How do you solve Minimum Swaps To Make Sequences Increasing on LeetCode?

Minimum Swaps To Make Sequences Increasing (LeetCode #801) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding the conditions for strict increase is crucial.

What is the time complexity of Minimum Swaps To Make Sequences Increasing?

The optimal solution for Minimum Swaps To Make Sequences Increasing runs in O(n) time and uses O(1) space. This complexity is linear because we only go through the arrays once, updating our swap counts based on previous states.

What is the brute force solution for Minimum Swaps To Make Sequences Increasing?

The brute force approach for Minimum Swaps To Make Sequences Increasing is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible combination of swaps to see if the sequences can be made strictly increasing. This is simple but inefficient as it explores all possibilities. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Swaps To Make Sequences Increasing?

Minimum Swaps To Make Sequences Increasing uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Array.

What are the interview tips for Minimum Swaps To Make Sequences Increasing?

Clearly explain your thought process as you work through the problem. Consider edge cases and how they might affect your solution. Practice dynamic programming problems to become more comfortable with the approach.

What are common mistakes when solving Minimum Swaps To Make Sequences Increasing?

Failing to consider both possibilities (swap and no swap) at each step. Not updating the swap counts correctly based on previous states.

#801

Minimum Swaps To Make Sequences Increasing

Hard

ArrayDynamic ProgrammingDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses dynamic programming to keep track of the minimum swaps needed to make both sequences strictly increasing. This is efficient and avoids unnecessary checks.

⚙️

Algorithm

4 steps

1Step 1: Initialize two variables to track the minimum swaps needed without and with a swap.
2Step 2: Iterate through both arrays, updating the swap counts based on the previous states.
3Step 3: Use conditions to decide whether to swap or not based on the increasing order.
4Step 4: Return the minimum swaps needed.

solution.py18 lines

1# Full working Python code
2
3def min_swaps(nums1, nums2):
4    n = len(nums1)
5    no_swap = 0
6    swap = 1
7    for i in range(1, n):
8        no_swap_temp = no_swap
9        swap_temp = swap
10        if nums1[i] > nums1[i - 1] and nums2[i] > nums2[i - 1]:
11            no_swap_temp = min(no_swap_temp, no_swap)
12            swap_temp = min(swap_temp, swap + 1)
13        if nums1[i] > nums2[i - 1] and nums2[i] > nums1[i - 1]:
14            no_swap_temp = min(no_swap_temp, swap)
15            swap_temp = min(swap_temp, no_swap + 1)
16        no_swap, swap = no_swap_temp, swap_temp
17    return min(no_swap, swap)
18

ℹ

Complexity note: This complexity is linear because we only go through the arrays once, updating our swap counts based on previous states.

1Understanding the conditions for strict increase is crucial.
2Dynamic programming can significantly reduce the number of checks needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.