LeetCode #3002 — MEDIUM

Maximum Size of a Set After Removals

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given two 0-indexed integer arrays nums1 and nums2 of even length n.

You must remove n / 2 elements from nums1 and n / 2 elements from nums2. After the removals, you insert the remaining elements of nums1 and nums2 into a set s.

Return the maximum possible size of the set s.

Example 1:

Input: nums1 = [1,2,1,2], nums2 = [1,1,1,1]
Output: 2
Explanation: We remove two occurences of 1 from nums1 and nums2. After the removals, the arrays become equal to nums1 = [2,2] and nums2 = [1,1]. Therefore, s = {1,2}.
It can be shown that 2 is the maximum possible size of the set s after the removals.

Example 2:

Input: nums1 = [1,2,3,4,5,6], nums2 = [2,3,2,3,2,3]
Output: 5
Explanation: We remove 2, 3, and 6 from nums1, as well as 2 and two occurrences of 3 from nums2. After the removals, the arrays become equal to nums1 = [1,4,5] and nums2 = [2,3,2]. Therefore, s = {1,2,3,4,5}.
It can be shown that 5 is the maximum possible size of the set s after the removals.

Example 3:

Input: nums1 = [1,1,2,2,3,3], nums2 = [4,4,5,5,6,6]
Output: 6
Explanation: We remove 1, 2, and 3 from nums1, as well as 4, 5, and 6 from nums2. After the removals, the arrays become equal to nums1 = [1,2,3] and nums2 = [4,5,6]. Therefore, s = {1,2,3,4,5,6}.
It can be shown that 6 is the maximum possible size of the set s after the removals.

Constraints:

n == nums1.length == nums2.length
1 <= n <= 2 * 10⁴
n is even.
1 <= nums1[i], nums2[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given two 0-indexed integer arrays nums1 and nums2 of even length n. You must remove n / 2 elements from nums1 and n / 2 elements from nums2. After the removals, you insert the remaining elements of nums1 and nums2 into a set s. Return the maximum possible size of the set s.

Baseline thinking

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

Pattern signal: Array · Hash Map · Greedy

Example 1

[1,2,1,2]
[1,1,1,1]

Example 2

[1,2,3,4,5,6]
[2,3,2,3,2,3]

Example 3

[1,1,2,2,3,3]
[4,4,5,5,6,6]

Core Insight

What unlocks the optimal approach

Removing <code>n / 2</code> elements from each array is the same as keeping <code>n / 2<code> elements in each array.
Think of a greedy algorithm.
For each array, we will greedily keep the elements that are only in that array. Once we run out of such elements, we will keep the elements that are common to both arrays.

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

Upper-end input sizes

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 #3002: Maximum Size of a Set After Removals
class Solution {
    public int maximumSetSize(int[] nums1, int[] nums2) {
        Set<Integer> s1 = new HashSet<>();
        Set<Integer> s2 = new HashSet<>();
        for (int x : nums1) {
            s1.add(x);
        }
        for (int x : nums2) {
            s2.add(x);
        }
        int n = nums1.length;
        int a = 0, b = 0, c = 0;
        for (int x : s1) {
            if (!s2.contains(x)) {
                ++a;
            }
        }
        for (int x : s2) {
            if (!s1.contains(x)) {
                ++b;
            } else {
                ++c;
            }
        }
        a = Math.min(a, n / 2);
        b = Math.min(b, n / 2);
        return Math.min(a + b + c, n);
    }
}

// Accepted solution for LeetCode #3002: Maximum Size of a Set After Removals
func maximumSetSize(nums1 []int, nums2 []int) int {
	s1 := map[int]bool{}
	s2 := map[int]bool{}
	for _, x := range nums1 {
		s1[x] = true
	}
	for _, x := range nums2 {
		s2[x] = true
	}
	a, b, c := 0, 0, 0
	for x := range s1 {
		if !s2[x] {
			a++
		}
	}
	for x := range s2 {
		if !s1[x] {
			b++
		} else {
			c++
		}
	}
	n := len(nums1)
	a = min(a, n/2)
	b = min(b, n/2)
	return min(a+b+c, n)
}

# Accepted solution for LeetCode #3002: Maximum Size of a Set After Removals
class Solution:
    def maximumSetSize(self, nums1: List[int], nums2: List[int]) -> int:
        s1 = set(nums1)
        s2 = set(nums2)
        n = len(nums1)
        a = min(len(s1 - s2), n // 2)
        b = min(len(s2 - s1), n // 2)
        return min(a + b + len(s1 & s2), n)

// Accepted solution for LeetCode #3002: Maximum Size of a Set After Removals
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3002: Maximum Size of a Set After Removals
// class Solution {
//     public int maximumSetSize(int[] nums1, int[] nums2) {
//         Set<Integer> s1 = new HashSet<>();
//         Set<Integer> s2 = new HashSet<>();
//         for (int x : nums1) {
//             s1.add(x);
//         }
//         for (int x : nums2) {
//             s2.add(x);
//         }
//         int n = nums1.length;
//         int a = 0, b = 0, c = 0;
//         for (int x : s1) {
//             if (!s2.contains(x)) {
//                 ++a;
//             }
//         }
//         for (int x : s2) {
//             if (!s1.contains(x)) {
//                 ++b;
//             } else {
//                 ++c;
//             }
//         }
//         a = Math.min(a, n / 2);
//         b = Math.min(b, n / 2);
//         return Math.min(a + b + c, n);
//     }
// }

// Accepted solution for LeetCode #3002: Maximum Size of a Set After Removals
function maximumSetSize(nums1: number[], nums2: number[]): number {
    const s1: Set<number> = new Set(nums1);
    const s2: Set<number> = new Set(nums2);
    const n = nums1.length;
    let [a, b, c] = [0, 0, 0];
    for (const x of s1) {
        if (!s2.has(x)) {
            ++a;
        }
    }
    for (const x of s2) {
        if (!s1.has(x)) {
            ++b;
        } else {
            ++c;
        }
    }
    a = Math.min(a, n >> 1);
    b = Math.min(b, n >> 1);
    return Math.min(a + b + c, n);
}

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(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.