The method should explore partitions and store results, so it returns void and uses parameters to track state.
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.
Final Answer:
void Backtrack(string s, int start, List<string> currentPartition, List result) -> Option D
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
Step 1: Identify palindrome partitions of "aab"
Possible palindrome partitions are ["a", "a", "b"] and ["aa", "b"].
Step 2: Count all palindrome partitions
There are 2 partitions, so result.Count will be 2 after backtracking.
Final Answer:
2 -> Option A
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
Step 1: Check substring length calculation
Substring(start, length) needs length = end - start + 1, but code uses end - start, missing last character.
Step 2: Verify other parts
IsPalindrome parameters and RemoveAt usage are correct; base case is correct as start == s.Length.
Final Answer:
Substring length calculation is incorrect -> Option B
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
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.
Step 2: Count all valid palindrome partitions
Counting carefully, there are 10 valid palindrome partitions in total.
Final Answer:
10 -> Option A
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
Recommended
Ready to prove it?
Pass the test and this pattern moves unknownstrongwhile earning +1 Grip toward your certificate.