2825. Make String a Subsequence Using Cyclic Increments

2825. Make String a Subsequence Using Cyclic Increments

Difficulty: Medium

Topics: Two Pointers, String

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.

Example 1:

• 
  Input: str1 = "abc", str2 = "ad"


• 
  Output: true


• 
  Explanation: Select index 2 in str1.
  
  
  
  
  • Increment str1[2] to become 'd'.
  
  
  • Hence, str1 becomes "abd" and str2 is now a subsequence. Therefore, true is returned.
  
  
  
  
  
  

Example 2:

• 
  Input: str1 = "zc", str2 = "ad"


• 
  Output: true


• 
  Explanation: Select indices 0 and 1 in str1.
  
  
  
  
  • 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.
  
  
  
  
  
  

Example 3:

• 
  Input: str1 = "ab", str2 = "d"


• 
  Output: false


• 
  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.
  
  
  
  
  • Therefore, false is returned.
  
  
  
  
  
  

Constraints:

• 1 <= str1.length <= 105


• 1 <= str2.length <= 105


• 
  str1 and str2 consist of only lowercase English letters.

Hint:

1. Consider the indices we will increment separately.


2. We can maintain two pointers: pointer i for str1 and pointer j for str2, while ensuring they remain within the bounds of the strings.


3. If both str1[i] and str2[j] match, or if incrementing str1[i] matches str2[j], we increase both pointers; otherwise, we increment only pointer i.


4. It is possible to make str2 a subsequence of str1 if j is at the end of str2, after we can no longer find a match.

Solution:

We need to check if we can make str2 a subsequence of str1 by performing at most one cyclic increment operation on any characters in str1:

Explanation:

• We will use two pointers, i for str1 and j for str2.


• If the character at str1[i] matches str2[j], we move both pointers forward.


• If str1[i] can be incremented to match str2[j] (cyclically), we try to match them and then move both pointers.


• If neither of the above conditions holds, we only move the pointer i for str1.


• Finally, if we can match all characters of str2, then it is possible to make str2 a subsequence of str1, otherwise not.

Let's implement this solution in PHP: 2825. Make String a Subsequence Using Cyclic Increments

<?php

/**

 * @param String $str1

 * @param String $str2

 * @return Boolean

 */

function canMakeSubsequence($str1, $str2) {

    ...

    ...

    ...

    /**

     * go to ./solution.php

     */

}



// Example Usage

$str1 = "abc";

$str2 = "ad";

echo canMakeSubsequence($str1, $str2) ? 'true' : 'false'; // Output: true



$str1 = "zc";

$str2 = "ad";

echo canMakeSubsequence($str1, $str2) ? 'true' : 'false'; // Output: true



$str1 = "ab";

$str2 = "d";

echo canMakeSubsequence($str1, $str2) ? 'true' : 'false'; // Output: false

?>

Explanation:

1. 
   Two Pointers: i and j are initialized to the start of str1 and str2, respectively.


2. 
   Matching Logic: Inside the loop, we check if the characters at str1[i] and str2[j] are the same or if we can increment str1[i] to match str2[j] cyclically.
   
   
   
   
   • The cyclic increment condition is handled using (ord($str1[$i]) + 1 - ord('a')) % 26 which checks if str1[i] can be incremented to match str2[j].
   
   
   
   


3. 
   Subsequence Check: If we have iterated through str2 completely (i.e., j == m), it means str2 is a subsequence of str1. Otherwise, it's not.

Time Complexity:

• The algorithm iterates through str1 once, and each character in str2 is checked only once, so the time complexity is O(n), where n is the length of str1.

Space Complexity:

• The space complexity is O(1) since we only use a few pointers and do not need extra space dependent on the input size.

This solution efficiently checks if it's possible to make str2 a subsequence of str1 with at most one cyclic increment operation.

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