algoadvance

You are given two integer arrays target and arr of the same length. In one step, you can select any non-empty subarray of arr and reverse it. You need to determine if you can make arr equal to target after performing any number of steps.

Clarifying Questions

1. Can the arrays contain duplicate elements?
   • Yes, the arrays can contain duplicate elements.
2. What is the range of the array elements and the length of the arrays?
   • Both arrays will have the same length n, where 1 <= target.length == arr.length <= 1000.
   • The elements of the arrays are integers between 1 and 1000.
3. Is the order of elements important?
   • No, the order of elements in arr and target is not important since we can reverse any subarray.

Strategy

1. Key Insight:
   • Since you can reverse any subarray, you have the freedom to rearrange the elements in any order. Therefore, the problem reduces to checking if both arrays have the same elements with the same frequency.
2. Approach:
   • Use a data structure like a Counter or dictionary to count the frequency of each element in both arrays.
   • Compare the two frequency counts. If they match, return True. Otherwise, return False.

Code

from collections import Counter

def canBeEqual(target, arr):
    return Counter(target) == Counter(arr)

# Example Usage
target = [1, 2, 3, 4]
arr = [2, 4, 1, 3]
print(canBeEqual(target, arr))  # Output: True

target = [7]
arr = [7]
print(canBeEqual(target, arr))  # Output: True

target = [3, 7, 9]
arr = [3, 7, 11]
print(canBeEqual(target, arr))  # Output: False

Time Complexity

• Time Complexity: O(n)
  • Creating Counter objects for both target and arr takes O(n) time each.
  • Comparing two Counters takes O(n) time.
  • Thus, the overall time complexity is O(n).
• Space Complexity: O(n)
  • Storing the counts for the elements of both arrays in the Counter objects requires O(n) space.

Try our interview co-pilot at AlgoAdvance.com

• Got blindsided by a question you didn’t expect?

• Spend too much time studying?

• Or simply don’t have the time to go over all 3000 questions?