Problem Description

You have two arrays nums1 and nums2, both with the same length n and 0-indexed. Your goal is to make sure that:

• The last element of nums1 (at index n-1) is the maximum value in nums1
• The last element of nums2 (at index n-1) is the maximum value in nums2

You can perform swap operations to achieve this goal. In each operation, you pick an index i (from 0 to n-1) and swap the values between nums1[i] and nums2[i].

The problem asks you to find the minimum number of operations needed to satisfy both conditions. If it's impossible to achieve both conditions, return -1.

For example, if you have:

• nums1 = [1, 2, 7] and nums2 = [4, 5, 3]
• The last elements are nums1[2] = 7 and nums2[2] = 3
• Currently, 7 is already the maximum in nums1, but 3 is not the maximum in nums2 (since 5 > 3)
• You could swap at index 2 to get nums1 = [1, 2, 3] and nums2 = [4, 5, 7]
• But now 3 is not the maximum in nums1
• So you'd need to check other swap combinations to find the minimum operations

The solution considers two main scenarios:

1. Keep the last elements as they are and swap other elements as needed
2. Swap the last elements first, then swap other elements as needed

For each scenario, it checks if swapping elements at positions 0 to n-2 can satisfy the conditions while ensuring all elements remain less than or equal to their respective array's last element.

Intuition

The key insight is that the last elements of both arrays must be the maximum values in their respective arrays. This means every other element in nums1 must be ≤ nums1[n-1] and every other element in nums2 must be ≤ nums2[n-1].

Since we can only swap elements at the same index between the two arrays, we need to think about what our final configuration should look like. There are only two possible end states for the last position:

1. Keep nums1[n-1] and nums2[n-1] as they are
2. Swap them so that nums1[n-1] becomes the original nums2[n-1] and vice versa

Why only these two cases? Because once we decide what values should be at the last position (which must be the maximums), all other decisions are forced upon us. For each position i from 0 to n-2, we have two values available: nums1[i] and nums2[i]. We need to distribute them such that one goes to the first array and one goes to the second array, ensuring they don't exceed their respective maximums.

For each position i (except the last), we check:

• If nums1[i] ≤ x and nums2[i] ≤ y (where x and y are our chosen last elements), we don't need to swap
• If nums1[i] ≤ y and nums2[i] ≤ x, we must swap at position i
• If neither condition holds, it's impossible to satisfy the requirements

The greedy approach works here because at each position, we have at most one valid choice (swap or don't swap). We're not making a choice that could be optimized later - either we must swap, must not swap, or the configuration is impossible.

By calculating the minimum swaps needed for both possible end configurations and taking the minimum (accounting for whether we swapped the last position), we get our answer.

Solution Approach

The solution implements a helper function f(x, y) that calculates the minimum number of swaps needed when we fix the last elements as x for nums1 and y for nums2.

Helper Function Logic: The function f(x, y) iterates through the first n-1 positions and counts how many swaps are needed:

• For each pair (a, b) from (nums1[i], nums2[i]) where i < n-1:
  • If a ≤ x and b ≤ y: No swap needed, continue to next position
  • If a ≤ y and b ≤ x: A swap is required, increment counter by 1
  • If neither condition is satisfied: Return -1 (impossible configuration)
• Return the total count of swaps needed

Main Algorithm:

1. Case 1 - No swap at last position:

   • Call a = f(nums1[-1], nums2[-1])
   • This calculates swaps needed when keeping the last elements as they are
   • The last elements are nums1[n-1] and nums2[n-1]

2. Case 2 - Swap at last position:

   • Call b = f(nums2[-1], nums1[-1])
   • This calculates swaps needed when the last elements are swapped
   • The last elements become nums2[n-1] in nums1 and nums1[n-1] in nums2
   • Note: If we choose this case, we need b + 1 total operations (the extra 1 for swapping the last position)

3. Result Determination:

   • If both a = -1 and b = -1 (meaning a + b = -2): Return -1 (impossible)
   • Otherwise: Return min(a, b + 1)
     • a represents operations for Case 1
     • b + 1 represents operations for Case 2 (including the swap at last position)

The time complexity is O(n) as we traverse the arrays twice (once for each case), and the space complexity is O(1) as we only use a few variables to track the counts.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Example: nums1 = [3, 7, 9] and nums2 = [8, 2, 4]

Our goal is to make nums1[2] the maximum of nums1 and nums2[2] the maximum of nums2.

