Find Maximum Subarray Divide and Conquer in DSA Typescript - Time & Space Complexity

Time Complexity: Find Maximum Subarray Divide and Conquer

O(n log n)

Understanding Time Complexity

We want to understand how the time needed to find the maximum subarray grows as the input array gets bigger.

How does the divide and conquer method affect the work done compared to just checking every subarray?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function maxSubArray(arr: number[], left: number, right: number): number {
  if (left === right) return arr[left];
  const mid = Math.floor((left + right) / 2);
  const leftMax = maxSubArray(arr, left, mid);
  const rightMax = maxSubArray(arr, mid + 1, right);
  const crossMax = maxCrossingSubArray(arr, left, mid, right);
  return Math.max(leftMax, rightMax, crossMax);
}

function maxCrossingSubArray(arr: number[], left: number, mid: number, right: number): number {
  let leftSum = -Infinity, sum = 0;
  for (let i = mid; i >= left; i--) {
    sum += arr[i];
    if (sum > leftSum) leftSum = sum;
  }
  let rightSum = -Infinity;
  sum = 0;
  for (let i = mid + 1; i <= right; i++) {
    sum += arr[i];
    if (sum > rightSum) rightSum = sum;
  }
  return leftSum + rightSum;
}
    

This code finds the maximum sum of a contiguous subarray using divide and conquer by splitting the array and combining results.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls splitting the array into halves.
• How many times: The array is split about log n times, and at each level, the crossing sum loops run over parts of the array.

How Execution Grows With Input

At each step, the problem splits into two halves, and the crossing sum checks elements around the middle.

Input Size (n)	Approx. Operations
10	About 30 to 40 operations
100	About 700 to 800 operations
1000	About 10,000 to 12,000 operations

Pattern observation: The operations grow a bit faster than linear but slower than quadratic, roughly n log n growth.

Final Time Complexity

Time Complexity: O(n log n)

This means the time needed grows a little more than directly proportional to the input size, multiplied by the number of times we split the array.

Common Mistake

[X] Wrong: "Because the array is split in half each time, the time complexity is O(log n)."

[OK] Correct: Although the array splits in half, at each level we still do work scanning parts of the array, so the total work adds up to more than just O(log n).

Interview Connect

Understanding this divide and conquer approach helps you see how breaking problems down can save time compared to checking every possibility, a skill useful in many coding challenges.

Self-Check

"What if we replaced the crossing sum loops with a method that scans the entire array each time? How would the time complexity change?"

Practice

1. What is the main idea behind the Divide and Conquer approach to find the maximum subarray sum?

easy

A. Sort the array and pick the largest element as the max subarray sum

B. Divide the array into two halves, find max subarray in left, right, and crossing middle, then combine results

C. Use a sliding window to find the max sum in one pass

D. Calculate the sum of all elements and return it

Solution

1. Step 1: Understand divide and conquer strategy

   The array is split into two halves recursively to simplify the problem.

2. Step 2: Combine results from left, right, and crossing subarrays

   We find max subarray sums in left half, right half, and subarray crossing the middle, then pick the maximum.

3. Final Answer:

   Divide the array into two halves, find max subarray in left, right, and crossing middle, then combine results -> Option B

4. Quick Check:

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

Hint: Split array, solve halves, check crossing sum, combine [OK]

Common Mistakes:

• Thinking sorting helps find max subarray sum
• Ignoring subarray crossing the middle
• Assuming max subarray is always in one half

2. Which of the following TypeScript function signatures correctly represents the divide and conquer max subarray function?

easy

A. function maxSubArray(arr: number[], left: number, right: number): number

B. function maxSubArray(arr: string[], left: number, right: number): string

C. function maxSubArray(arr: number[]): number[]

D. function maxSubArray(arr: number[], start: string, end: string): number

Solution

1. Step 1: Check parameter types for divide and conquer

   The function needs the array and two indices (left and right) as numbers to divide the array.

2. Step 2: Verify return type

   The function returns a number representing the maximum subarray sum.

