801. Minimum Swaps To Make Sequences Increasing

Problem Description

In this problem, we have two integer arrays nums1 and nums2 of the same length. Our goal is to make both arrays strictly increasing by swapping elements at the same index in both arrays. To clarify, nums1[i] can be swapped with nums2[i] for any i. What we're trying to find is the minimum number of such swaps required to achieve our goal. An array is strictly increasing if each element is larger than the one before it. The challenge assures us that there is always a way to make the arrays strictly increasing through such swaps.

Intuition

To solve this problem, dynamic programming is employed to keep track of the minimum number of swaps needed. At each index i, we have two choices:

1. Do not swap elements at index i: We choose this if the current elements of nums1 and nums2 are both greater than the previous elements of their respective arrays, which keeps both arrays strictly increasing without a swap.

2. Swap elements at index i: If the current elements are not greater than the previous ones, we must swap to maintain the strictly increasing order. In some cases, even when it's not required to maintain the strictly increasing order, swapping might still minimize the total number of swaps.

We maintain two variables:

• a: The minimum number of swaps needed up to the current index i without swapping nums1[i] and nums2[i].
• b: The minimum number of swaps needed if we swap nums1[i] and nums2[i].

After considering whether a swap is required or beneficial at each index, and updating a and b accordingly, the solution returns the minimum of a and b after iterating through the arrays, which represents the minimum number of swaps necessary to make both nums1 and nums2 strictly increasing.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution uses a dynamic programming approach to minimize the number of swaps required to make both arrays strictly increasing. Let's walk through the implementation details:

• We initialize two variables, a and b. Here, a represents the minimum number of swaps required if we do not swap at the current position i, and b is the minimum number of swaps required if we do swap at the current position.
• a is initially set to 0, because no swaps are performed at the start, and b is set to 1 since performing a swap at the first position would count as one operation.

The main loop of the algorithm starts at index 1 and goes until the end of the arrays as follows:

• For each index i, we first store the old values of a and b into temporary variables x and y. This is to prevent updating a and b simultaneously in the loop, which would give incorrect results, as the calculation of b may depend on the old value of a.
• There are conditions that determine if a swap is necessary or if we can move to the next index without a swap:
  • If nums1[i-1] >= nums1[i] or nums2[i-1] >= nums2[i], then the arrays are not strictly increasing, and a swap is necessary. In this case, we set a to y (the old value of b) since we have to swap the previous position, and b to x + 1 because we are performing a swap at the current position, which adds to the swap count.
  • If the arrays are already strictly increasing without a swap, a remains unchanged, and b becomes y + 1 since we continue from the previous swap scenario.
• Additionally, if nums1[i-1] < nums2[i] and nums2[i-1] < nums1[i], then swapping may yield a better result even if it’s not necessary. In this case, we update a to the minimum of its current value and y, and b to the minimum of its current value and x + 1.

In the end, we return the minimum of a and b, as this would represent either performing no swap or swapping at the last index, whichever has the minimum swaps overall to make the arrays strictly increasing.

By using dynamic programming, we avoid re-computation for each index, evaluating the minimum number of swaps needed at each step. This approach ensures that the final complexity of the algorithm is O(n), where n is the length of the given arrays.

Example Walkthrough

Let's consider two small integer arrays nums1 and nums2:

1nums1: [1, 3, 5, 7]
2nums2: [1, 2, 3, 4]

According to the problem, we want to make both arrays strictly increasing by swapping elements at the same index where necessary.

Initialization:

• No swaps at the start, so a = 0.
• If we swap at the first position, it would count as one swap, so b = 1.

We start our loop from index 1 and go up to the last index.

At index 1:

• Current elements are nums1[1] = 3 and nums2[1] = 2.
• Previous elements are nums1[0] = 1 and nums2[0] = 1.
• Arrays are already strictly increasing (1 < 3 and 1 < 2), so no swap needed here.
• We set tempA = a and tempB = b.
• Checking further, though, we see that swapping might be beneficial for future positions because nums2[1] > nums1[0] and nums1[1] > nums2[0].
• We update a to the minimum of a and tempB which are both 0.
• We update b to the minimum of b and tempA + 1, which is 1.

Now our arrays look like this (no change because no swap was done):

1nums1: [1, 3, 5, 7]
2nums2: [1, 2, 3, 4]

At index 2:

• Current elements are nums1[2] = 5 and nums2[2] = 3.
• Previous elements are nums1[1] = 3 and nums2[1] = 2.
• We cannot move to the next index without a swap since nums2[2] is not greater than nums2[1].
• Therefore, a swap is necessary for nums2.
• We again set tempA = a and tempB = b (here tempA = 0 and tempB = 1).
• We set a to tempB since we are continuing from the previous decision to swap.
• We set b to tempA + 1 which equals 1.
• After swapping, our arrays looks like this:

1nums1: [1, 3, 3, 7]
2nums2: [1, 2, 5, 4]

• But now we have a problem as nums1 is no longer strictly increasing. We must also swap at index 1 to correct this.

Result after also swapping at index 1:

1nums1: [1, 2, 3, 7]
2nums2: [1, 3, 5, 4]

At index 3:

• Arrays are nums1: [1, 2, 3, 7] and nums2: [1, 3, 5, 4].
• Current elements are fine as 3 < 7 and 5 < 4 requires a swap.
• We set tempA = a and tempB = b.
• Since the arrays without the current elements being swapped are increasing, we do not update a.
• To minimize total swaps, we choose not to swap here, so b = tempB + 1.

Finally, our arrays look like this, which are both strictly increasing:

1nums1: [1, 2, 3, 7]
2nums2: [1, 3, 5, 4]

The minimum number of swaps required is min(a, b). Given the values of a = 0 and b = 2, the minimum number of swaps is 0.

This small example illustrated the solution approach using dynamic programming, where we made both nums1 and nums2 strictly increasing while minimizing the total number of swaps necessary.

Solution Implementation

Time and Space Complexity

The code provided is a dynamic programming solution designed to determine the minimum number of swaps needed to make two lists strictly increasing.

Time Complexity

To analyze the time complexity of this approach, we need to consider the operations performed with each iteration of the loop that runs len(nums1) times.

• Within each iteration, the algorithm performs constant-time comparisons and assignments, regardless of the size of the lists.
• The loop runs for every element of the two lists starting from index 1, which means it runs n-1 times, where n is the number of elements in each list.

As the time taken per iteration does not depend on the size of the dataset, the overall time complexity of the algorithm is O(n), where n is the length of the lists.

Space Complexity

For the space complexity:

• Two variables, a and b, are used to keep track of the minimum number of swaps when the ith element is not swapped and when it is swapped, respectively.
• Additionally, temporary variables x and y are used within the loop.

Since the space used does not increase with the size of the input lists, this algorithm has a constant space complexity, thus the space complexity is O(1).

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