Why Math and Number Theory Appear in DSA Problems in DSA Typescript - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the remainder when dividing a number by 5.

DSA Typescript

const remainder = num [1] 5;

Drag options to blanks, or click blank then click option'

A+

B-

C*

D%

Attempts:

3 left

💡 Hint

Common Mistakes

Using + or * instead of the remainder operator

Confusing subtraction with remainder

The % operator gives the remainder of division, which is key in many number theory problems.

2fill in blank

medium

Complete the code to check if a number is even using modulus.

DSA Typescript

if (num [1] 2 === 0) {
  console.log('Even number');
}

Drag options to blanks, or click blank then click option'

A+

B%

C-

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using + or - instead of %

Checking for remainder 1 instead of 0

The modulus operator % with 2 checks if a number is even by seeing if the remainder is zero.

3fill in blank

hard

Fix the error in the code to compute the greatest common divisor (GCD) using Euclid's algorithm.

DSA Typescript

function gcd(a: number, b: number): number {
  if (b === 0) return a;
  return gcd(b, a [1] b);
}

Drag options to blanks, or click blank then click option'

A%

B-

C+

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of modulus

Using addition or multiplication incorrectly

Euclid's algorithm uses the remainder operator % to find the GCD.

4fill in blank

hard

Fill both blanks to create a function that returns a list of prime numbers up to n.

DSA Typescript

function primesUpTo(n: number): number[] {
  const primes = [];
  for (let i = 2; i [1]= n; i++) {
    let isPrime = true;
    for (let j = 2; j [2] i; j++) {
      if (i % j === 0) {
        isPrime = false;
        break;
      }
    }
    if (isPrime) primes.push(i);
  }
  return primes;
}

Drag options to blanks, or click blank then click option'

A<=

B<

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > or >= in loops causing infinite loops or wrong checks

Using <= in inner loop causing false negatives

The outer loop runs from 2 up to and including n (<=), and the inner loop checks divisors less than i (<).

5fill in blank

hard

Fill all three blanks to create a function that returns a dictionary of numbers and their squares for numbers greater than 3.

DSA Typescript

function squaresMap(nums: number[]): {[key: number]: number} {
  return nums.reduce((acc, num) => {
    if (num [1] 3) {
      acc[num] = num [2] num;
    }
    return acc;
  }, {} as {[key: number]: number});
}

Drag options to blanks, or click blank then click option'

A>

B*

C<

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of >

Using + instead of * for square calculation

The condition checks if the number is greater than 3 (>), and the square is calculated by multiplying the number by itself (*).

Practice

1. Why do many Data Structures & Algorithms problems involve concepts from Math and Number Theory?

easy

A. Because they help solve problems involving patterns, divisibility, and optimization

B. Because Math is easier than programming

C. Because Number Theory is unrelated to algorithms

D. Because Math concepts slow down the algorithm

Solution

1. Step 1: Understand the role of Math in DSA

   Math and Number Theory provide tools to analyze patterns, divisibility, and properties of numbers which are common in algorithm problems.

2. Step 2: Connect Math to problem solving

   These concepts help optimize solutions and solve problems efficiently, such as finding primes, gcd, or modular arithmetic.

3. Final Answer:

   Because they help solve problems involving patterns, divisibility, and optimization -> Option A

4. Quick Check:

   Math aids pattern and optimization in DSA = Because they help solve problems involving patterns, divisibility, and optimization [OK]

Hint: Math helps find patterns and optimize algorithms fast [OK]

Common Mistakes:

• Thinking Math is unrelated to algorithms
• Assuming Math makes problems slower
• Ignoring Number Theory concepts

2. Which of the following is the correct TypeScript syntax to calculate the greatest common divisor (GCD) of two numbers a and b using Euclid's algorithm?

easy

A. function gcd(a: number, b: number): number { return Math.max(a, b); }

B. function gcd(a: number, b: number): number { while (a !== 0) { let temp = b; a = b % a; b = temp; } return b; }

C. function gcd(a: number, b: number): number { return a + b; }

D. function gcd(a: number, b: number): number { return b === 0 ? a : gcd(b, a % b); }

Solution

1. Step 1: Identify Euclid's algorithm structure

   Euclid's algorithm uses recursion and modulo: gcd(a, b) = gcd(b, a % b) until b is zero.

