Palindrome Partitioning Using Backtracking in DSA C - 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 for remaining string

↓

If no string left

→Save current partition as solution

↓

Backtrack: remove last substring

↩Back to choose next substring

We try all substrings from the start, check palindrome, add if yes, recurse on rest, and backtrack to explore all partitions.

Execution Sample

DSA C

void backtrack(char* s, int start) {
  if (start == strlen(s)) save_partition();
  for (int end = start; end < strlen(s); end++) {
    if (isPalindrome(s, start, end)) {
      addSubstring(s, start, end);
      backtrack(s, end + 1);
      removeLastSubstring();
    }
  }
}

This code tries all substrings starting at 'start', adds palindrome substrings to partition, recurses, then backtracks.

Execution Table

Step	Operation	Substring Chosen	Palindrome?	Current Partition	Action	Visual State
1	Choose substring s[0:0]	"a"	Yes	[]	Add 'a'	["a"]
2	Recurse from index 1			["a"]	Continue	["a"]
3	Choose substring s[1:1]	"b"	Yes	["a"]	Add 'b'	["a", "b"]
4	Recurse from index 2			["a", "b"]	Continue	["a", "b"]
5	Choose substring s[2:2]	"a"	Yes	["a", "b"]	Add 'a'	["a", "b", "a"]
6	Recurse from index 3			["a", "b", "a"]	Save partition	["a", "b", "a"]
7	Backtrack			["a", "b", "a"]	Remove 'a'	["a", "b"]
8	Backtrack			["a", "b"]	Remove 'b'	["a"]
9	Choose substring s[1:2]	"ba"	No	["a"]	Discard	["a"]
10	Backtrack			["a"]	Remove 'a'	[]
11	Choose substring s[0:1]	"ab"	No	[]	Discard	[]
12	Choose substring s[0:2]	"aba"	Yes	[]	Add 'aba'	["aba"]
13	Recurse from index 3			["aba"]	Save partition	["aba"]
14	Backtrack			["aba"]	Remove 'aba'	[]
15	End			[]	All partitions found	[]

💡 All substrings tried and all palindrome partitions saved.

Variable Tracker

Variable	Start	After Step 1	After Step 3	After Step 5	After Step 6	After Step 7	After Step 8	After Step 10	After Step 12	After Step 13	Final
start index	0	1	2	3	3	3	2	1	3	3	end of string
current partition	[]	["a"]	["a", "b"]	["a", "b", "a"]	["a", "b", "a"]	["a", "b"]	["a"]	[]	["aba"]	["aba"]	[]

Key Moments - 3 Insights

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

Because after exploring partitions with that substring, we must remove it to try other substrings starting at the same position (see steps 6 to 8 in execution_table).

Why do we only add substrings that are palindromes?

Because the problem requires partitions where every substring is a palindrome. Non-palindromes are discarded immediately (see step 9).

How do we know when to save a partition?

When the start index reaches the string length, meaning the whole string is partitioned (see step 6 and 13).

Visual Quiz - 3 Questions

Look at the execution_table at step 5, what is the current partition?

A ["a", "b"]

B ["a"]

C ["a", "b", "a"]

D []

💡 Hint

Check the 'Visual State' column at step 5 in execution_table.

At which step does the algorithm discard a non-palindrome substring?

A Step 9

B Step 12

C Step 3

D Step 1

💡 Hint

Look for 'Palindrome?' column with 'No' and 'Action' as 'Discard' in execution_table.

If the input string was "aa", how would the partitions change?

A ["a"] only

B ["a", "a"] and ["aa"]

C ["aa"] only

D No partitions

💡 Hint

Consider palindrome substrings of "aa" and how backtracking explores all partitions.

Concept Snapshot

Palindrome Partitioning Using Backtracking:
- Try all substrings from current start
- Check if substring is palindrome
- If yes, add to current partition
- Recurse on remaining string
- If end reached, save partition
- Backtrack by removing last substring
- Explore all palindrome partitions

Full Transcript

Palindrome Partitioning using backtracking tries all possible substrings starting from the current index. For each substring, it checks if it is a palindrome. If yes, it adds the substring to the current partition and recursively continues with the remaining string. When the start index reaches the end of the string, it saves the current partition as a valid solution. After recursion, it backtracks by removing the last added substring to try other possibilities. This process continues until all palindrome partitions are found.

Practice

1. What is the main goal of palindrome partitioning using backtracking on a string s?

easy

A. To find the longest palindrome substring in s

B. To split s into all possible substrings where each substring is a palindrome

C. To count the number of palindromes in s

D. To reverse the string s using recursion

Solution

1. Step 1: Understand palindrome partitioning

   Palindrome partitioning means dividing the string into parts where each part reads the same forwards and backwards.

