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

Minimum Operations to Make Array Equal to Target (LeetCode #3229) 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: Understanding the difference between the two arrays allows for a more efficient solution.

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

The optimal solution for Minimum Operations to Make Array Equal to Target runs in O(n) time and uses O(n) space. The complexity is linear because we only traverse the array a couple of times, once to compute differences and once to count segments.

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

The brute force approach for Minimum Operations to Make Array Equal to Target is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subarray and adjusting it to match the target. This is simple but inefficient, as it can take a long time for larger arrays. The optimal Optimal Solution approach improves this to O(n).

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

Minimum Operations to Make Array Equal to Target uses the following concepts: Array, Dynamic Programming, Stack, Greedy, Monotonic Stack. The recommended patterns to study are: Greedy, Array.

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

Always consider the simplest approach first; it can lead to insights for optimization. Be mindful of edge cases, such as when nums is already equal to target. Practice explaining your thought process clearly, as communication is key in interviews.

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

Not recognizing that contiguous segments can be adjusted together. Overcomplicating the problem by trying to adjust individual elements instead of segments.

#3229

Minimum Operations to Make Array Equal to Target

Hard

ArrayDynamic ProgrammingStackGreedyMonotonic StackGreedyArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages the concept of differences between the two arrays and uses a greedy approach to minimize operations by adjusting contiguous segments of the array.

⚙️

Algorithm

3 steps

1Step 1: Calculate the difference array, diff[i] = nums[i] - target[i].
2Step 2: Traverse the difference array and count the number of contiguous segments where the difference is not zero.
3Step 3: Each contiguous segment represents one operation, so return the count of segments.

solution.py9 lines

1def minOperations(nums, target):
2    diff = [nums[i] - target[i] for i in range(len(nums))]
3    ops = 0
4    for i in range(len(diff)):
5        if diff[i] != 0:
6            ops += 1
7            while i < len(diff) and diff[i] != 0:
8                i += 1
9    return ops

ℹ

Complexity note: The complexity is linear because we only traverse the array a couple of times, once to compute differences and once to count segments.

1Understanding the difference between the two arrays allows for a more efficient solution.
2Contiguous segments of differences can be adjusted in one operation.

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