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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 subarrays 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 1

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      private long m;
      private int[] nums;
  
      public int medianOfUniquenessArray(int[] nums) {
          int n = nums.length;
          this.nums = nums;
          m = (1L + n) * n / 2;
          int l = 0, r = n;
          while (l < r) {
              int mid = (l + r) >> 1;
              if (check(mid)) {
                  r = mid;
              } else {
                  l = mid + 1;
              }
          }
          return l;
      }
  
      private boolean check(int mx) {
          Map<Integer, Integer> cnt = new HashMap<>();
          long k = 0;
          for (int l = 0, r = 0; r < nums.length; ++r) {
              int x = nums[r];
              cnt.merge(x, 1, Integer::sum);
              while (cnt.size() > mx) {
                  int y = nums[l++];
                  if (cnt.merge(y, -1, Integer::sum) == 0) {
                      cnt.remove(y);
                  }
              }
              k += r - l + 1;
              if (k >= (m + 1) / 2) {
                  return true;
              }
          }
          return false;
      }
  }
  

• class Solution {
  public:
      int medianOfUniquenessArray(vector<int>& nums) {
          int n = nums.size();
          using ll = long long;
          ll m = (1LL + n) * n / 2;
          int l = 0, r = n;
          auto check = [&](int mx) -> bool {
              unordered_map<int, int> cnt;
              ll k = 0;
              for (int l = 0, r = 0; r < n; ++r) {
                  int x = nums[r];
                  ++cnt[x];
                  while (cnt.size() > mx) {
                      int y = nums[l++];
                      if (--cnt[y] == 0) {
                          cnt.erase(y);
                      }
                  }
                  k += r - l + 1;
                  if (k >= (m + 1) / 2) {
                      return true;
                  }
              }
              return false;
          };
          while (l < r) {
              int mid = (l + r) >> 1;
              if (check(mid)) {
                  r = mid;
              } else {
                  l = mid + 1;
              }
          }
          return l;
      }
  };
  

• class Solution:
      def medianOfUniquenessArray(self, nums: List[int]) -> int:
          def check(mx: int) -> bool:
              cnt = defaultdict(int)
              k = l = 0
              for r, x in enumerate(nums):
                  cnt[x] += 1
                  while len(cnt) > mx:
                      y = nums[l]
                      cnt[y] -= 1
                      if cnt[y] == 0:
                          cnt.pop(y)
                      l += 1
                  k += r - l + 1
                  if k >= (m + 1) // 2:
                      return True
              return False
  
          n = len(nums)
          m = (1 + n) * n // 2
          return bisect_left(range(n), True, key=check)
  
  

• func medianOfUniquenessArray(nums []int) int {
  	n := len(nums)
  	m := (1 + n) * n / 2
  	return sort.Search(n, func(mx int) bool {
  		cnt := map[int]int{}
  		l, k := 0, 0
  		for r, x := range nums {
  			cnt[x]++
  			for len(cnt) > mx {
  				y := nums[l]
  				cnt[y]--
  				if cnt[y] == 0 {
  					delete(cnt, y)
  				}
  				l++
  			}
  			k += r - l + 1
  			if k >= (m+1)/2 {
  				return true
  			}
  		}
  		return false
  	})
  }
  
  

• function medianOfUniquenessArray(nums: number[]): number {
      const n = nums.length;
      const m = Math.floor(((1 + n) * n) / 2);
      let [l, r] = [0, n];
      const check = (mx: number): boolean => {
          const cnt = new Map<number, number>();
          let [l, k] = [0, 0];
          for (let r = 0; r < n; ++r) {
              const x = nums[r];
              cnt.set(x, (cnt.get(x) || 0) + 1);
              while (cnt.size > mx) {
                  const y = nums[l++];
                  cnt.set(y, cnt.get(y)! - 1);
                  if (cnt.get(y) === 0) {
                      cnt.delete(y);
                  }
              }
              k += r - l + 1;
              if (k >= Math.floor((m + 1) / 2)) {
                  return true;
              }
          }
          return false;
      };
      while (l < r) {
          const mid = (l + r) >> 1;
          if (check(mid)) {
              r = mid;
          } else {
              l = mid + 1;
          }
      }
      return l;
  }
  
  

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