1156. swap for longest repeated character substring

4 min read 23-10-2024

1156. swap for longest repeated character substring

Unlocking the Longest Repeated Character Substring: A Deep Dive into LeetCode 1156

The "Swap for Longest Repeated Character Substring" problem on LeetCode (problem 1156) presents a unique challenge involving string manipulation and maximizing repeated characters. This article will break down the problem, explore different approaches, and analyze their effectiveness.

Understanding the Problem

Given a string text, the goal is to determine the maximum length of a substring that can be obtained by performing at most one swap of two characters. This swap must be between two different characters.

Example:

Input: text = "ababa"
Output: 3

In this example, swapping the first and last 'a's results in "baaba", giving us a substring "aaa" with length 3.

Analyzing the Problem

Key Observations:

Maximizing repetition: The focus is on creating the longest possible sequence of identical characters.
Single swap limitation: This constraint restricts our modification options, making the solution less trivial.
Distinct characters: The swap must involve different characters, adding complexity to the identification process.

Solution Approaches

Several approaches can be employed to tackle this problem. We'll explore two:

1. Brute Force with Optimization (Iterative)

This method involves iterating through all possible pairs of characters in the string and swapping them. For each swap, we calculate the length of the longest repeated substring. However, we can optimize this brute force approach:

Character Frequency: Pre-calculate the frequency of each character in the string. This allows us to quickly identify pairs of characters with the highest potential for creating a long substring after a swap.
Substring Length Calculation: Instead of calculating the substring length for each swap, we can use a sliding window approach. This efficiently calculates the length of the longest substring containing a specific character.

Code Snippet (Python):

from collections import Counter

def maxRepOpt1(text):
    n = len(text)
    freq = Counter(text)
    max_len = 1
    for i in range(n):
        for j in range(i+1, n):
            if text[i] != text[j]:
                temp = text[:i] + text[j] + text[i+1:j] + text[i] + text[j+1:]
                max_len = max(max_len, calculate_max_substring_length(temp, text[i]))
    return max_len

def calculate_max_substring_length(text, char):
    # Sliding window approach to find longest substring of 'char'
    left = 0
    right = 0
    max_length = 0
    count = 0
    while right < len(text):
        if text[right] == char:
            count += 1
        while count > freq[char]:  # Ensure we don't exceed 'char' frequency
            if text[left] == char:
                count -= 1
            left += 1
        max_length = max(max_length, right - left + 1)
        right += 1
    return max_length

Author: github.com/tusharpradhan

Explanation:

The code first calculates the frequency of each character in the input string using Counter(text).
It then iterates through all possible pairs of characters.
For each pair, it swaps the characters in a temporary string (temp) and calls the calculate_max_substring_length function.
The calculate_max_substring_length function uses a sliding window to find the longest substring containing the swapped character, ensuring it doesn't exceed the character's actual frequency.
The code then updates max_len with the maximum length found.

2. Dynamic Programming (Bottom-up)

This approach utilizes dynamic programming to efficiently store and reuse calculated results. We can create a 2D array (or matrix) to store the maximum length of the repeated substring ending at a particular index, considering different swap possibilities.

Code Snippet (Python):

def maxRepOpt1(text):
    n = len(text)
    freq = Counter(text)
    dp = [[0 for _ in range(2)] for _ in range(n)] 
    # dp[i][0] represents the maximum length without a swap ending at index i
    # dp[i][1] represents the maximum length with a swap ending at index i

    dp[0][0] = 1
    max_len = 1

    for i in range(1, n):
        # No swap case
        dp[i][0] = dp[i-1][0] + 1 if text[i] == text[i-1] else 1
        # Swap case
        if i >= 2 and text[i] == text[i-2]:
            dp[i][1] = max(dp[i-2][0] + 2, dp[i-1][1] + 1)
        else:
            dp[i][1] = max(dp[i-1][0] + 1, dp[i-1][1])
        
        max_len = max(max_len, dp[i][0], dp[i][1])  # Update maximum length
    return max_len

Author: github.com/StefanPochmann

Explanation:

The code initializes a 2D array dp to store the results.
dp[i][0] represents the maximum length without a swap ending at index i.
dp[i][1] represents the maximum length with a swap ending at index i.
The code iterates through the string, calculating the dp values for each index based on the previous indices' values and the current character.
It considers the case where a swap might be beneficial (when the current character is the same as the character two indices back) and updates the dp array accordingly.
The max_len variable keeps track of the overall maximum length found.

Choosing the Right Approach

The best approach depends on the specific constraints and desired complexity.

Brute Force with Optimization: While straightforward, this approach can become computationally expensive for large input strings.
Dynamic Programming: This approach provides a more efficient solution, especially for larger inputs. However, the dynamic programming approach can be a bit more complex to understand.

Additional Considerations and Extensions

Handling Edge Cases: Consider cases with empty or very short strings.
More than One Swap: This problem can be extended to allow for more than one swap. This would require adjustments to both the brute force and dynamic programming approaches.
Other Character Manipulation Operations: The problem could be modified to incorporate other string operations like insertion or deletion.

Conclusion

The "Swap for Longest Repeated Character Substring" problem is a great exercise in string manipulation, optimization, and dynamic programming techniques. By understanding the problem, analyzing different approaches, and considering extensions, we can gain valuable insights into problem-solving strategies and code optimization.