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Introduction

LeetCode 131: Palindrome Partitioning is a problem where the goal is to partition a given string into all possible palindrome partitions. A palindrome is a string that reads the same backward as forward. The problem requires finding all possible ways to partition the string such that every substring in the partition is a palindrome.

Problem Statement

Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s.

A palindrome is defined as a string that reads the same forward and backward.

Example:

go

Input: "aab"
Output: [
  ["a", "a", "b"],
  ["aa", "b"]
]

Explanation:

• The string "aab" can be partitioned in two ways: ["a", "a", "b"] and ["aa", "b"].

Approach:

Backtracking Approach

The problem is typically solved using backtracking, where we explore all possible ways to partition the string into substrings and check whether each substring is a palindrome.

Here’s the detailed step-by-step approach:

1. Backtrack Through the String:

• Start by recursively partitioning the string.
• For each partition, check if the current substring is a palindrome.

1. Palindrome Check:

• If the substring is a palindrome, continue partitioning the remaining string.
• If not, discard that partition and backtrack to try the next possible substring.

1. Recursive Call:

• At each step, for each possible substring of the string, we recursively partition the remainder of the string.
• If the entire string is processed, store the current valid partition.

Optimization:

• Palindrome Check: A helper function to check if a substring is a palindrome can be used to optimize the process.

Edge Case:

• If the string is empty, return an empty list.

Go Implementation

Solution Using Backtracking

go

package main

import "fmt"

// Helper function to check if a string is a palindrome
func isPalindrome(s string) bool {
    left, right := 0, len(s)-1
    for left < right {
        if s[left] != s[right] {
            return false
        }
        left++
        right--
    }
    return true
}

// Backtracking function to find all palindrome partitions
func backtrack(s string, start int, path []string, result *[][]string) {
    // If we've reached the end of the string, add the current partition to the result
    if start == len(s) {
        // Append a copy of path to the result
        temp := make([]string, len(path))
        copy(temp, path)
        *result = append(*result, temp)
        return
    }

    // Try every possible substring starting at 'start'
    for end := start + 1; end <= len(s); end++ {
        substring := s[start:end]
        if isPalindrome(substring) {
            // If the substring is a palindrome, add it to the path and recurse
            path = append(path, substring)
            backtrack(s, end, path, result)
            path = path[:len(path)-1] // Backtrack
        }
    }
}

// Main function to find all palindrome partitions
func partition(s string) [][]string {
    var result [][]string
    backtrack(s, 0, []string{}, &result)
    return result
}

func main() {
    s := "aab"
    result := partition(s)
    fmt.Println("Palindrome partitions:", result)
}

Explanation:

1. isPalindrome Function:

• This helper function checks whether a given string is a palindrome by comparing characters from the left and right ends of the string.

1. backtrack Function:

• This function recursively partitions the string. Starting from the start index, it generates all possible substrings and checks if they are palindromes.
• If a substring is a palindrome, it is added to the current path, and the function recurses on the remaining part of the string.
• Once the string is fully partitioned (i.e., start == len(s)), the current valid partition is added to the result.

1. partition Function:

• This is the main function that initializes the result variable and starts the backtracking process from the first character of the string.

1. Example in main:

• The string "aab" is passed to the partition function, and the possible palindrome partitions are printed.

Time and Space Complexity

MetricComplexityTime ComplexityO(2^n)Space ComplexityO(n)

Time Complexity:

• O(2^n) where n is the length of the string. The time complexity arises because at each step, we can choose either to include or exclude each character, leading to exponential growth. Additionally, we check each substring for being a palindrome.

Space Complexity:

• O(n) where n is the length of the string. This is because, in the worst case, the recursion depth can be as deep as the length of the string. The space is also used to store intermediate results in the recursion stack.

Edge Cases

1. Edge Case: Empty String

• Input: ""
• Output: []
• Explanation: An empty string has no valid palindrome partitions.

1. Edge Case: String with No Palindromes

• Input: "abc"
• Output: [["a", "b", "c"]]
• Explanation: Every individual character is a palindrome, so the only valid partition is splitting the string into individual characters.

1. Edge Case: Single Character

• Input: "a"
• Output: [["a"]]
• Explanation: A single character is always a palindrome.

Conclusion

LeetCode 131: Palindrome Partitioning is a challenging problem that can be efficiently solved using backtracking. By recursively exploring all substrings and checking if they are palindromes, we can generate all possible valid partitions. With an optimal solution leveraging recursive calls, the problem becomes manageable despite its exponential complexity. This approach is both intuitive and flexible for solving similar problems involving string partitioning and palindrome checks.