Palindrome Partitioning Using Backtracking in DSA Rust - Execution Trace

Concept Flow - Palindrome Partitioning Using Backtracking

Start with empty partition

↓

Choose substring from start

↓

Check if substring is palindrome?

No→Discard substring, try next

Yes↓

Add substring to current partition

↓

Recurse on remaining string

↓

If no string left

→Save current partition as solution

↓

Backtrack: remove last substring

↩Back to choose next substring

Start with empty partitions, pick substrings from the start, check palindrome, add if yes, recurse on rest, save when done, then backtrack to try other splits.

Execution Sample

DSA Rust

fn is_palindrome(s: &str) -> bool {
    s.chars().eq(s.chars().rev())
}

fn backtrack(s: &str, start: usize, path: &mut Vec<String>, res: &mut Vec<Vec<String>>) {
    if start == s.len() {
        res.push(path.clone());
        return;
    }
    for end in start+1..=s.len() {
        let substr = &s[start..end];
        if is_palindrome(substr) {
            path.push(substr.to_string());
            backtrack(s, end, path, res);
            path.pop();
        }
    }
}

This code tries all substrings from start, adds palindrome substrings to path, recurses, and backtracks to find all palindrome partitions.

Execution Table

Step	Operation	Substring Chosen	Is Palindrome?	Current Partition	Action	Resulting Partitions
1	Start			[]	Begin backtracking at index 0	[]
2	Choose substring	"a"	Yes	["a"]	Add and recurse from index 1	[]
3	Choose substring	"b"	Yes	["a", "b"]	Add and recurse from index 2	[]
4	Choose substring	"c"	Yes	["a", "b", "c"]	Add and recurse from index 3	[["a", "b", "c"]]
5	Backtrack			["a", "b"]	Remove last substring	[["a", "b", "c"]]
6	Backtrack			["a"]	Remove last substring (no more from index 2)	[["a", "b", "c"]]
7	Choose substring	"bc"	No	["a"]	Discard and try next	[["a", "b", "c"]]
8	Backtrack			[]	Remove last substring (no more from index 1)	[["a", "b", "c"]]
9	Choose substring	"ab"	No	[]	Discard and try next	[["a", "b", "c"]]
10	Choose substring	"abc"	No	[]	Discard and try next	[["a", "b", "c"]]
11	End			[]	All substrings tried, backtracking complete	[["a", "b", "c"]]

💡 All substrings from each start index checked; backtracking finished with all palindrome partitions found.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 7	Final
start	0	1	2	3	2	1	0
path	[]	["a"]	["a", "b"]	["a", "b", "c"]	["a", "b"]	["a"]	[]
res	[]	[]	[]	[["a", "b", "c"]]	[["a", "b", "c"]]	[["a", "b", "c"]]	[["a", "b", "c"]]

Key Moments - 3 Insights

Why do we backtrack by removing the last substring after recursion?

Because after exploring partitions with the last substring, we must remove it to try other substrings starting at the same position, as shown in steps 5 and 6 in the execution_table.

Why do we check if the substring is a palindrome before adding it?

Only palindrome substrings are valid partitions. Checking prevents invalid partitions and unnecessary recursion, as seen in step 7 where "bc" is discarded.

What happens when start equals the string length?

It means we have partitioned the entire string successfully, so we save the current partition to results, as in step 4.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what is the current partition?

A ["a", "b"]

B ["a", "b", "c"]

C ["a"]

D []

💡 Hint

Check the 'Current Partition' column at step 4 in the execution_table.

At which step does the algorithm add the first complete partition to the results?

A Step 2

B Step 4

C Step 6

D Step 16

💡 Hint

Look for when 'Resulting Partitions' first shows a partition in the execution_table.

If the input string was "aa", how would the partitions change compared to the current execution?

A More partitions because "aa" is palindrome and "a" substrings too

B Fewer partitions because "aa" is not palindrome

C Same partitions as "abc"

D No partitions because string is too short

💡 Hint

Check how palindrome substrings affect partitions in the variable_tracker and execution_table.

Concept Snapshot

Palindrome Partitioning Using Backtracking:
- Start from index 0 with empty path
- For each substring from start, check palindrome
- If palindrome, add to path and recurse from end index
- If start reaches string length, save path as a partition
- Backtrack by removing last substring to try others
- Finds all palindrome partitions of the string

Full Transcript

Palindrome Partitioning uses backtracking to find all ways to split a string into palindrome substrings. We start with an empty partition and pick substrings from the start. Each substring is checked if it is a palindrome. If yes, it is added to the current partition path and the function recurses on the remaining string. When the start index reaches the string length, the current partition is saved as a valid solution. Then the algorithm backtracks by removing the last substring to try other possible substrings. This process continues until all partitions are found. The execution table shows each step, substring chosen, palindrome check, current partition, and results. Variable tracker shows how start index, path, and results change over time. Key moments clarify why backtracking is needed, why palindrome check is important, and what happens when the string is fully partitioned. The visual quiz tests understanding of partition states and effects of input changes.

