What is the time complexity of Minimum Operations to Convert All Elements to Zero?

The optimal solution for Minimum Operations to Convert All Elements to Zero runs in O(n log n) time and uses O(n) space. Sorting the unique values takes O(n log n), and counting segments is O(n).

What is the brute force solution for Minimum Operations to Convert All Elements to Zero?

The brute force approach for Minimum Operations to Convert All Elements to Zero is Brute Force, with O(n²) time complexity and O(n) space complexity. We can try every possible subarray and count operations for each unique minimum value. This approach is simple but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Operations to Convert All Elements to Zero?

Minimum Operations to Convert All Elements to Zero uses the following concepts: Array, Hash Table, Stack, Greedy, Monotonic Stack. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Operations to Convert All Elements to Zero?

Explain your thought process clearly. Use examples to illustrate your approach. Practice similar problems to build intuition.

What are common mistakes when solving Minimum Operations to Convert All Elements to Zero?

Not considering contiguous segments. Overcounting operations for duplicates.

#3542

Minimum Operations to Convert All Elements to Zero

Medium

ArrayHash TableStackGreedyMonotonic StackHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(n)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

By processing values in sorted order and identifying contiguous segments, we can minimize operations effectively.

⚙️

Algorithm

3 steps

1Step 1: Sort the unique values in the array, excluding 0.
2Step 2: For each unique value, find its contiguous segments in the original array.
3Step 3: Count the segments for each unique value to determine operations.

solution.py15 lines

1def minOperations(nums):
2    from collections import defaultdict
3    segments = defaultdict(int)
4    nums = sorted(set(nums))
5    for num in nums:
6        if num == 0: continue
7        in_segment = False
8        for n in nums:
9            if n == num:
10                if not in_segment:
11                    segments[num] += 1
12                    in_segment = True
13            else:
14                in_segment = False
15    return sum(segments.values())

ℹ

Complexity note: Sorting the unique values takes O(n log n), and counting segments is O(n).

1Identify unique values to minimize operations.
2Contiguous segments can be zeroed in one operation.

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