Divide and Conquer Strategy and Recurrence Relations in DSA Javascript - Build from Scratch

Divide and Conquer Strategy and Recurrence Relations

📖 Scenario: Imagine you have a large list of numbers representing daily sales in a store. You want to find the maximum sale amount quickly. Using the divide and conquer strategy, you can split the list into smaller parts, find the maximum in each part, and then combine the results.

🎯 Goal: Build a JavaScript program that uses the divide and conquer strategy to find the maximum number in an array. You will also see how the recurrence relation describes the time taken by this approach.

📋 What You'll Learn

Create an array called sales with exact values

Create a variable called n to store the length of the array

Write a recursive function called findMax that uses divide and conquer to find the maximum value

Print the maximum sale value found by findMax

💡 Why This Matters

🌍 Real World

Divide and conquer is used in many real-world problems like sorting large datasets, searching efficiently, and solving complex problems by breaking them down.

💼 Career

Understanding divide and conquer and recurrence relations is essential for software engineers to write efficient algorithms and analyze their performance.

1

Create the sales data array

Create an array called sales with these exact values: [120, 450, 300, 700, 150, 900, 400]

DSA Javascript

# Your code here

Hint

Use const sales = [120, 450, 300, 700, 150, 900, 400]; to create the array.

2

Store the length of the sales array

Create a variable called n and set it to the length of the sales array using sales.length

DSA Javascript

const sales = [120, 450, 300, 700, 150, 900, 400];
# Your code here

Hint

Use const n = sales.length; to get the array size.

3

Write the divide and conquer function to find max

Write a recursive function called findMax that takes arr, start, and end as parameters. If start equals end, return arr[start]. Otherwise, find the middle index, call findMax on the left half and right half, then return the larger of the two results.

DSA Javascript

const sales = [120, 450, 300, 700, 150, 900, 400];
const n = sales.length;
# Your code here

Hint

Use recursion to split the array until one element remains, then compare left and right halves.

4

Print the maximum sale value

Use console.log to print the result of calling findMax with sales, 0, and n - 1 as arguments.

DSA Javascript

const sales = [120, 450, 300, 700, 150, 900, 400];
const n = sales.length;

function findMax(arr, start, end) {
    if (start === end) {
        return arr[start];
    }
    const mid = Math.floor((start + end) / 2);
    const leftMax = findMax(arr, start, mid);
    const rightMax = findMax(arr, mid + 1, end);
    return leftMax > rightMax ? leftMax : rightMax;
}
# Your code here

Hint

Call findMax(sales, 0, n - 1) and print the result using console.log.

Practice

1. What is the main idea behind the Divide and Conquer strategy in algorithms?

easy

A. Solve the problem by trying all possible solutions one by one

B. Use a single loop to iterate through all elements

C. Break the problem into smaller subproblems, solve them independently, then combine results

D. Store all intermediate results to avoid repeated calculations

Solution

1. Step 1: Understand the divide and conquer approach

   It splits a big problem into smaller parts that are easier to solve.

2. Step 2: Combine the solutions of smaller parts

   After solving subproblems, their results are merged to get the final answer.

3. Final Answer:

   Break the problem into smaller subproblems, solve them independently, then combine results -> Option C

4. Quick Check:

   Divide and conquer = split, solve, combine [OK]

Hint: Divide and conquer means split, solve, then merge results [OK]

Common Mistakes:

• Confusing divide and conquer with brute force
• Thinking it uses only one loop
• Mixing it with dynamic programming

2. Which of the following is the correct syntax for the recurrence relation of Merge Sort?

easy

A. T(n) = 3T(n/3) + O(log n)

B. T(n) = 2T(n/2) + O(n)

C. T(n) = T(n/2) + O(n^2)

D. T(n) = T(n-1) + O(1)

Solution

1. Step 1: Recall Merge Sort recurrence

   Merge Sort splits array into two halves, sorts each half, then merges in O(n).

2. Step 2: Write recurrence form

   It calls itself twice on half size (2T(n/2)) plus linear merge time O(n).

3. Final Answer:

   T(n) = 2T(n/2) + O(n) -> Option B

4. Quick Check:

   Merge Sort recurrence = 2T(n/2) + n [OK]

Hint: Merge Sort splits in two, merges in linear time [OK]

Common Mistakes:

• Using T(n-1) instead of T(n/2)
• Adding O(n^2) instead of O(n)
• Confusing with Quick Sort recurrence

3. What is the output of the following JavaScript function that uses divide and conquer?

function sumArray(arr) {
  if (arr.length === 1) return arr[0];
  const mid = Math.floor(arr.length / 2);
  return sumArray(arr.slice(0, mid)) + sumArray(arr.slice(mid));
}
console.log(sumArray([2, 4, 6, 8]));

medium

A. 20

B. 24

C. 10

D. Error

Solution

1. Step 1: Understand the function behavior

   The function splits the array into halves recursively and sums elements.

2. Step 2: Trace the calls for [2,4,6,8]

   Split into [2,4] and [6,8], sumArray([2,4]) = 6, sumArray([6,8]) = 14, total 20.

3. Final Answer:

   20 -> Option A

4. Quick Check:

   Sum of 2+4+6+8 = 20 [OK]

Hint: Divide array, sum halves, add results [OK]

Common Mistakes:

• Confusing slice indices causing wrong sums
• Assuming function returns array instead of sum
• Forgetting base case for single element

4. Identify the error in this divide and conquer function to find the maximum element:

function findMax(arr) {
  if (arr.length === 1) return arr[0];
  const mid = Math.floor(arr.length / 2);
  const leftMax = findMax(arr.slice(0, mid));
  const rightMax = findMax(arr.slice(mid));
  return leftMax > rightMax ? leftMax : rightMax;
}
console.log(findMax([]));

medium

A. Fails on empty array input causing infinite recursion

B. Incorrect comparison operator used

C. No error, works correctly

D. Slices do not split array properly

Solution

1. Step 1: Check base case for empty array

   The function does not handle empty array input, so arr.length === 0 is not checked.

2. Step 2: Understand recursion on empty array

   Calling findMax([]) leads to infinite recursion because base case is only for length 1.

3. Final Answer:

   Fails on empty array input causing infinite recursion -> Option A

4. Quick Check:

   Empty input needs base case to avoid infinite calls [OK]

Hint: Always check for empty input base case [OK]

Common Mistakes:

• Ignoring empty array edge case
• Assuming slice always returns non-empty array
• Thinking comparison operator is wrong

5. Given the recurrence relation T(n) = 3T(n/3) + n, what is the time complexity using the Master Theorem?

hard

A. O(log n)

B. O(n^2)

C. O(n)

D. O(n log n)

Solution

1. Step 1: Identify parameters for Master Theorem

   Here, a=3, b=3, f(n)=n. Calculate log_b(a) = log_3(3) = 1.

2. Step 2: Compare f(n) with n^log_b(a)

   f(n) = n and n^log_b(a) = n^1 = n, so f(n) = Theta(n^log_b(a)).

3. Step 3: Apply Master Theorem case 2

   When f(n) = Theta(n^log_b(a)), T(n) = Theta(n log n).

4. Final Answer:

   O(n log n) -> Option D

5. Quick Check:

   Master Theorem case 2 = n log n [OK]

Hint: Compare f(n) to n^log_b(a) for Master Theorem [OK]

Common Mistakes:

• Confusing log base or parameters a and b
• Choosing wrong Master Theorem case
• Ignoring the extra n term in recurrence