What is the time complexity of Equal Sum Arrays With Minimum Number of Operations?

The optimal solution for Equal Sum Arrays With Minimum Number of Operations runs in O(n) time and uses O(1) space. This complexity is linear because we only traverse the arrays a couple of times and use a fixed-size frequency array.

What is the brute force solution for Equal Sum Arrays With Minimum Number of Operations?

The brute force approach for Equal Sum Arrays With Minimum Number of Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we try all possible combinations of changing elements in both arrays to see if we can make their sums equal. This is straightforward but inefficient as it checks every possibility. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Equal Sum Arrays With Minimum Number of Operations?

Equal Sum Arrays With Minimum Number of Operations uses the following concepts: Array, Hash Table, Greedy, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Equal Sum Arrays With Minimum Number of Operations?

Always start with a clear understanding of the problem and edge cases. Explain your thought process clearly while coding; it shows your understanding. Practice similar problems to recognize patterns and improve your problem-solving speed.

What are common mistakes when solving Equal Sum Arrays With Minimum Number of Operations?

Failing to check if the sums are already equal before performing operations. Not considering the constraints of the problem, such as the range of numbers.

#1775

Equal Sum Arrays With Minimum Number of Operations

Medium

ArrayHash TableGreedyCountingHash MapArray

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 focuses on calculating the difference between the two sums and using a greedy approach to minimize the number of operations needed to equalize them. We can use a counting technique to track how many changes we can make effectively.

⚙️

Algorithm

5 steps

1Step 1: Calculate the sums of both arrays, sum1 and sum2.
2Step 2: Determine the difference, diff = abs(sum1 - sum2).
3Step 3: Create a frequency array to count occurrences of numbers 1 to 6 in both arrays.
4Step 4: Calculate the maximum possible increase or decrease for each number and use it to minimize operations.
5Step 5: Iterate through the frequency array to find the minimum number of operations needed to cover the difference.

solution.py20 lines

1# Full working Python code
2
3def min_operations(nums1, nums2):
4    sum1 = sum(nums1)
5    sum2 = sum(nums2)
6    if sum1 == sum2:
7        return 0
8    diff = abs(sum1 - sum2)
9    count = [0] * 7
10    for num in nums1:
11        count[num] += 1
12    for num in nums2:
13        count[num] -= 1
14    operations = 0
15    for i in range(6, 0, -1):
16        while count[i] > 0 and diff > 0:
17            diff -= i
18            operations += 1
19            count[i] -= 1
20    return operations if diff <= 0 else -1

ℹ

Complexity note: This complexity is linear because we only traverse the arrays a couple of times and use a fixed-size frequency array.

1Understanding the difference between sums is crucial for determining how to adjust the arrays.
2Using a frequency count allows us to quickly assess how many operations we can perform.

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