LeetCode 3810: Minimum Operations to Reach Target Array ·

Table of Contents

Meta Data
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Difficulty: medium First Attempt: 2026-01-17

Submissions: Accepted (first try)

Approach
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Choosing x changes every position whose value is x to target, whether or not they are contiguous.

Upper bound: one operation for each value appearing at a mismatched position fixes every mismatch. Touching an already-correct position is harmless (it becomes target again).
Lower bound: a mismatched position changes only when its current value is chosen. That value is nums[i] until the position changes, so every value at a mismatched position must be chosen at least once.

They match, so the answer is the count of distinct values (Greedy Template 4).

My Solution
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class Solution:
    def minOperations(self, nums: List[int], target: List[int]) -> int:
        return len({a for a, b in zip(nums, target) if a != b})

I originally wrote len(diff) - (len(diff) - len(set(diff))), which simplifies to len(set(diff)).

Complexity
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Time O(n), space O(n).

Mistakes to Avoid
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The “contiguous segments” wording only describes which positions change; all segments of one value change together in one operation.

Interview Explanation
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Choosing a value rewrites every position holding it, so one operation per value at a mismatched position is enough. A mismatched position only changes when its current value is chosen, so each such value needs at least one operation. The answer is the number of distinct values at mismatched positions. Time O(n).