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

Minimum Operations to Make Array Equal II (LeetCode #2541) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: You can only balance the arrays if the total excess equals the total deficit.

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

The optimal solution for Minimum Operations to Make Array Equal II runs in O(n) time and uses O(1) space. The complexity is O(n) because we only need a single pass through the arrays to calculate the excess and deficit.

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

The brute force approach for Minimum Operations to Make Array Equal II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying every possible pair of indices to perform the operations needed to make nums1 equal to nums2. This is straightforward but inefficient, as it checks all combinations of operations. 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 II?

Minimum Operations to Make Array Equal II uses the following concepts: Array, Math, Greedy. The recommended patterns to study are: Greedy, Math.

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

Always check for edge cases like k = 0 before diving into the main logic. Think about how you can reduce the problem size with mathematical observations instead of brute force. Explain your thought process clearly; it shows your understanding and can help you get hints if you're stuck.

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

Assuming that any combination of operations will work without checking total excess and deficit. Not considering the case where k = 0 and the arrays are not equal.

#2541

Minimum Operations to Make Array Equal II

Medium

ArrayMathGreedyGreedyMath

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution leverages the fact that we can only increment and decrement by k. By calculating the total excess and deficit, we can determine the minimum number of operations needed without brute-forcing through pairs.

⚙️

Algorithm

3 steps

1Step 1: Calculate the total excess (sum of positive differences) and deficit (sum of negative differences) between nums1 and nums2.
2Step 2: If the total excess is not equal to the total deficit, return -1 (impossible to balance).
3Step 3: Return the total excess divided by k, as each operation can adjust both excess and deficit by k.

solution.py13 lines

1def minOperations(nums1, nums2, k):
2    if k == 0:
3        return -1 if nums1 != nums2 else 0
4    excess = 0
5    deficit = 0
6    for a, b in zip(nums1, nums2):
7        if a > b:
8            excess += (a - b)
9        elif a < b:
10            deficit += (b - a)
11    if excess != deficit:
12        return -1
13    return excess // k

ℹ

Complexity note: The complexity is O(n) because we only need a single pass through the arrays to calculate the excess and deficit.

1You can only balance the arrays if the total excess equals the total deficit.
2The operations can be optimized by calculating the total differences instead of simulating each operation.

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