUM + 1});
  const isWithinRange = (left, right, target) => {
    const indices = numIndices[target];
    if (!indices) return false;
    let start = 0;
    let end = indices.length - 1;

    while (start <= end) {
      const mid = Math.floor((start + end) / 2);

      if (indices[mid] >= left && indices[mid] <= right) return true;
      indices[mid] > right ? (end = mid - 1) : (start = mid + 1);
    }
    return false;
  };

  nums.forEach((num, index) => {
    const indices = numIndices[num] ?? [];

    numIndices[num] = indices;
    indices.push(index);
  });

  return queries.map(([left, right]) => {
    let minimum = MAX_NUM + 1;
    let previous = 0;

    for (let num = 1; num <= MAX_NUM; num++) {
      if (!isWithinRange(left, right, num)) continue;
      if (previous) minimum = Math.min(num - previous, minimum);
      previous = num;
    }
    return minimum > MAX_NUM ? -1 : minimum;
  });
};