Trie Node Design and Initialization in DSA C - Time & Space Complexity
We analyze how much work it takes to create and set up a trie node.
We want to know how the time grows when we add more children to the node.
Analyze the time complexity of the following code snippet.
// Define a trie node structure
typedef struct TrieNode {
struct TrieNode* children[26];
bool isEndOfWord;
} TrieNode;
// Initialize a new trie node
TrieNode* getNode() {
TrieNode* node = malloc(sizeof(TrieNode));
node->isEndOfWord = false;
for (int i = 0; i < 26; i++) {
node->children[i] = NULL;
}
return node;
}
This code creates a new trie node and sets all its children pointers to NULL.
Identify the loops, recursion, array traversals that repeat.
- Primary operation: The for-loop that sets each child pointer to NULL.
- How many times: It runs 26 times, once for each possible child.
Each time we create a node, we do a fixed number of steps based on the number of children.
| Input Size (n) | Approx. Operations |
|---|---|
| 26 (English alphabet) | 26 steps to initialize children |
| 52 (uppercase + lowercase) | 52 steps to initialize children |
| 100 (larger alphabet) | 100 steps to initialize children |
Pattern observation: The work grows linearly with the number of children per node.
Time Complexity: O(k)
This means the time to create a node grows linearly with the number of possible children (k).
[X] Wrong: "Creating a trie node is always O(1) time regardless of children."
[OK] Correct: Because initializing each child pointer takes time proportional to the number of children, so it depends on k.
Understanding this helps you explain how tries handle alphabets of different sizes efficiently.
"What if we used a dynamic data structure like a hash map instead of a fixed array for children? How would the time complexity change?"
Practice
Solution
Step 1: Understand Trie node role
A Trie node represents a character and links to children nodes for next characters in strings.Step 2: Identify correct purpose
It stores children nodes to build paths for strings, not integer values or lengths.Final Answer:
To store children nodes representing characters of strings -> Option DQuick Check:
Trie node = children nodes [OK]
- Thinking Trie nodes store whole strings
- Confusing with storing string length
- Assuming nodes only point to parents
Solution
Step 1: Check children declaration
Children must be pointers to TrieNode, so use struct TrieNode* children[26];Step 2: Check end of word flag type
Using int for boolean is common in C; bool requires stdbool.h and is less common.Final Answer:
typedef struct TrieNode { struct TrieNode* children[26]; int isEndOfWord; } TrieNode; -> Option AQuick Check:
Children pointers array + int flag = typedef struct TrieNode { struct TrieNode* children[26]; int isEndOfWord; } TrieNode; [OK]
- Declaring children as struct array instead of pointers
- Using bool without including stdbool.h
- Declaring children as single pointer instead of array
typedef struct TrieNode {
struct TrieNode* children[26];
int isEndOfWord;
} TrieNode;
TrieNode* getNode() {
TrieNode* node = (TrieNode*)malloc(sizeof(TrieNode));
node->isEndOfWord = 0;
for (int i = 0; i < 26; i++)
node->children[i] = NULL;
return node;
}
int main() {
TrieNode* root = getNode();
printf("%d\n", root->isEndOfWord);
printf("%p\n", root->children[0]);
return 0;
}Solution
Step 1: Analyze getNode initialization
isEndOfWord is set to 0 explicitly; children array elements set to NULL (0).Step 2: Check main output
Printing isEndOfWord prints 0; printing children[0] prints NULL pointer (0x0).Final Answer:
0 0x0 (NULL pointer address) -> Option CQuick Check:
Initialized zero and NULL = 0 0x0 (NULL pointer address) [OK]
- Assuming uninitialized pointers have zero value
- Confusing isEndOfWord default value
- Expecting children pointers to have non-null addresses
typedef struct TrieNode {
struct TrieNode* children[26];
int isEndOfWord;
} TrieNode;
TrieNode* getNode() {
TrieNode* node = malloc(sizeof(TrieNode));
node->isEndOfWord = 0;
for (int i = 0; i <= 26; i++)
node->children[i] = NULL;
return node;
}Solution
Step 1: Check loop bounds
Loop runs from i = 0 to i <= 26, which means 27 iterations, but children array size is 26, causing out-of-bounds access.Step 2: Identify correct fix
Loop should run i < 26 to stay within array bounds and avoid memory errors.Final Answer:
Loop condition should be i < 26 to avoid out-of-bounds access -> Option AQuick Check:
Array size 26 means max index 25 [OK]
- Using <= instead of < in loops
- Ignoring array bounds causing runtime errors
- Assuming malloc cast is mandatory in C
Solution
Step 1: Check children and end flag
Children array of pointers size 26 is correct; isEndOfWord as int is standard in C.Step 2: Add frequency field correctly
Frequency should be an int to count occurrences; pointer to int is unnecessary complexity.Final Answer:
typedef struct TrieNode { struct TrieNode* children[26]; int isEndOfWord; int frequency; } TrieNode; -> Option BQuick Check:
Use int for frequency and isEndOfWord [OK]
- Using pointer for frequency unnecessarily
- Omitting isEndOfWord flag
- Using bool without stdbool.h
Ready to prove it?
Pass the test and this pattern moves unknown
