Description
https://leetcode.com/problems/make-two-arrays-equal-by-reversing-sub-arrays/
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.
Example 5:
Input: target = [1,1,1,1,1], arr = [1,1,1,1,1] Output: true
Constraints:
target.length == arr.length
1 <= target.length <= 1000
1 <= target[i] <= 1000
1 <= arr[i] <= 1000
Explanation
Just check if the occurrence of two list are the same.
Python Solution
class Solution:
def canBeEqual(self, target: List[int], arr: List[int]) -> bool:
counter1 = {}
counter2 = {}
for i in target:
counter1[i] = counter1.get(i, 0) + 1
for i in arr:
counter2[i] = counter2.get(i, 0) + 1
return counter1 == counter2
- Time Complexity: O(N).
- Space Complexity: O(N).