Palindrome Partitioning Using Backtracking in DSA C# - Execution Trace

Concept Flow - Palindrome Partitioning Using Backtracking

Start with empty partition

↓

Choose substring from current index

↓

Check if substring is palindrome?

No→Discard substring, try next

Yes↓

Add substring to current partition

↓

Recurse from next index

↓

If end of string reached

↓

Save current partition as a solution

↓

Backtrack: remove last substring

↩Back to choose next substring

The process tries all substrings from the current position, checks if they are palindromes, adds them to the current partition, recurses forward, and backtracks to explore all partitions.

Execution Sample

DSA C#

void Backtrack(string s, int start, List<string> current, List<List<string>> result) {
  if (start == s.Length) {
    result.Add(new List<string>(current));
    return;
  }
  for (int end = start; end < s.Length; end++) {
    if (IsPalindrome(s, start, end)) {
      current.Add(s.Substring(start, end - start + 1));
      Backtrack(s, end + 1, current, result);
      current.RemoveAt(current.Count - 1);
    }
  }
}

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

Execution Table

Step	Operation	Substring Checked	Is Palindrome?	Current Partition	Action	Visual State
1	Start	-	-	[]	Begin backtracking at index 0	[]
2	Check substring	"a" (0-0)	Yes	[]	Add "a" to partition	["a"]
3	Recurse	Start=1	-	["a"]	Recurse from index 1	["a"]
4	Check substring	"b" (1-1)	Yes	["a"]	Add "b" to partition	["a", "b"]
5	Recurse	Start=2	-	["a", "b"]	Recurse from index 2	["a", "b"]
6	Check substring	"a" (2-2)	Yes	["a", "b"]	Add "a" to partition	["a", "b", "a"]
7	Recurse	Start=3	-	["a", "b", "a"]	Reached end, save partition	[["a", "b", "a"]]
8	Backtrack	Remove last substring	-	["a", "b", "a"]	Remove "a"	["a", "b"]
9	Check substring	"ba" (1-2)	No	["a"]	Discard substring	["a"]
10	Backtrack	Remove last substring	-	["a"]	Remove "a"	[]
11	Check substring	"ab" (0-1)	No	[]	Discard substring	[]
12	Check substring	"aba" (0-2)	Yes	[]	Add "aba" to partition	["aba"]
13	Recurse	Start=3	-	["aba"]	Reached end, save partition	[["a", "b", "a"], ["aba"]]
14	Backtrack	Remove last substring	-	["aba"]	Remove "aba"	[]
15	End	-	-	[]	All substrings checked, done	[["a", "b", "a"], ["aba"]]

💡 All substrings from all positions checked, all palindrome partitions found.

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 6	After Step 7	After Step 8	After Step 12	After Step 13	Final
start	0	1	1	2	3	3	0	3	3
current partition	[]	["a"]	["a", "b"]	["a", "b", "a"]	["a", "b", "a"]	["a", "b"]	["aba"]	["aba"]	[]
result	[]	[]	[]	[]	[["a", "b", "a"]]	[["a", "b", "a"]]	[["a", "b", "a"]]	[["a", "b", "a"], ["aba"]]	[["a", "b", "a"], ["aba"]]

Key Moments - 3 Insights

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

After exploring partitions with the last added substring, we remove it to try other substrings starting at the same position, as shown in steps 7 to 8 and 13 to 14 in the execution_table.

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

Only palindrome substrings can be part of the partition. Checking avoids invalid partitions. This is shown in steps 2, 4, 6, 9, and 11 where non-palindromes are discarded.

What happens when the start index reaches the string length?

It means we have a complete partition covering the whole string, so we save it as a solution. This occurs at steps 7 and 13.

Visual Quiz - 3 Questions

Look at the execution_table, what is the current partition at step 6?

A ["a", "b", "a"]

B ["a", "b"]

C ["a"]

D []

💡 Hint

Check the 'Current Partition' column at step 6 in the execution_table.

At which step does the algorithm add the substring "aba" to the partition?

A Step 9

B Step 12

C Step 4

D Step 2

💡 Hint

Look for the operation 'Add "aba" to partition' in the execution_table.

If the substring "ba" (1-2) was palindrome, how would the current partition change at step 9?

A It would add "ba" to the partition

B It would remove "a" from the partition

C It would skip adding any substring

D It would end the recursion

💡 Hint

Step 9 shows 'Discard substring' because "ba" is not palindrome; if it was, it would be added.

