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Palindrome Partitioning (All)

Backtracking
O(n × 2^n) time, O(n) space
Advanced

Partition string such that every substring is palindrome. Return all possible partitions using backtracking.

Prerequisites:
Backtracking
Palindrome checking
String manipulation
Recursion

Visualization

Interactive visualization for Palindrome Partitioning (All)

Palindrome Partitioning:

  • • Backtracking to find all partitions
  • • Each part must be palindrome

Interactive visualization with step-by-step execution

Implementation

Language:
1function partition(s: string): string[][] {
2  const result: string[][] = [];
3  const current: string[] = [];
4  
5  function isPalindrome(str: string, left: number, right: number): boolean {
6    while (left < right) {
7      if (str[left++] !== str[right--]) return false;
8    }
9    return true;
10  }
11  
12  function backtrack(start: number): void {
13    if (start === s.length) {
14      result.push([...current]);
15      return;
16    }
17    
18    for (let end = start; end < s.length; end++) {
19      if (isPalindrome(s, start, end)) {
20        current.push(s.substring(start, end + 1));
21        backtrack(end + 1);
22        current.pop();
23      }
24    }
25  }
26  
27  backtrack(0);
28  return result;
29}

Deep Dive

Theoretical Foundation

For each position, try all possible palindromic substrings starting from that position. Recursively partition remaining string. Time: O(n × 2^n) - 2^n partitions, O(n) to check palindrome and copy.

Complexity

Time

Best

O(n × 2^n)

Average

O(n × 2^n)

Worst

O(n × 2^n)

Space

Required

O(n)

Applications

Industry Use

1

Text processing and analysis

2

String segmentation problems

3

Natural language processing

4

DNA sequence analysis

5

Compression algorithm preprocessing

6

Pattern recognition in strings

7

Linguistic analysis tools

Use Cases

String segmentation
Text processing
Palindrome analysis

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