# 2825. Make String a Subsequence Using Cyclic Increments

## Problem Statement

<br>

You are given two **0-indexed** strings `str1` and `str2`.

In an operation, you select a **set** of indices in `str1`, and for each index `i` in the set, increment `str1[i]` to the next character **cyclically**. That is `'a'` becomes `'b'`, `'b'` becomes `'c'`, and so on, and `'z'` becomes `'a'`.

Return `true` *if it is possible to make* `str2` *a subsequence of* `str1` *by performing the operation **at most once***, *and* `false` *otherwise*.

**Note:** A subsequence of a string is a new string that is formed from the original string by deleting some (possibly none) of the characters without disturbing the relative positions of the remaining characters.

&#x20;

**Example 1:**

<pre><code><strong>Input: str1 = "abc", str2 = "ad"
</strong><strong>Output: true
</strong><strong>Explanation: Select index 2 in str1.
</strong>Increment str1[2] to become 'd'. 
Hence, str1 becomes "abd" and str2 is now a subsequence. Therefore, true is returned.
</code></pre>

**Example 2:**

<pre><code><strong>Input: str1 = "zc", str2 = "ad"
</strong><strong>Output: true
</strong><strong>Explanation: Select indices 0 and 1 in str1. 
</strong>Increment str1[0] to become 'a'. 
Increment str1[1] to become 'd'. 
Hence, str1 becomes "ad" and str2 is now a subsequence. Therefore, true is returned.
</code></pre>

**Example 3:**

<pre><code><strong>Input: str1 = "ab", str2 = "d"
</strong><strong>Output: false
</strong><strong>Explanation: In this example, it can be shown that it is impossible to make str2 a subsequence of str1 using the operation at most once. 
</strong>Therefore, false is returned.
</code></pre>

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**Constraints:**

* `1 <= str1.length <= 105`
* `1 <= str2.length <= 105`
* `str1` and `str2` consist of only lowercase English letters.

## Intuition

```
Just match the string and check if, not matches exactly , incr by 1 and try to match
```

### Links

<https://leetcode.com/problems/make-string-a-subsequence-using-cyclic-increments/description/>

### Video Links

### Approach 1:

```
Earlier 
```

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

```cpp
class Solution {
public:
    bool canMakeSubsequence(string s, string t) {
        int i = s.size()-1;
        int j = t.size()-1;

        while(i>=0 and j>=0){
            if(s[i] == t[j] or s[i]+1 == t[j] or (s[i]=='z' and t[j]=='a')){
                i--;
                j--;
            }

            else
                i--;
        }

        if(j<0)
            return true;

        return false;
    }
};
```

{% endcode %}

### Approach 2:

```
New
```

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

```cpp
class Solution {
public:
    bool canMakeSubsequence(string s, string t) {
        int i = s.size()-1;
        int j = t.size()-1;

        while(i>=0 and j>=0){
            if(s[i] == t[j]){
                j--;
            }

            else{
                int idx = s[i]-'a';
                int new_idx = (idx + 1)%26;

                char ch = char(new_idx + 'a');

                if(ch == t[j])
                    j--;    
            }

            i--;
        }

        return j<0;
    }
};
```

{% endcode %}

### Approach 3:

```
```

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

```cpp
```

{% endcode %}

### Approach 4:

```
```

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

```cpp
```

{% endcode %}

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