1460. Make Two Arrays Equal by Reversing Sub-arrays

LeetCode 1460. Make Two Arrays Equal by Reversing Sub-arrays

Description

Given two integer arrays of equal length target and arr.

In one step, you can select any non-empty sub-array of arr and reverse it. You are allowed to make any number of steps.

Return True if you can make arr equal to target, or False otherwise.

Example 1:

Input: target = [1,2,3,4], arr = [2,4,1,3]
Output: true
Explanation: You can follow the next steps to convert arr to target:
1- Reverse sub-array [2,4,1], arr becomes [1,4,2,3]
2- Reverse sub-array [4,2], arr becomes [1,2,4,3]
3- Reverse sub-array [4,3], arr becomes [1,2,3,4]
There are multiple ways to convert arr to target, this is not the only way to do so.

Example 2:

Input: target = [7], arr = [7]
Output: true
Explanation: arr is equal to target without any reverses.

Example 3:

Input: target = [1,12], arr = [12,1]
Output: true

Example 4:

Input: target = [3,7,9], arr = [3,7,11]
Output: false
Explanation: arr doesn't have value 9 and it can never be converted to target.

Constraints:

• target.length == arr.length

• 1 <= target.length <= 1000

• 1 <= target[i] <= 1000

• 1 <= arr[i] <= 1000

Tags

Array

Solution

Essentially, this problem is equivalent to check if both arrays share the same elements.

Solution 1:

Sort both arrays and check each pair of values.

Solution 2:

We can use a hash table to check if there is an element that only appears once. Return false if so.

Complexity

Solution 1:

• Time complexity: $O(n\log(n))$

• Space complexity: $O(\log(n))$, which is equivalent to the space complexity of quick sort algorithm.

Solution 2:

• Time complexity: $O(n)$

• Space complexity: $O(1)$, as the range of element is given.

Code

Solution 1:

func canBeEqual(target []int, arr []int) bool {
	var m [1001]int
	for i := 0; i < len(target); i++ {
		m[target[i]]++
		m[arr[i]]--
	}
	for _, n := range m {
		if n != 0 {
			return false
		}
	}
	return true
}

Solution 2:

func canBeEqual1(target []int, arr []int) bool {
	sort.Ints(target)
	sort.Ints(arr)
	for i := 0; i < len(target); i++ {
		if target[i] != arr[i] {
			return false
		}
	}
	return true
}