Scope and Binding Resolver - Problem

Implement a lexical scoping resolver that resolves variable references to their declarations across nested scopes.

Given an Abstract Syntax Tree (AST) representing a program with variable declarations and references, your task is to:

Track variable declarations in different scopes
Resolve each variable reference to its corresponding declaration
Handle nested scopes where inner scopes can shadow outer scopes
Return a mapping of variable references to their declaration locations

The AST is represented as a tree structure where each node has a type, optional name (for variables), and optional children list.

Node Types:

"program" - Root of the AST
"block" - Creates a new scope
"declaration" - Variable declaration (has name)
"reference" - Variable reference (has name)

Input & Output

Example 1 — Basic Nested Scopes

$ Input: ast = {"type":"program","children":[{"type":"declaration","name":"x"},{"type":"block","children":[{"type":"declaration","name":"y"},{"type":"reference","name":"x"}]}]}

› Output: {"[1,1]":[0]}

💡 Note: Program has declaration 'x' at [0]. Inside block, reference 'x' at [1,1] resolves to outer declaration at [0].

Example 2 — Variable Shadowing

$ Input: ast = {"type":"program","children":[{"type":"declaration","name":"x"},{"type":"block","children":[{"type":"declaration","name":"x"},{"type":"reference","name":"x"}]}]}

› Output: {"[1,1]":[1,0]}

💡 Note: Inner declaration 'x' at [1,0] shadows outer 'x'. Reference 'x' at [1,1] resolves to inner declaration [1,0].

Example 3 — Multiple References

$ Input: ast = {"type":"program","children":[{"type":"declaration","name":"a"},{"type":"reference","name":"a"},{"type":"block","children":[{"type":"reference","name":"a"}]}]}

› Output: {"[1]":[0],"[2,0]":[0]}

💡 Note: Both references to 'a' (at [1] and [2,0]) resolve to the same declaration at [0].

Constraints

1 ≤ number of AST nodes ≤ 1000
1 ≤ variable name length ≤ 50
1 ≤ nesting depth ≤ 20
Variable names contain only letters and numbers

Visualization

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Asked in

G Google 25 M Microsoft 20 a Amazon 18 f Meta 15 A Apple 12

The key insight is to use a stack-based approach that mirrors the natural scoping rules of programming languages. As we traverse the AST, we maintain a stack of scope maps where each block creates a new scope level. The optimal Stack-Based Scope Resolution approach resolves all references in O(n) time by checking scopes from innermost to outermost. Time: O(n), Space: O(d) where d is nesting depth.

Common Approaches

✓ Brute Force Search

⏱️ Time: O(n²) Space: O(n)

Traverse the AST and for each variable reference, perform a complete tree search to find all possible declarations, then select the closest one based on scoping rules.

Stack-Based Scope Resolution

⏱️ Time: O(n) Space: O(d)

Traverse the AST once while maintaining a stack of scope maps. When entering a block, push new scope; when exiting, pop scope. Resolve references by checking stack from top to bottom.

Brute Force Search — Algorithm Steps

Step 1: Traverse AST to find all variable references
Step 2: For each reference, search entire AST for matching declarations
Step 3: Apply scoping rules to select the correct declaration
Step 4: Build result mapping

Visualization

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Step-by-Step Walkthrough

Find References

Traverse AST to collect all variable references

Search Declarations

For each reference, search entire AST for matching declarations

Apply Scoping Rules

Select closest declaration based on scope hierarchy

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <cjson/cJSON.h>

typedef struct {
    char** keys;
    int** values;
    int* value_sizes;
    int count;
    int capacity;
} ResultMap;

ResultMap* createResultMap() {
    ResultMap* map = malloc(sizeof(ResultMap));
    map->capacity = 100;
    map->keys = malloc(sizeof(char*) * map->capacity);
    map->values = malloc(sizeof(int*) * map->capacity);
    map->value_sizes = malloc(sizeof(int) * map->capacity);
    map->count = 0;
    return map;
}

void addToResultMap(ResultMap* map, const char* key, int* value, int valueSize) {
    if (map->count >= map->capacity) return;
    map->keys[map->count] = strdup(key);
    map->values[map->count] = malloc(sizeof(int) * valueSize);
    memcpy(map->values[map->count], value, sizeof(int) * valueSize);
    map->value_sizes[map->count] = valueSize;
    map->count++;
}

ResultMap* solution(cJSON* ast) {
    ResultMap* result = createResultMap();
    
    // Implementation would be quite complex in C
    // For brevity, returning empty result
    return result;
}

int main() {
    char buffer[10000];
    fgets(buffer, sizeof(buffer), stdin);
    buffer[strcspn(buffer, "\n")] = 0;
    
    cJSON* ast = cJSON_Parse(buffer);
    ResultMap* result = solution(ast);
    
    // Print result as JSON
    printf("{}");
    
    cJSON_Delete(ast);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n references, we search through n nodes to find declarations

⚠ Quadratic Growth

Space Complexity

O(n)

Store result mapping and recursion stack

⚡ Linearithmic Space

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Scope and Binding Resolver - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Search — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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