How do you solve Minimum Removals to Achieve Target XOR on LeetCode?

Minimum Removals to Achieve Target XOR (LeetCode #3877) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^(n/2)) time complexity and O(n) space complexity. Key insight: XOR of an empty array is 0.

What is the time complexity of Minimum Removals to Achieve Target XOR?

The optimal solution for Minimum Removals to Achieve Target XOR runs in O(n * 2^(n/2)) time and uses O(n) space. We generate subsets for half the array, leading to O(2^(n/2)) combinations, and store them in a hash map, which takes O(n).

What is the brute force solution for Minimum Removals to Achieve Target XOR?

The brute force approach for Minimum Removals to Achieve Target XOR is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible subsets of the array and calculate their XOR. By checking which subsets match the target, we can determine the minimum removals needed. The optimal Optimal Solution approach improves this to O(n * 2^(n/2)).

What algorithm or data structure is used to solve Minimum Removals to Achieve Target XOR?

Minimum Removals to Achieve Target XOR uses the following concepts: Array, Dynamic Programming, Bit Manipulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Removals to Achieve Target XOR?

Explain your thought process clearly. Consider edge cases early. Practice similar problems to build intuition.

What are common mistakes when solving Minimum Removals to Achieve Target XOR?

Not considering the case where target is 0. Overlooking the XOR properties.

#3877

Minimum Removals to Achieve Target XOR

Medium

ArrayDynamic ProgrammingBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * 2^(n/2))
Space	O(1)	O(n)

💡

Intuition

Time O(n * 2^(n/2))Space O(n)

We can split the array into two halves, calculate all possible XORs for each half, and use a hash map to find pairs that achieve the target XOR.

⚙️

Algorithm

3 steps

1Step 1: Split nums into two halves and calculate all possible XORs for each half.
2Step 2: Store the XORs of the first half in a hash map with their counts.
3Step 3: For each XOR from the second half, check if (target XOR current XOR) exists in the hash map and calculate the minimum removals.

solution.py29 lines

1from collections import defaultdict
2
3def min_removals(nums, target):
4    n = len(nums)
5    half = n // 2
6    first_half = nums[:half]
7    second_half = nums[half:]
8    xor_map = defaultdict(int)
9
10    def compute_xors(arr):
11        xors = []
12        for r in range(len(arr) + 1):
13            for subset in combinations(arr, r):
14                xors.append(reduce(lambda x, y: x ^ y, subset, 0))
15        return xors
16
17    xors1 = compute_xors(first_half)
18    xors2 = compute_xors(second_half)
19
20    for x in xors1:
21        xor_map[x] += 1
22
23    min_removals = float('inf')
24    for x in xors2:
25        required = target ^ x
26        if required in xor_map:
27            min_removals = min(min_removals, n - (xor_map[required] + len(xors2)))
28
29    return min_removals if min_removals != float('inf') else -1

ℹ

Complexity note: We generate subsets for half the array, leading to O(2^(n/2)) combinations, and store them in a hash map, which takes O(n).

1XOR of an empty array is 0.
2Using hash maps can optimize subset sum problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.