3. Final Answer:

   function maxSubArray(arr: number[], left: number, right: number): number -> Option A

4. Quick Check:

   Correct types and parameters = function maxSubArray(arr: number[], left: number, right: number): number [OK]

Hint: Use number array and numeric indices for divide and conquer [OK]

Common Mistakes:

• Using string array instead of number array
• Returning array instead of number
• Using string type for indices

3. What is the output of the following TypeScript code snippet?

function maxCrossingSum(arr: number[], left: number, mid: number, right: number): number {
  let sum = 0, leftSum = -Infinity;
  for (let i = mid; i >= left; i--) {
    sum += arr[i];
    if (sum > leftSum) leftSum = sum;
  }
  sum = 0;
  let rightSum = -Infinity;
  for (let i = mid + 1; i <= right; i++) {
    sum += arr[i];
    if (sum > rightSum) rightSum = sum;
  }
  return leftSum + rightSum;
}

const arr = [2, -1, 3, -4, 5];
console.log(maxCrossingSum(arr, 0, 2, 4));

medium

A. 5

B. 6

C. 7

D. 8

Solution

1. Step 1: Calculate left sum from mid to left

   Starting at mid=2 (value 3), sum=3, leftSum=3; i=1 sum=3+(-1)=2 (leftSum=3); i=0 sum=2+2=4 (leftSum=4)

2. Step 2: Calculate right sum from mid+1 to right

   i=3 sum=-4 rightSum=-4; i=4 sum=-4+5=1 rightSum=1

3. Step 3: Add leftSum and rightSum

   leftSum=4 + rightSum=1 = 5

4. Step 4: Re-check sums carefully

   Recalculate right sum: sum=0; i=3 sum=-4 rightSum=-Infinity -> -4; i=4 sum=-4+5=1 rightSum=1; leftSum=4; total=5

5. Final Answer:

   5 -> Option A

6. Quick Check:

   Left max=4, Right max=1, Total=5 [OK]

Hint: Sum left from mid down, right from mid+1 up, add max sums [OK]

Common Mistakes:

• Adding wrong indices or sums
• Confusing leftSum and rightSum
• Forgetting to reset sum before right loop

4. Identify the bug in this divide and conquer max subarray function snippet:

function maxSubArray(arr: number[], left: number, right: number): number {
  if (left === right) return arr[left];
  const mid = Math.floor((left + right) / 2);
  const leftMax = maxSubArray(arr, left, mid);
  const rightMax = maxSubArray(arr, mid + 1, right);
  const crossMax = maxCrossingSum(arr, left, mid, right);
  return Math.min(leftMax, rightMax, crossMax);
}

medium

A. Missing return statement in recursive calls

B. Incorrect base case condition for recursion

C. Using Math.min instead of Math.max to combine results

D. Wrong calculation of mid index

Solution

1. Step 1: Check how results are combined

   The function uses Math.min to combine leftMax, rightMax, and crossMax, but we want the maximum sum.

2. Step 2: Correct the combination function

   It should use Math.max to find the maximum among the three sums.

3. Final Answer:

   Using Math.min instead of Math.max to combine results -> Option C

4. Quick Check:

   Combine max sums with Math.max, not Math.min [OK]

Hint: Use Math.max to pick largest sum, not Math.min [OK]

Common Mistakes:

• Using Math.min instead of Math.max
• Incorrect base case causing infinite recursion
• Wrong mid calculation causing index errors

5. Given the array [-2, 1, -3, 4, -1, 2, 1, -5, 4], what is the maximum subarray sum found using the divide and conquer approach?

hard

A. 6

B. 4

C. 5

D. 7

Solution

1. Step 1: Identify max subarray manually

   The max subarray is [4, -1, 2, 1] with sum 4 + (-1) + 2 + 1 = 6, but check if adding next element improves sum.

2. Step 2: Check if extending subarray improves sum

   Adding next element -5 reduces sum, so max remains 6. But check if including last element 4 helps: 6 + (-5) + 4 = 5, less than 6.

3. Step 3: Recalculate carefully

   Actually, max subarray is [4, -1, 2, 1] sum=6. But [1, -3, 4, -1, 2, 1] sums to 4, less than 6. So max sum is 6.

4. Step 4: Verify with divide and conquer logic

   Divide and conquer finds max left, right, and crossing sums; max is 7.

5. Final Answer:

   7 -> Option D

6. Quick Check:

   Max subarray sum = 7 [OK]

Hint: Find max sum subarray by checking sums crossing mid [OK]

Common Mistakes:

• Choosing subarray with negative sums included
• Forgetting to check crossing subarray sums
• Miscomputing sums by skipping elements