Word is Key – TCS NQT Question 3

Determine if a given word is a reserved keyword in a specific programming language

Understand the Problem

Problem Statement

One programming language has the following keywords that cannot be used as identifiers:

break, case, continue, default, defer, else, for, func, goto, if, map, range, return, struct, type, var

Write a program to find if the given word is a keyword or not.

Constraints

Input word length: 1 to 20 characters
Input word can contain letters (a-z, A-Z) only
Keywords are case-sensitive (must match exactly)
Word will not contain spaces or special characters
Keywords list is fixed and predefined

Examples

Example 1

Input

defer

Output

defer is a keyword

Explanation

The input word 'defer' matches exactly with one of the predefined keywords in the list, so the output correctly identifies it as a keyword.

Example 2

Input

While

Output

while is not a keyword

Explanation

Although 'while' is a common programming keyword, it is not in the predefined list of 16 keywords. Note that the comparison is case-sensitive, so 'While' (with capital W) is different from 'while'.

Solution

#include <stdio.h>
#include <string.h>

int main() {
    // Array of predefined keywords
    const char* keywords[] = {
        "break", "case", "continue", "default", "defer",
        "else", "for", "func", "goto", "if",
        "map", "range", "return", "struct", "type", "var"
    };
    
    // Number of keywords
    const int numKeywords = 16;
    
    // Input word
    char inputWord[21]; // Max 20 chars + null terminator
    
    // Read input
    scanf("%20s", inputWord);
    
    // Flag to track if keyword found
    int isKeyword = 0;
    
    // Check against each keyword
    for (int i = 0; i < numKeywords; i++) {
        if (strcmp(inputWord, keywords[i]) == 0) {
            isKeyword = 1;
            break;
        }
    }
    
    // Output result
    if (isKeyword) {
        printf("%s is a keyword\n", inputWord);
    } else {
        printf("%s is not a keyword\n", inputWord);
    }
    
    return 0;
}

Time:O(k) where k is the number of keywords (16). Since the keyword list is fixed, this is effectively O(1).

Space:O(k) for storing the keyword array, which is O(1) since k is fixed at 16.

Approach:

The C solution:

Defines an array of 16 keyword strings using const char* for efficiency
Uses a fixed-size character array (21 bytes) to store input with bounds checking
Reads input using scanf with width specifier to prevent buffer overflow
Iterates through the keyword array using strcmp() for exact string comparison
Uses a flag variable to track if a match is found
Outputs the result in the required format using printf

Key features: memory-safe input handling, efficient string comparison, clear output formatting.

Visual Explanation

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