2. Step 2: Role of backtracking

   Backtracking helps explore all possible ways to split the string into palindrome substrings.

3. Final Answer:

   To split s into all possible substrings where each substring is a palindrome -> Option B

4. Quick Check:

   Palindrome partitioning = all palindrome substrings [OK]

Hint: Backtracking explores all palindrome splits [OK]

Common Mistakes:

• Confusing palindrome partitioning with longest palindrome substring
• Thinking it only counts palindromes instead of listing partitions
• Assuming it reverses the string

2. Which of the following is the correct base case condition in a backtracking function for palindrome partitioning of string s starting at index start?

easy

A. if (start == s.length()) add current partition to result and return;

B. if (start > s.length()) return false;

C. if (start == s.length()) return true;

D. if (start < 0) return;

Solution

1. Step 1: Identify when to stop recursion

   When start reaches the length of the string, it means the entire string is partitioned.

2. Step 2: What to do at base case

   At this point, add the current list of palindrome substrings to the result list and return to explore other partitions.

3. Final Answer:

   if (start == s.length()) add current partition to result and return; -> Option A

4. Quick Check:

   Base case adds partition when start reaches string end [OK]

Hint: Add partition when start equals string length [OK]

Common Mistakes:

• Returning true/false instead of adding partition
• Checking start > length instead of == length
• Using invalid index checks like start < 0

3. What will be the output partitions for the string "aab" using palindrome partitioning with backtracking?

medium

A. [["aab"]]

B. [["aa", "ab"]]

C. [["a", "ab"]]

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

Solution

1. Step 1: Check all palindrome substrings

   "a" is palindrome, "aa" is palindrome, "b" is palindrome, but "ab" and "aab" are not.

2. Step 2: List all palindrome partitions

   Partitions are ["a", "a", "b"] and ["aa", "b"].

3. Final Answer:

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

4. Quick Check:

   Valid palindrome partitions = [["a", "a", "b"], ["aa", "b"]] [OK]

Hint: Check substrings palindrome before partitioning [OK]

Common Mistakes:

• Including non-palindrome substrings like "ab"
• Missing the partition ["aa", "b"]
• Returning the whole string as one partition

4. Given the following code snippet for palindrome partitioning, what is the error?

void backtrack(char* s, int start, char** current, int currentSize, char*** result, int* returnSize) {
  if (start == strlen(s)) {
    result[*returnSize] = current;
    (*returnSize)++;
    return;
  }
  for (int end = start; end < strlen(s); end++) {
    if (isPalindrome(s, start, end)) {
      current[currentSize] = substring(s, start, end);
      backtrack(s, end + 1, current, currentSize + 1, result, returnSize);
    }
  }
}

medium

A. Assigning current directly to result causes shared reference issues

B. The base case condition is incorrect

C. The loop should run from end = 0 to end < strlen(s)

D. Missing return statement after recursive call

Solution

1. Step 1: Analyze base case assignment

   Assigning current pointer directly to result[*returnSize] stores the same reference for all partitions.

2. Step 2: Understand backtracking data handling

   We must copy the current partition array to avoid overwriting when backtracking continues.

3. Final Answer:

   Assigning current directly to result causes shared reference issues -> Option A

4. Quick Check:

   Copy partitions to result to avoid shared references [OK]

Hint: Copy current partition before adding to result [OK]

Common Mistakes:

• Not copying current partition array
• Incorrect loop range for end index
• Ignoring base case correctness

5. How can you optimize palindrome checking in backtracking palindrome partitioning for string s of length n?

hard

A. Skip palindrome checking and add all substrings to partitions

B. Check palindrome by reversing substring every time during recursion

C. Precompute a 2D boolean table dp where dp[i][j] is true if substring s[i..j] is palindrome

D. Use a hash map to store all substrings and their palindrome status

Solution

1. Step 1: Understand palindrome checking cost

   Checking palindrome by reversing substring each time costs O(n) per check, leading to O(n^3) overall.

2. Step 2: Use dynamic programming optimization

   Precompute a 2D table dp where dp[i][j] stores palindrome status for substring s[i..j]. This reduces palindrome checks to O(1).

3. Step 3: Benefit in backtracking

   Using dp speeds up backtracking by avoiding repeated palindrome checks.

4. Final Answer:

   Precompute a 2D boolean table dp where dp[i][j] is true if substring s[i..j] is palindrome -> Option C

5. Quick Check:

   DP table for palindrome checking optimizes backtracking [OK]

Hint: Use DP table to check palindrome in O(1) time [OK]

Common Mistakes:

• Reversing substring every time causing slow performance
• Using hash map adds unnecessary overhead
• Skipping palindrome checks leads to wrong partitions