Divide and Conquer Strategy and Recurrence Relations in DSA Typescript - 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 total sales quickly by splitting the list into smaller parts, adding each part, and then combining the results.

🎯 Goal: Build a simple divide and conquer function in TypeScript that sums numbers in an array by splitting it into halves recursively. Understand how the problem breaks down and how the recurrence relation describes the time complexity.

📋 What You'll Learn

Create an array called sales with exactly these numbers: [10, 20, 30, 40, 50, 60, 70, 80]

Create a variable called threshold and set it to 2

Write a recursive function called sumSales that takes an array of numbers and returns their sum using divide and conquer:

sumSales should return the sum directly if the array length is less than or equal to threshold

Otherwise, split the array into two halves, recursively sum each half, and return their sum

Print the result of calling sumSales on the sales array

💡 Why This Matters

🌍 Real World

Divide and conquer is used in many algorithms like sorting, searching, and multiplying large numbers efficiently.

💼 Career

Understanding divide and conquer helps in designing efficient algorithms and solving complex problems in software development.

1

Create the sales data array

Create an array called sales with these exact numbers: [10, 20, 30, 40, 50, 60, 70, 80]

DSA Typescript

# Your code here

Hint

Use const sales: number[] = [...] to create the array with the exact numbers.

2

Set the threshold for direct sum

Create a variable called threshold and set it to 2

DSA Typescript

const sales: number[] = [10, 20, 30, 40, 50, 60, 70, 80];
# Your code here

Hint

Use const threshold: number = 2; to set the threshold.

3

Write the divide and conquer sum function

Write a recursive function called sumSales that takes an array of numbers called arr and returns their sum using divide and conquer. If arr.length is less than or equal to threshold, return the sum directly using a loop. Otherwise, split arr into two halves, recursively call sumSales on each half, and return their sum.

DSA Typescript

const sales: number[] = [10, 20, 30, 40, 50, 60, 70, 80];
const threshold: number = 2;

# Your code here

Hint

Use recursion with a base case for small arrays and split the array for larger ones.

4

Print the total sales using sumSales

Print the result of calling sumSales on the sales array using console.log

DSA Typescript

const sales: number[] = [10, 20, 30, 40, 50, 60, 70, 80];
const threshold: number = 2;

function sumSales(arr: number[]): number {
    if (arr.length <= threshold) {
        let total = 0;
        for (let i = 0; i < arr.length; i++) {
            total += arr[i];
        }
        return total;
    } else {
        const mid = Math.floor(arr.length / 2);
        const leftSum = sumSales(arr.slice(0, mid));
        const rightSum = sumSales(arr.slice(mid));
        return leftSum + rightSum;
    }
}

# Your code here

Hint

Use console.log(sumSales(sales)); to print the total sum.

Practice

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

easy

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

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

C. Use a single loop to iterate through all elements

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 A

4. Quick Check:

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

Hint: Divide big problems into smaller parts, solve, then merge [OK]

Common Mistakes:

• Confusing divide and conquer with brute force
• Thinking it uses only loops without recursion
• Mixing it up with dynamic programming

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

easy

A. T(n) = 3T(n/3) + O(n^2)

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

C. T(n) = T(n/2) + O(1)

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

Solution

1. Step 1: Recall Merge Sort process

   Merge Sort splits array into two halves, sorts each half, then merges.

2. Step 2: Write recurrence relation

   It calls itself twice on half size (2T(n/2)) plus merging cost 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)
• Confusing merge cost with recursive calls
• Choosing cubic or quadratic terms incorrectly

3. What is the output of the following TypeScript function that uses divide and conquer to find the maximum element?

function findMax(arr: number[], left: number, right: number): number {
  if (left === right) return arr[left];
  const mid = Math.floor((left + right) / 2);
  const maxLeft = findMax(arr, left, mid);
  const maxRight = findMax(arr, mid + 1, right);
  return maxLeft > maxRight ? maxLeft : maxRight;
}

const array = [3, 5, 1, 8, 2];
console.log(findMax(array, 0, array.length - 1));

medium

A. 3

B. 5

C. 8

D. 2

Solution

1. Step 1: Understand the function logic

   The function splits the array into halves recursively and compares max values.

2. Step 2: Trace the recursive calls on array [3,5,1,8,2]

   Max in left half [3,5,1] is 5; max in right half [8,2] is 8; final max is 8.

3. Final Answer:

   8 -> Option C

4. Quick Check:

   Max element found by divide and conquer = 8 [OK]

Hint: Divide array, find max in halves, pick bigger [OK]

Common Mistakes:

• Returning min instead of max
• Incorrect mid calculation causing infinite recursion
• Mixing indices causing wrong subarray

4. Identify the error in the following recurrence relation and fix it:

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

This recurrence is supposed to represent the time complexity of a divide and conquer algorithm similar to Merge Sort.

medium

A. Change O(n^2) to O(n) because merging takes linear time

B. Change 2T(n/2) to T(n-1) because it is not dividing

C. Change O(n^2) to O(log n) because merging is logarithmic

D. No error, the recurrence is correct

Solution

1. Step 1: Analyze the merging cost in divide and conquer

   In Merge Sort-like algorithms, merging two halves takes O(n) time, not O(n^2).

2. Step 2: Correct the recurrence relation

   Replace O(n^2) with O(n) to reflect the linear merging cost.

3. Final Answer:

   Change O(n^2) to O(n) because merging takes linear time -> Option A

4. Quick Check:

   Merging cost = O(n), not O(n^2) [OK]

Hint: Merging in divide and conquer is linear, not quadratic [OK]

Common Mistakes:

• Assuming merging is quadratic
• Changing recursive calls incorrectly
• Ignoring the cost of merging step

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

hard

A. O(log n)

B. O(n^2)

C. O(n^{\log_3 3}) = 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_b(a)} log n) = Theta(n log n).

4. Final Answer:

   O(n log n) -> Option D

5. Quick Check:

   Master Theorem case 2 gives O(n log n) [OK]

Hint: Use Master Theorem: compare f(n) with n^{log_b(a)} [OK]

Common Mistakes:

• Ignoring the log factor in case 2
• Confusing log base or parameters a and b
• Choosing linear time incorrectly