LeetCode #3134 — HARD

Find the Median of the Uniqueness Array

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

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

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Binary Search · Sliding Window

Example 1

[1,2,3]

Example 2

[3,4,3,4,5]

Example 3

[4,3,5,4]

Related Problems

Find K-th Smallest Pair Distance (find-k-th-smallest-pair-distance)
Total Appeal of A String (total-appeal-of-a-string)

Step 02

Core Insight

What unlocks the optimal approach

Binary search over the answer.
For a given <code>x</code>, you need to check if <code>x</code> is the median, to the left of the median, or to the right of the median. You can do that by counting the number of sub-arrays <code>nums[i…j]</code> such that <code>distinct(num[i…j]) <= x</code>.
Use the sliding window to solve the counting problem in the hint above.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3134: Find the Median of the Uniqueness Array
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;
    }
}

// Accepted solution for LeetCode #3134: Find the Median of the Uniqueness Array
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
	})
}

# Accepted solution for LeetCode #3134: Find the Median of the Uniqueness Array
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)

// Accepted solution for LeetCode #3134: Find the Median of the Uniqueness Array
/**
 * [3134] Find the Median of the Uniqueness Array
 */
pub struct Solution {}

// submission codes start here
use std::collections::HashMap;

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

        // 检查数组中不同元素数目小于等于t的连续子数组是否大于等于median
        let check = |t: i64| -> bool {
            let mut map = HashMap::new();
            let mut total = 0;

            let mut j = 0;
            for i in 0..n {
                let entry = map.entry(nums[i as usize]).or_insert(0);
                *entry += 1;

                while map.len() as i64 > t {
                    let entry = map.get_mut(&nums[j]).unwrap();
                    *entry -= 1;

                    if *entry == 0 {
                        map.remove(&nums[j]);
                    }

                    j += 1;
                }

                total += i - j as i64 + 1;
            }

            total >= median
        };

        let mut result = 0;
        let (mut left, mut right) = (1, n);

        while left <= right {
            let middle = (left + right) / 2;

            if check(middle) {
                result = middle;
                right = middle - 1;
            } else {
                left = middle + 1;
            }
        }

        result as i32
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3134() {
        assert_eq!(1, Solution::median_of_uniqueness_array(vec![1, 2, 3]));
        assert_eq!(2, Solution::median_of_uniqueness_array(vec![3, 4, 3, 4, 5]));
    }
}

// Accepted solution for LeetCode #3134: Find the Median of the Uniqueness Array
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.