Find the median of the uniqueness array

Array Hash Table Binary Search Sliding Window

29.9% acceptance

Feb 24, 2026

176

You are given an integer array nums. The uniqueness array of nums is the sort ed 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 sub array 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 arra y when it is sorted in non-decreasing order. If there are two choices for a median, the smaller of the two values is taken.

Solution

Rust

Time O(n²)

Space O(n)

LeetCode

solution.rs

impl Solution {
  pub fn median_of_uniqueness_array(nums: Vec<i32>) -> i32 {
    let n = nums.len();
    let total = (n as i64) * (n as i64 + 1) / 2;
    let target = (total + 1) / 2; // median position (1-indexed)

    // count_at_most(k): number of subarrays with <= k distinct elements
    let count_at_most = |k: i32| -> i64 {
      use std::collections::HashMap;
      let mut freq: HashMap<i32, i32> = HashMap::new();
      let mut left = 0usize;
      let mut distinct = 0i32;
      let mut count = 0i64;
      for right in 0..n {
        let e = freq.entry(nums[right]).or_insert(0);
        if *e == 0 { distinct += 1; }
        *e += 1;
        while distinct > k {
          let entry = freq.entry(nums[left]).or_insert(0);
          *entry -= 1;
          if *entry == 0 { distinct -= 1; }
          left += 1;
        }
        count += (right - left + 1) as i64;
      }
      count
    };

    // Binary search: smallest k where count_at_most(k) >= target
    let mut lo = 1i32;
    let mut hi = n as i32;
    while lo < hi {
      let mid = lo + (hi - lo) / 2;
      if count_at_most(mid) >= target {
        hi = mid;
      } else {
        lo = mid + 1;
      }
    }
    lo
  }
}