3134. Find the Median of the Uniqueness Array

Description

You are given an integer array nums. The uniqueness array of nums is the sorted array that contains the number of distinct elements of all the of nums. In other words, it is a sorted array consisting of distinct(nums[i..j]), for all 0 <= i <= j < nums.length.

Here, distinct(nums[i..j]) denotes the number of distinct elements in the subarray that starts at index i and ends at index j.

Return the median of the uniqueness array of nums.

Note that the median of an array is defined as the middle element of the array when it is sorted in non-decreasing order. If there are two choices for a median, the smaller of the two values is taken.

 

Example 1:

Input: nums = [1,2,3]

Output: 1

Explanation:

The uniqueness array of nums is [distinct(nums[0..0]), distinct(nums[1..1]), distinct(nums[2..2]), distinct(nums[0..1]), distinct(nums[1..2]), distinct(nums[0..2])] which is equal to [1, 1, 1, 2, 2, 3]. The uniqueness array has a median of 1. Therefore, the answer is 1.

Example 2:

Input: nums = [3,4,3,4,5]

Output: 2

Explanation:

The uniqueness array of nums is [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3]. The uniqueness array has a median of 2. Therefore, the answer is 2.

Example 3:

Input: nums = [4,3,5,4]

Output: 2

Explanation:

The uniqueness array of nums is [1, 1, 1, 1, 2, 2, 2, 3, 3, 3]. The uniqueness array has a median of 2. Therefore, the answer is 2.

 

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 105

 

Solutions

Solution: Binary Search + Sliding Window

• Time complexity: O(nlogn)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
const medianOfUniquenessArray = function (nums) {
  const n = nums.length;
  const subarrayCount = (n * (n + 1)) / 2;
  const midianCount = Math.floor((subarrayCount + 1) / 2);

  const getMostKDistinctSubarrayCount = k => {
    const numMap = new Map();
    let distinct = 0;
    let left = 0;
    let result = 0;

    for (let index = 0; index < n; index++) {
      const num = nums[index];
      const count = numMap.get(num) ?? 0;

      if (!count) {
        distinct += 1;
      }

      numMap.set(num, count + 1);

      while (distinct > k) {
        const leftNum = nums[left];
        const leftCount = numMap.get(leftNum);

        if (leftCount === 1) {
          numMap.delete(leftNum);
          distinct -= 1;
        } else {
          numMap.set(leftNum, leftCount - 1);
        }

        left += 1;
      }

      result += index - left + 1;
    }

    return result;
  };

  let left = 1;
  let right = n;

  while (left <= right) {
    const mid = Math.floor((left + right) / 2);

    if (getMostKDistinctSubarrayCount(mid) >= midianCount) {
      right = mid - 1;
    } else {
      left = mid + 1;
    }
  }

  return left;
};