Concept Snapshot

Palindrome Partitioning Using Backtracking:
- Start from index 0 with empty partition
- For each substring from current index, check palindrome
- If palindrome, add to partition and recurse from next index
- If end reached, save partition
- Backtrack by removing last substring to try others
- Explore all partitions this way

Full Transcript

Palindrome Partitioning Using Backtracking tries all ways to split a string into palindrome substrings. It starts with an empty partition and checks every substring from the current position. If a substring is a palindrome, it adds it to the current partition and recurses from the next index. When the end of the string is reached, it saves the current partition as a valid solution. Then it backtracks by removing the last 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?

easy

A. To find the longest palindrome substring in a string

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

C. To split a string into all possible palindrome substrings

D. To count the number of palindromes in a string

Solution

1. Step 1: Understand palindrome partitioning

   Palindrome partitioning means dividing a string into parts where each part is a palindrome.

2. Step 2: Role of backtracking

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

3. Final Answer:

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

4. Quick Check:

   Palindrome partitioning = all palindrome splits [OK]

Hint: Backtracking explores all palindrome splits of the string [OK]

Common Mistakes:

• Confusing palindrome partitioning with finding longest palindrome
• Thinking palindrome partitioning only counts palindromes
• Assuming palindrome partitioning reverses the string

2. Which of the following C# method signatures is correct for a backtracking function that partitions a string into palindromes?

easy

A. string Backtrack(string s, int start, List<string> currentPartition, List result)

B. int Backtrack(string s, int start, List<string> currentPartition)

C. bool Backtrack(string s, int start, List<string> currentPartition, List result)

D. void Backtrack(string s, int start, List<string> currentPartition, List result)

Solution

1. Step 1: Understand backtracking method requirements

   The method should explore partitions and store results, so it returns void and uses parameters to track state.

2. Step 2: Check method signatures

   void Backtrack(string s, int start, List<string> currentPartition, List result) returns void and takes parameters for string, start index, current partition, and result list, matching backtracking needs.

3. Final Answer:

   void Backtrack(string s, int start, List<string> currentPartition, List result) -> Option D

4. Quick Check:

   Backtracking method returns void with state parameters [OK]

Hint: Backtracking methods usually return void and use parameters to track state [OK]

Common Mistakes:

• Using return types other than void for backtracking
• Missing the result list parameter to store partitions
• Confusing return type with boolean or string

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

var result = new List();
Backtrack("aab", 0, new List<string>(), result);
Console.WriteLine(result.Count);

Assuming Backtrack partitions the string into all palindrome substrings correctly.

medium

A. 2

B. 3

C. 1

D. 4

Solution

1. Step 1: Identify palindrome partitions of "aab"

   Possible palindrome partitions are ["a", "a", "b"] and ["aa", "b"].

2. Step 2: Count all palindrome partitions

   There are 2 partitions, so result.Count will be 2 after backtracking.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Palindrome partitions of "aab" = 2 [OK]

Hint: Count all palindrome partitions generated by backtracking [OK]

Common Mistakes:

• Counting substrings instead of partitions
• Forgetting "aa" is a palindrome
• Assuming single partition only

4. Identify the bug in this backtracking code snippet for palindrome partitioning:

void Backtrack(string s, int start, List<string> current, List result) {
  if (start == s.Length) {
    result.Add(new List<string>(current));
    return;
  }
  for (int end = start; end < s.Length; end++) {
    if (IsPalindrome(s, start, end)) {
      current.Add(s.Substring(start, end - start));
      Backtrack(s, end + 1, current, result);
      current.RemoveAt(current.Count - 1);
    }
  }
}

medium

A. Base case condition should be start > s.Length

B. Substring length calculation is incorrect

C. RemoveAt should remove from index 0

D. IsPalindrome method parameters are wrong

Solution

1. Step 1: Check substring length calculation

   Substring(start, length) needs length = end - start + 1, but code uses end - start, missing last character.

2. Step 2: Verify other parts

   IsPalindrome parameters and RemoveAt usage are correct; base case is correct as start == s.Length.

3. Final Answer:

   Substring length calculation is incorrect -> Option B

4. Quick Check:

   Substring length = end - start + 1 [OK]

Hint: Substring length must include end index +1 [OK]

Common Mistakes:

• Using substring length as end - start instead of end - start + 1
• Removing wrong index from current list
• Incorrect base case condition

5. Given the string "racecar", how many palindrome partitions will the backtracking algorithm find? Consider all possible partitions where each substring is a palindrome.

hard

A. 10

B. 6

C. 8

D. 4

Solution

1. Step 1: List all palindrome partitions of "racecar"

   Partitions include ["r", "a", "c", "e", "c", "a", "r"], ["r", "a", "cec", "a", "r"], ["r", "aceca", "r"], ["racecar"], and others.

2. Step 2: Count all valid palindrome partitions

   Counting carefully, there are 10 valid palindrome partitions in total.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   Palindrome partitions of "racecar" = 10 [OK]

Hint: Count all palindrome splits including nested palindromes [OK]

Common Mistakes:

• Counting only single character partitions
• Missing longer palindrome substrings like "cec" or "aceca"
• Assuming only 1 or 2 partitions