#1775

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Equal Sum Arrays With Minimum Number of Operations

Solve the problem of balancing the sums of two integer arrays with minimal operations, using array scanning and hash lookups.

array hash table greedy counting

Problem Statement

You are given two arrays of integers nums1 and nums2, which may have different lengths. The values in the arrays range from 1 to 6. In each operation, you can change any integer in either array to any value between 1 and 6, inclusive. The goal is to make the sums of both arrays equal with the minimum number of operations. If it's impossible, return -1.

The challenge is to determine the minimum number of operations needed to achieve equal sums. If it's not possible, return -1. The key to solving this problem efficiently involves balancing the sum of the larger array by decreasing its values, while increasing the sum of the smaller array. The solution involves array scanning and hash table lookups to optimize the process.

Examples

Example 1

Input: nums1 = [1,2,3,4,5,6], nums2 = [1,1,2,2,2,2]

Output: 3

You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed.

Change nums2[0] to 6. nums1 = [1,2,3,4,5,6], nums2 = [6,1,2,2,2,2].
Change nums1[5] to 1. nums1 = [1,2,3,4,5,1], nums2 = [6,1,2,2,2,2].
Change nums1[2] to 2. nums1 = [1,2,2,4,5,1], nums2 = [6,1,2,2,2,2].

Example 2

Input: nums1 = [1,1,1,1,1,1,1], nums2 = [6]

Output: -1

There is no way to decrease the sum of nums1 or to increase the sum of nums2 to make them equal.

Example 3

Input: nums1 = [6,6], nums2 = [1]

Output: 3

You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed.

Change nums1[0] to 2. nums1 = [2,6], nums2 = [1].
Change nums1[1] to 2. nums1 = [2,2], nums2 = [1].
Change nums2[0] to 4. nums1 = [2,2], nums2 = [4].

Constraints

1 <= nums1.length, nums2.length <= 105
1 <= nums1[i], nums2[i] <= 6

Solution Approach

Identify the sum difference

First, calculate the sum of both arrays. If the sums are already equal, no operations are needed. If not, identify the difference between the two sums and determine if the difference can be adjusted through valid operations on the arrays.

Use hash tables for efficient tracking

Track the frequencies of numbers in both arrays using hash tables. This allows efficient lookups and helps in determining the most optimal elements to modify. Adjusting the sum can be achieved by either increasing the sum of the smaller array or decreasing the sum of the larger one.

Greedily modify the arrays

Greedily modify the elements of the arrays in a way that minimizes the number of operations. If the sum of one array is greater, reduce the values in that array, while increasing values in the smaller array. Each modification should maximize the change in sum per operation.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity of this solution depend on the approach used for array scanning and hash table lookups. The hash table enables efficient element tracking, reducing the need for nested loops and improving performance for large arrays, making the solution scalable to the input size constraints.

What Interviewers Usually Probe

Evaluate how efficiently the candidate uses hash tables to track frequencies of array elements.
Check the candidate's ability to apply a greedy approach for minimal operations.
Assess if the candidate considers edge cases where the sums cannot be equalized.

Common Pitfalls or Variants

Common pitfalls

Not considering cases where it is impossible to balance the sums, which should return -1.
Using an inefficient method to find the best elements to modify, resulting in higher-than-necessary operation counts.
Neglecting to check if the arrays' sums are already equal before proceeding with operations.

Follow-up variants

What if the arrays have significantly larger sizes or different value ranges?
How would the solution change if the value range of elements was increased beyond 1 to 6?
Can this problem be solved by minimizing operations in a different order or sequence of changes?

FAQ

How do I approach the Equal Sum Arrays With Minimum Number of Operations problem?

You should focus on balancing the sums of the two arrays using hash tables to track frequencies and applying a greedy approach to minimize operations.

What is the primary pattern for solving this problem?

The primary pattern involves array scanning plus hash lookup to track and modify array elements efficiently.

Can the problem be solved without hash tables?

While possible, using hash tables significantly optimizes the solution by reducing redundant checks and improving performance with larger input sizes.

What should I do if the arrays have different lengths?

The length difference doesn't affect the approach, but the sum difference must be considered when balancing the arrays.

What are some edge cases to consider for this problem?

Consider cases where it's impossible to balance the sums, such as when one array has much larger or smaller values and no valid operations can achieve equality.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
        s1, s2 = sum(nums1), sum(nums2)
        if s1 == s2:
            return 0
        if s1 > s2:
            return self.minOperations(nums2, nums1)
        arr = [6 - v for v in nums1] + [v - 1 for v in nums2]
        d = s2 - s1
        for i, v in enumerate(sorted(arr, reverse=True), 1):
            d -= v
            if d <= 0:
                return i
        return -1

Solution 2

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class Solution:
    def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
        s1, s2 = sum(nums1), sum(nums2)
        if s1 == s2:
            return 0
        if s1 > s2:
            return self.minOperations(nums2, nums1)
        arr = [6 - v for v in nums1] + [v - 1 for v in nums2]
        d = s2 - s1
        for i, v in enumerate(sorted(arr, reverse=True), 1):
            d -= v
            if d <= 0:
                return i
        return -1

Continue Practicing

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