# 2840. Check if Strings Can be Made Equal With Operations II

## Problem Statement

<br>

You are given two strings `s1` and `s2`, both of length `n`, consisting of **lowercase** English letters.

You can apply the following operation on **any** of the two strings **any** number of times:

* Choose any two indices `i` and `j` such that `i < j` and the difference `j - i` is **even**, then **swap** the two characters at those indices in the string.

Return `true` *if you can make the strings* `s1` *and* `s2` *equal, and* `false` *otherwise*.

&#x20;

**Example 1:**

<pre><code><strong>Input: s1 = "abcdba", s2 = "cabdab"
</strong><strong>Output: true
</strong><strong>Explanation: We can apply the following operations on s1:
</strong>- Choose the indices i = 0, j = 2. The resulting string is s1 = "cbadba".
- Choose the indices i = 2, j = 4. The resulting string is s1 = "cbbdaa".
- Choose the indices i = 1, j = 5. The resulting string is s1 = "cabdab" = s2.
</code></pre>

**Example 2:**

<pre><code><strong>Input: s1 = "abe", s2 = "bea"
</strong><strong>Output: false
</strong><strong>Explanation: It is not possible to make the two strings equal.
</strong></code></pre>

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**Constraints:**

* `n == s1.length == s2.length`
* `1 <= n <= 105`
* `s1` and `s2` consist only of lowercase English letters.

## Intuition

```
Approach:
Even index alphabets can come only on even, and odd on odd

So we sort the even and odd for both strings

Compare both strings in end
```

### Links

<https://leetcode.com/problems/check-if-strings-can-be-made-equal-with-operations-ii/>

### Video Links

### Approach 1:

```
Sorting
```

{% code title="C++" lineNumbers="true" %}

```cpp
class Solution {
public:
    string find(string &s1){
        string t1_even = "", t1_odd = "";

        for(int i=0; i<s1.size(); i++){
            if(i%2==0)
                t1_even += s1[i];
            else    
                t1_odd += s1[i];
        }

        sort(t1_even.begin(), t1_even.end());
        sort(t1_odd.begin(), t1_odd.end());

        string temp_s1="";
        int i=0,j=0;

        while(i<t1_even.size() or j<t1_odd.size()){
            if(t1_even[i])
                temp_s1 += t1_even[i++];
            if(t1_odd[j])
                temp_s1 += t1_odd[j++];
        }

        return temp_s1;
    }

    bool checkStrings(string s1, string s2) {
        string t1 = find(s1);
        string t2 = find(s2);

        return t1 == t2;
    }
};
```

{% endcode %}

### Approach 2:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 4:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

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