Problem Statement

You are given two strings pattern and text. It is given that all characters of the pattern string are unique.

Rearrange the characters of the text string based on the characters' order in pattern string. In other words, if a character a occurs before a character b in the pattern string, then a should occur before b in the output string.

Characters in text that do not appear in pattern should be appended at the end of the rearranged string in their original order.

Examples

1. Example 1:

   • Input: pattern = "xy", text = "yyzx"
   • Expected Output: "xyyz"
   • Justification: In 'pattern', the order is 'x', 'y'. In the text, 'x' appears once, 'y' twice, and 'z' doesn't appear. So, we put z at the end of the string. Thus, the output string is xzyy.

2. Example 2:

   • Input: pattern = "abc", text = "aabbcc"
   • Expected Output: "aabbcc"
   • Justification: Here, the 'text' already follows the order 'a', 'b', 'c' as specified in 'pattern', so the order remains 'aabbcc'.

3. Example 3:

   • Input: pattern = "mno", text = "onomon"
   • Expected Output: "mnnooo"
   • Justification: According to 'pattern', 'm' comes first, then 'n', then 'o'. Rearranging 'text' in this order results in 'm' once, 'n' twice, followed by 'o' thrice, hence 'mnnooo'.

Solution

To solve this problem, we will create a custom sorting mechanism based on the 'pattern' string. The key idea is to prioritize the characters in 'text' that appear in 'pattern', arranging them in the order they appear in 'pattern', and then appending the remaining characters from 'text' in their original order. This approach works because it respects the specified order in 'pattern' while maintaining the sequence of characters in 'text' that are not in 'pattern'.

Our approach involves first counting the frequency of each character of the text string. We'll use a hashmap for this purpose. Then, we'll iterate over each character in 'pattern', and if that character exists in text, we'll append it to our result string as many times as it appears in text. Finally, we'll go through text again and append any characters to the result string that were not in pattern. This ensures that characters in text are first ordered according to pattern and then the remaining characters retain their original order.

Step-by-Step Algorithm

• Create a HashMap to store the frequency of each character in 'text'.
• Iterate over 'text' and update the frequency in the hashmap for each character.
• Initialize an empty string result for the final sorted string.
• Iterate over pattern:
  • For each character in 'pattern', check if it exists in the hashmap.
  • If it exists, append it to result as many times as its frequency, and then remove it from the hashmap.
• Iterate over 'text' again:
  • For each character, if it still exists in the hashmap (meaning it was not in 'pattern'), append it to result.
• Return result as the final output.

Algorithm Walkthrough

Input: pattern = "mno", text = "onomon"

1. Create frequency hashmap for 'text': {o: 3, n: 2, m: 1}
2. Initialize result as an empty string.
3. Iterate over 'pattern':
   • 'm' found in the map, append to result -> result = "m"
   • 'n' found in the map, append twice -> result = "mnn"
   • 'o' found in the map, append thrice -> result = "mnnooo"
4. 'm', 'n', and 'o' are removed from hashmap.
5. No additional characters to append from 'text'.
6. Final output is "mnnooo".

Code

Complexity Analysis

Time Complexity

1. HashMap Creation: O(N), where N is the length of 'text'. This is for iterating over 'text' to create the frequency hashmap.
2. Iterating Over 'Pattern': O(M), where M is the length of 'pattern'. This covers iterating over 'pattern' and appending characters to the result.
3. Final Iteration Over 'Text': O(N) for appending remaining characters not in 'pattern'.

The overall time complexity is O(N + M), considering both the iterations over 'text' and 'pattern'.

Space Complexity

1. HashMap: O(N), where N is the number of distinct characters in 'text'. In the worst case, this could be equal to the length of 'text' if all characters are unique.
2. Result String: O(N), equivalent to the length of 'text', as in the worst case, all characters from 'text' are added to the result.

The overall space complexity is O(N), considering the space for the hashmap and the result string.