Step 1: Analyze current state

• Last elements: nums1[2] = 9, nums2[2] = 4
• Max in nums1: max(3, 7, 9) = 9 ✓ (already satisfied)
• Max in nums2: max(8, 2, 4) = 8 ✗ (not satisfied, since 8 > 4)

Step 2: Try Case 1 - Keep last elements as they are

• Target: nums1 elements ≤ 9, nums2 elements ≤ 4
• Check position 0: nums1[0] = 3 ≤ 9 ✓ and nums2[0] = 8 ≤ 4 ✗
  • Can we swap? After swap: 8 ≤ 9 ✓ and 3 ≤ 4 ✓
  • Yes! Need to swap position 0
• Check position 1: nums1[1] = 7 ≤ 9 ✓ and nums2[1] = 2 ≤ 4 ✓
  • No swap needed
• Result for Case 1: 1 swap (at position 0)
• Arrays become: nums1 = [8, 7, 9], nums2 = [3, 2, 4]

Step 3: Try Case 2 - Swap last elements first

• After swapping last position: nums1[2] = 4, nums2[2] = 9
• Target: nums1 elements ≤ 4, nums2 elements ≤ 9
• Check position 0: nums1[0] = 3 ≤ 4 ✓ and nums2[0] = 8 ≤ 9 ✓
  • No swap needed
• Check position 1: nums1[1] = 7 ≤ 4 ✗ and nums2[1] = 2 ≤ 9 ✓
  • Can we swap? After swap: 2 ≤ 4 ✓ and 7 ≤ 9 ✓
  • Yes! Need to swap position 1
• Result for Case 2: 1 swap (at position 1) + 1 swap (at last position) = 2 swaps
• Arrays become: nums1 = [3, 2, 4], nums2 = [8, 7, 9]

Step 4: Determine minimum

• Case 1: 1 swap
• Case 2: 2 swaps
• Answer: 1 (minimum of the two cases)

The final configuration using Case 1 is:

• nums1 = [8, 7, 9] where max = 9 ✓
• nums2 = [3, 2, 4] where max = 4 ✓

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the arrays nums1 and nums2. The function f is called twice, and each call iterates through the arrays once using zip(nums1[:-1], nums2[:-1]), which processes n-1 elements. Since we iterate through the arrays a constant number of times (twice), the overall time complexity remains O(n).

The space complexity is O(1). The algorithm only uses a constant amount of extra space for variables like cnt, a, and b. The zip function creates an iterator that doesn't consume additional space proportional to the input size, and no additional data structures are created that scale with the input size.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Forgetting to Add 1 for the Last Swap in Case 2

One of the most common mistakes is forgetting that when we swap the last elements (Case 2), we need to count that swap as an additional operation.

Incorrect Implementation:

# WRONG: Forgetting to add 1 for the last swap
return min(swaps_without_last_swap, swaps_with_last_swap)  # Missing +1

Correct Implementation:

# CORRECT: Account for swapping the last elements
return min(swaps_without_last_swap, swaps_with_last_swap + 1)

2. Incorrect Handling of Impossible Cases

Another pitfall is not properly handling the edge cases when one or both scenarios return -1.

Incorrect Implementation:

# WRONG: This will return incorrect results when one case is -1
return min(swaps_without_last_swap, swaps_with_last_swap + 1)
# If swaps_without_last_swap = -1, min(-1, valid_number) returns -1 incorrectly

Correct Implementation:

# CORRECT: Check for impossible cases first
if swaps_without_last_swap == -1 and swaps_with_last_swap == -1:
    return -1
if swaps_without_last_swap == -1:
    return swaps_with_last_swap + 1
if swaps_with_last_swap == -1:
    return swaps_without_last_swap
return min(swaps_without_last_swap, swaps_with_last_swap + 1)

3. Confusing the Swap Logic in the Helper Function

It's easy to mix up the conditions for when a swap is needed versus when it's impossible.

Incorrect Logic:

# WRONG: Checking wrong conditions
if num1 <= max_val1 or num2 <= max_val2:  # Using OR instead of AND
    continue

Correct Logic:

# CORRECT: Both conditions must be satisfied without swap
if num1 <= max_val1 and num2 <= max_val2:
    continue
# Check if swap would help
if num1 <= max_val2 and num2 <= max_val1:
    swap_count += 1
else:
    return -1  # Impossible even with swap

4. Including the Last Elements in the Iteration

The helper function should only check elements from index 0 to n-2, not including the last elements.

Incorrect Implementation:

# WRONG: Including last elements in the check
for num1, num2 in zip(nums1, nums2):  # Iterates through ALL elements
    # This would incorrectly process the last pair

Correct Implementation:

# CORRECT: Exclude the last elements
for num1, num2 in zip(nums1[:-1], nums2[:-1]):
    # Only processes elements before the last position

These pitfalls can lead to incorrect results or runtime errors. Always remember to carefully consider the swap counting logic and edge case handling when implementing this type of algorithm.