Introduction

In this blog post, we will explore how to solve the LeetCode problem "801" 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 integer arrays of the same length nums1 and nums2. In one operation, you are allowed to swap nums1[i] with nums2[i]. For example, if nums1 = [1,2,3,8], and nums2 = [5,6,7,4], you can swap the element at i = 3 to obtain nums1 = [1,2,3,4] and nums2 = [5,6,7,8]. Return the minimum number of needed operations to make nums1 and nums2 strictly increasing. The test cases are generated so that the given input always makes it possible. An array arr is strictly increasing if and only if arr[0] < arr[1] < arr[2] < ... < arr[arr.length - 1]. Example 1: Input: nums1 = [1,3,5,4], nums2 = [1,2,3,7] Output: 1 Explanation: Swap nums1[3] and nums2[3]. Then the sequences are: nums1 = [1, 3, 5, 7] and nums2 = [1, 2, 3, 4] which are both strictly increasing. Example 2: Input: nums1 = [0,3,5,8,9], nums2 = [2,1,4,6,9] Output: 1 Constraints: 2 <= nums1.length <= 105 nums2.length == nums1.length 0 <= nums1[i], nums2[i] <= 2 * 105

Explanation

Here's the approach and solution:

• Dynamic Programming: The core idea is to use dynamic programming to track the minimum swaps needed to make both arrays strictly increasing up to index i.
• Two States: For each index i, we maintain two states: swap (minimum swaps needed if we swap nums1[i] and nums2[i]) and no_swap (minimum swaps needed if we don't swap).

• Transitions: We calculate the swap and no_swap values based on the previous index's states and whether swapping at the current index is valid given the previous state (swapped or not).

• Runtime Complexity: O(n) & Storage Complexity: O(n) can be further optimized to O(1).

Code

    def min_swap(nums1, nums2):
    """
    Calculates the minimum number of swaps needed to make nums1 and nums2 strictly increasing.

    Args:
        nums1: The first list of integers.
        nums2: The second list of integers.

    Returns:
        The minimum number of swaps needed.
    """

    n = len(nums1)
    swap = [float('inf')] * n
    no_swap = [float('inf')] * n

    swap[0] = 1
    no_swap[0] = 0

    for i in range(1, n):
        # Case 1: No swap at i-1
        if nums1[i] > nums1[i - 1] and nums2[i] > nums2[i - 1]:
            no_swap[i] = no_swap[i - 1]
            swap[i] = swap[i - 1] + 1

        if nums1[i] > nums2[i - 1] and nums2[i] > nums1[i - 1]:
            no_swap[i] = min(no_swap[i], swap[i - 1])
            swap[i] = min(swap[i], no_swap[i - 1] + 1)

    return min(swap[n - 1], no_swap[n - 1])