How do you solve Minimum Operations to Reach Target Array on LeetCode?

Minimum Operations to Reach Target Array (LeetCode #3810) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Focus on differences between arrays.

What is the time complexity of Minimum Operations to Reach Target Array?

The optimal solution for Minimum Operations to Reach Target Array runs in O(n) time and uses O(n) space. We traverse the arrays once to find differing positions and store distinct values, leading to linear complexity.

What is the brute force solution for Minimum Operations to Reach Target Array?

The brute force approach for Minimum Operations to Reach Target Array is Brute Force, with O(n²) time complexity and O(1) space complexity. This method checks every possible value of x and updates segments iteratively. It's straightforward but inefficient due to repeated checks. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Operations to Reach Target Array?

Minimum Operations to Reach Target Array uses the following concepts: Array, Hash Table, Greedy. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Operations to Reach Target Array?

Clarify the problem requirements. Discuss edge cases like empty arrays. Explain your thought process clearly.

What are common mistakes when solving Minimum Operations to Reach Target Array?

Overcounting operations by not using distinct values. Ignoring unchanged segments.

#3810

Minimum Operations to Reach Target Array

Medium

ArrayHash TableGreedyHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

Focus on the positions where nums differs from target. The number of distinct values in nums at these positions determines the operations needed.

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Algorithm

3 steps

1Step 1: Identify indices where nums[i] != target[i].
2Step 2: Create a set of distinct values from nums at these indices.
3Step 3: The size of this set is the minimum number of operations required.

solution.py2 lines

1def minOperations(nums, target):
2    return len(set(nums[i] for i in range(len(nums)) if nums[i] != target[i]))

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Complexity note: We traverse the arrays once to find differing positions and store distinct values, leading to linear complexity.

1Focus on differences between arrays.
2Distinct values dictate the number of operations.

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