Practice

1. What is the main goal of palindrome partitioning using backtracking?

easy

A. To split a string into all possible palindrome substrings

B. To find the longest palindrome substring in a string

C. To count the number of palindromes in a string

D. To reverse the string and check if it is a palindrome

Solution

1. Step 1: Understand palindrome partitioning using backtracking

   Palindrome partitioning means dividing a string into parts where each part is a palindrome. Backtracking helps explore all possible ways to split the string into palindrome parts.

2. Final Answer:

   To split a string into all possible palindrome substrings -> Option A

3. Quick Check:

   Palindrome partitioning = all palindrome splits [OK]

Hint: Backtracking explores all palindrome splits [OK]

Common Mistakes:

• Confusing palindrome partitioning with longest palindrome substring
• Thinking it only counts palindromes
• Assuming it just reverses the string

2. Which Rust syntax correctly defines a function to check if a string slice is a palindrome?

easy

A. fn is_palindrome(s: &str) -> bool { s.chars().eq(s.chars().rev()) }

B. fn is_palindrome(s: String) -> bool { s == s.reverse() }

C. fn is_palindrome(s: &str) -> bool { s == s.reverse() }

D. fn is_palindrome(s: String) -> bool { s.chars().eq(s.chars()) }

Solution

1. Step 1: Check function signature and verify palindrome logic

   Using &str is efficient for string slices without ownership transfer. Comparing characters iterator with its reverse checks palindrome correctly.

2. Final Answer:

   fn is_palindrome(s: &str) -> bool { s.chars().eq(s.chars().rev()) } -> Option A

3. Quick Check:

   Correct Rust palindrome check syntax = fn is_palindrome(s: &str) -> bool { s.chars().eq(s.chars().rev()) } [OK]

Hint: Use &str and compare chars with reversed chars [OK]

Common Mistakes:

• Using String instead of &str unnecessarily
• Calling reverse() method which does not exist on String
• Comparing string to itself without reversing

3. What will be the output of the following Rust code snippet?

fn main() {
    let s = "aab";
    let mut result = vec![];
    fn backtrack(s: &str, start: usize, path: &mut Vec<String>, result: &mut Vec) {
        if start == s.len() {
            result.push(path.clone());
            return;
        }
        for end in start+1..=s.len() {
            let substr = &s[start..end];
            if substr.chars().eq(substr.chars().rev()) {
                path.push(substr.to_string());
                backtrack(s, end, path, result);
                path.pop();
            }
        }
    }
    backtrack(s, 0, &mut vec![], &mut result);
    println!("{:?}", result);
}

medium

A. [["a", "ab"]]

B. [["aab"]]

C. [["a", "a", "b"], ["aa", "b"]]

D. [["aa", "bb"]]

Solution

1. Step 1: Trace backtracking on "aab"

   Partitions are checked for palindrome substrings starting from index 0. "a" is palindrome, then "a" again, then "b"; also "aa" is palindrome, then "b". Output: [["a", "a", "b"], ["aa", "b"]].

2. Final Answer:

   [["a", "a", "b"], ["aa", "b"]] -> Option C

3. Quick Check:

   All palindrome splits of "aab" = [["a", "a", "b"], ["aa", "b"]] [OK]

Hint: Check all palindrome substrings from each start index [OK]

Common Mistakes:

• Missing the "aa" palindrome substring
• Including non-palindrome substrings like "ab"
• Confusing output format

4. Identify the bug in this Rust backtracking function for palindrome partitioning:

fn backtrack(s: &str, start: usize, path: &mut Vec<String>, result: &mut Vec) {
    if start > s.len() {
        result.push(path.clone());
        return;
    }
    for end in start+1..=s.len() {
        let substr = &s[start..end];
        if substr.chars().eq(substr.chars().rev()) {
            path.push(substr.to_string());
            backtrack(s, end, path, result);
            path.pop();
        }
    }
}

medium

A. The for loop range should be start..=s.len(), not start+1..=s.len()

B. The path vector should not be cloned when pushing to result

C. The substring check should use bytes instead of chars

D. The base case condition should be start == s.len(), not start > s.len()

Solution

1. Step 1: Analyze base case and why it fails

   Backtracking ends when start reaches string length exactly, not greater. If start > s.len(), it means out of bounds, so base case never triggers correctly.

2. Final Answer:

   The base case condition should be start == s.len(), not start > s.len() -> Option D

3. Quick Check:

   Base case must match string length exactly [OK]

Hint: Base case triggers at start == string length [OK]

Common Mistakes:

• Using start > s.len() causes missed base case
• Changing loop range incorrectly
• Unnecessary cloning or wrong substring checks

5. Given the string "aaaaa", how many palindrome partitions will the backtracking algorithm find if it returns all possible palindrome partitions?

hard

A. 8

B. 16

C. 20

D. 12

Solution

1. Step 1: Analyze all-'a' string partitions

   Since all characters are identical 'a', every substring is a palindrome. The number of partitions equals the ways to place cuts in the 4 gaps between 5 characters: 2^4 = 16.

2. Final Answer:

   16 -> Option B

3. Quick Check:

   2^{n-1} = 16 for n=5 [OK]

Hint: For identical chars string, 2^{n-1} partitions [OK]

Common Mistakes:

• Counting only large palindrome substrings
• Ignoring smaller palindrome partitions
• Underestimating total partition count