Introduction

In this blog post, we will explore how to solve the LeetCode problem "2449" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given two positive integer arrays nums and target, of the same length. In one operation, you can choose any two distinct indices i and j where 0 <= i, j < nums.length and: set nums[i] = nums[i] + 2 and set nums[j] = nums[j] - 2. Two arrays are considered to be similar if the frequency of each element is the same. Return the minimum number of operations required to make nums similar to target. The test cases are generated such that nums can always be similar to target. Example 1: Input: nums = [8,12,6], target = [2,14,10] Output: 2 Explanation: It is possible to make nums similar to target in two operations: - Choose i = 0 and j = 2, nums = [10,12,4]. - Choose i = 1 and j = 2, nums = [10,14,2]. It can be shown that 2 is the minimum number of operations needed. Example 2: Input: nums = [1,2,5], target = [4,1,3] Output: 1 Explanation: We can make nums similar to target in one operation: - Choose i = 1 and j = 2, nums = [1,4,3]. Example 3: Input: nums = [1,1,1,1,1], target = [1,1,1,1,1] Output: 0 Explanation: The array nums is already similiar to target. Constraints: n == nums.length == target.length 1 <= n <= 105 1 <= nums[i], target[i] <= 106 It is possible to make nums similar to target.

Explanation

Here's the breakdown of the approach, complexity, and the code:

• Key Approach:

  • Separate even and odd numbers in both nums and target arrays.
  • Sort the even and odd number lists separately for both arrays. This ensures that we're comparing and adjusting corresponding elements to minimize operations.
  • Calculate the total operations needed by summing the absolute differences between the sorted even lists and the sorted odd lists, dividing the total sum by 4 (since each operation involves adding 2 and subtracting 2).

• Complexity:

  • Runtime: O(n log n) due to sorting.
  • Storage: O(n) to store the even and odd number lists.

Code

    def solve():
    n = int(input())  # Or len(nums) if the input is an array
    nums = list(map(int, input().split()))
    target = list(map(int, input().split()))

    nums_even = sorted([x for x in nums if x % 2 == 0])
    nums_odd = sorted([x for x in nums if x % 2 != 0])
    target_even = sorted([x for x in target if x % 2 == 0])
    target_odd = sorted([x for x in target if x % 2 != 0])

    operations = 0
    for i in range(len(nums_even)):
        operations += abs(nums_even[i] - target_even[i])

    for i in range(len(nums_odd)):
        operations += abs(nums_odd[i] - target_odd[i])

    return operations // 4

# Example Usage (with direct array inputs instead of taking input):

def similar_arrays(nums, target):
    nums_even = sorted([x for x in nums if x % 2 == 0])
    nums_odd = sorted([x for x in nums if x % 2 != 0])
    target_even = sorted([x for x in target if x % 2 == 0])
    target_odd = sorted([x for x in target if x % 2 != 0])

    operations = 0
    for i in range(len(nums_even)):
        operations += abs(nums_even[i] - target_even[i])

    for i in range(len(nums_odd)):
        operations += abs(nums_odd[i] - target_odd[i])

    return operations // 4


if __name__ == "__main__":
    nums1 = [8, 12, 6]
    target1 = [2, 14, 10]
    print(similar_arrays(nums1, target1))  # Output: 2

    nums2 = [1, 2, 5]
    target2 = [4, 1, 3]
    print(similar_arrays(nums2, target2))  # Output: 1

    nums3 = [1, 1, 1, 1, 1]
    target3 = [1, 1, 1, 1, 1]
    print(similar_arrays(nums3, target3))  # Output: 0

    nums4 = [5, 3, 1]
    target4 = [1, 3, 5]
    print(similar_arrays(nums4, target4)) #output 0

    #Example using input from stdin:
    # n = int(input())
    # nums = list(map(int, input().split()))
    # target = list(map(int, input().split()))
    # print(similar_arrays(nums,target))