1460. Make Two Arrays Equal by Reversing Subarrays 

Description

You are given two integer arrays of equal length target and arr. In one step, you can select any non-empty subarray 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 subarray [2,4,1], arr becomes [1,4,2,3]
2- Reverse subarray [4,2], arr becomes [1,2,4,3]
3- Reverse subarray [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 = [3,7,9], arr = [3,7,11]
Output: false
Explanation: arr does not 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

Solutions

Solution 1: Sorting

If two arrays are equal after sorting, then they can be made equal by reversing sub-arrays.

Therefore, we only need to sort the two arrays and then check if the sorted arrays are equal.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$, where $n$ is the length of the array $arr$.

Python3

class Solution:
    def canBeEqual(self, target: List[int], arr: List[int]) -> bool:
        return sorted(target) == sorted(arr)

Java

class Solution {
    public boolean canBeEqual(int[] target, int[] arr) {
        Arrays.sort(target);
        Arrays.sort(arr);
        return Arrays.equals(target, arr);
    }
}

C++

class Solution {
public:
    bool canBeEqual(vector<int>& target, vector<int>& arr) {
        sort(target.begin(), target.end());
        sort(arr.begin(), arr.end());
        return target == arr;
    }
};

Go

func canBeEqual(target []int, arr []int) bool {
	sort.Ints(target)
	sort.Ints(arr)
	return reflect.DeepEqual(target, arr)
}

TypeScript

function canBeEqual(target: number[], arr: number[]): boolean {
    target.sort();
    arr.sort();
    return target.every((x, i) => x === arr[i]);
}

JavaScript

function canBeEqual(target, arr) {
    target.sort();
    arr.sort();
    return target.every((x, i) => x === arr[i]);
}

Rust

impl Solution {
    pub fn can_be_equal(mut target: Vec<i32>, mut arr: Vec<i32>) -> bool {
        target.sort();
        arr.sort();
        target == arr
    }
}

PHP

class Solution {
    /**
     * @param Integer[] $target
     * @param Integer[] $arr
     * @return Boolean
     */
    function canBeEqual($target, $arr) {
        sort($target);
        sort($arr);
        return $target === $arr;
    }
}

C

int compare(const void* a, const void* b) {
    return (*(int*) a - *(int*) b);
}

bool canBeEqual(int* target, int targetSize, int* arr, int arrSize) {
    qsort(target, targetSize, sizeof(int), compare);
    qsort(arr, arrSize, sizeof(int), compare);
    for (int i = 0; i < targetSize; ++i) {
        if (target[i] != arr[i]) {
            return false;
        }
    }
    return true;
}

Solution 2: Counting

We note that the range of the array elements given in the problem is $1 \sim 1000$. Therefore, we can use two arrays cnt1 and cnt2 of length $1001$ to record the number of times each element appears in the arrays target and arr respectively. Finally, we just need to check if the two arrays are equal.

We can also use only one array cnt. We traverse the arrays target and arr. For target[i], we increment cnt[target[i]], and for arr[i], we decrement cnt[arr[i]]. In the end, we check if all elements in the array cnt are $0$.

The time complexity is $O(n + M)$, and the space complexity is $O(M)$. Here, $n$ is the length of the array arr, and $M$ is the range of the array elements. In this problem, $M = 1001$.