grid47 Exploring patterns and algorithms
Sep 23, 2024
6 min read
Solution to LeetCode 443: String Compression Problem
Given an array of characters, compress it by replacing consecutive repeating characters with the character followed by the count of repetitions. If the character repeats only once, just include the character. The result should modify the input array and return the new length of the array.
Problem
Approach
Steps
Complexity
Input: The input is an array of characters where each character is a lowercase or uppercase letter, digit, or symbol.
Example: ["a","a","b","b","c","c","c"]
Constraints:
• 1 <= chars.length <= 2000
• chars[i] is a lowercase English letter, uppercase English letter, digit, or symbol.
Output: The output should be the new length of the modified array, with the array itself being compressed in place.
Example: 6
Constraints:
• The modified array should be stored in the input array itself.
Goal: Efficiently compress the character array by grouping consecutive repeating characters and replacing them with the character followed by the count.
Steps:
• 1. Iterate over the characters in the input array.
• 2. For each group of consecutive repeating characters, append the character to the result.
• 3. If the group contains more than one character, append the length of the group as well.
• 4. Modify the input array in place to store the compressed version.
• 5. Return the new length of the array.
Goal: The input array should be compressed in place and no additional space should be used other than constant extra space.
Steps:
• 1 <= chars.length <= 2000
• The space complexity must be constant.
Assumptions:
• The input will always contain at least one character.
• Input: ["a","a","b","b","c","c","c"]
• Explanation: In this case, the consecutive characters 'a' and 'b' appear twice, while 'c' appears three times. The output compresses to ['a','2','b','2','c','3'].
• Input: ["x"]
• Explanation: The array has only one character, so no compression is necessary and the length remains 1.
Approach: The solution uses a greedy approach to iterate through the input array, compressing consecutive character groups in place without using extra space.
Observations:
• This problem requires modifying the array in place, so careful management of indices is necessary.
• We can use two pointers: one to track the current character and another to track the position where we will store the compressed result.
Steps:
• 1. Initialize two pointers, i for traversing the input array and j for filling the result in the same array.
• 2. Traverse the input array and count consecutive repetitions of the current character.
• 3. Append the character to the result, and if the repetition count is greater than 1, append the count as well.
• 4. Update the pointer j to reflect the new length of the array.
Empty Inputs:
• If the input array is empty, return 0.
Large Inputs:
• For large input arrays with many consecutive repeating characters, the solution must run efficiently in O(n) time.
Special Values:
• Consider cases where all characters are unique or all characters are the same.
Constraints:
• Ensure that no additional space is used beyond constant space for character count storage.