2. Step 2: Match syntax with algorithm

   function gcd(a: number, b: number): number { return b === 0 ? a : gcd(b, a % b); } correctly implements recursion with base case b === 0 returning a.

3. Final Answer:

   function gcd(a: number, b: number): number { return b === 0 ? a : gcd(b, a % b); } -> Option D

4. Quick Check:

   Recursive gcd with modulo = function gcd(a: number, b: number): number { return b === 0 ? a : gcd(b, a % b); } [OK]

Hint: Use recursion and modulo for GCD in TypeScript [OK]

Common Mistakes:

• Using addition instead of modulo
• Returning max instead of gcd
• Incorrect loop conditions

3. What is the output of the following TypeScript code that uses modular arithmetic to find the remainder of a large exponentiation?

const mod = 1000000007;
function modExp(base: number, exp: number, mod: number): number {
  if (exp === 0) return 1;
  let half = modExp(base, Math.floor(exp / 2), mod);
  let result = (half * half) % mod;
  if (exp % 2 !== 0) result = (result * base) % mod;
  return result;
}
console.log(modExp(2, 10, mod));

medium

A. 1000000007

B. 512

C. 1024

D. 0

Solution

1. Step 1: Understand modular exponentiation

   The function calculates (2^10) % 1000000007 efficiently using recursion and squaring.

2. Step 2: Calculate 2^10 modulo 1000000007

   2^10 = 1024, which is less than mod, so result is 1024.

3. Final Answer:

   1024 -> Option C

4. Quick Check:

   2^10 mod 1000000007 = 1024 [OK]

Hint: Use fast exponentiation to compute large powers modulo [OK]

Common Mistakes:

• Confusing exponentiation with multiplication
• Ignoring modulo operation
• Incorrect base cases in recursion

4. Identify the bug in the following TypeScript function that tries to check if a number is prime:

function isPrime(n: number): boolean {
  if (n <= 1) return false;
  for (let i = 2; i <= n / 2; i++) {
    if (n % i === 0) return false;
  }
  return true;
}

medium

A. The loop should run until i <= Math.sqrt(n)

B. The modulo operator % is used incorrectly

C. The function should return true for n <= 1

D. The loop should run until i < n, not n / 2

Solution

1. Step 1: Analyze the loop range for prime checking

   Checking divisors up to n/2 is inefficient; checking up to sqrt(n) is enough.

2. Step 2: Correct the loop condition

   Change loop condition to i <= Math.sqrt(n) to improve performance and correctness.

3. Final Answer:

   The loop should run until i <= Math.sqrt(n) -> Option A

4. Quick Check:

   Prime check loop limit = sqrt(n) [OK]

Hint: Check divisors only up to square root of n [OK]

Common Mistakes:

• Looping too far causing inefficiency
• Returning true for non-prime numbers
• Misusing modulo operator

5. You want to find the number of integers between 1 and 1000 that are divisible by 3 or 5 but not both. Which approach using Number Theory concepts is best?

hard

A. Count multiples of 3 and 5, then subtract the count of multiples of 15

B. Count multiples of 3 and 5, then subtract twice the count of multiples of 15

C. Count multiples of 3 only

D. Count multiples of 5 only

Solution

1. Step 1: Use inclusion-exclusion principle

   Count multiples of 3 and 5 separately, then subtract multiples of their LCM (15) to avoid double counting.

2. Step 2: Adjust for numbers divisible by both

   Since we want numbers divisible by 3 or 5 but not both, subtract twice the count of multiples of 15.

3. Final Answer:

   Count multiples of 3 and 5, then subtract twice the count of multiples of 15 -> Option B

4. Quick Check:

   Use inclusion-exclusion with double subtraction for both divisibility [OK]

Hint: Use inclusion-exclusion and subtract twice for both divisors [OK]

Common Mistakes:

• Subtracting multiples of 15 only once
• Counting only one divisor's multiples
• Ignoring overlap of multiples