Find Maximum Subarray Divide and Conquer in DSA C - Execution Trace

Concept Flow - Find Maximum Subarray Divide and Conquer

Start with full array

↓

Divide array into two halves

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Find max subarray in left half

↓

Find max crossing subarray

↓

Compare left, right, crossing max

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Return max subarray

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Done

Divide the array into halves, find max subarray in left, right, and crossing middle, then return the max of these three.

Execution Sample

DSA C

int maxCrossingSum(int arr[], int left, int mid, int right) {
  int sum = 0, left_sum = INT_MIN;
  for (int i = mid; i >= left; i--) {
    sum += arr[i];
    if (sum > left_sum) left_sum = sum;
  }
  sum = 0; int right_sum = INT_MIN;
  for (int i = mid + 1; i <= right; i++) {
    sum += arr[i];
    if (sum > right_sum) right_sum = sum;
  }
  return left_sum + right_sum;
}

This function finds the maximum subarray sum that crosses the middle point.

Execution Table

Step	Operation	Subarray Range	Max Left Sum	Max Right Sum	Cross Sum	Max Subarray Sum	Visual State
1	Divide full array	[0..7]	-	-	-	-	[2, -1, 3, 4, -2, 1, 5, -3]
2	Divide left half	[0..3]	-	-	-	-	[2, -1, 3, 4]
3	Divide left-left half	[0..1]	-	-	-	-	[2, -1]
4	Base case single element	[0..0]	2	-	-	2	[2]
5	Base case single element	[1..1]	-1	-	-	-1	[-1]
6	Max crossing sum [0..1]	[0..1]	2	-1	1	2	[2, -1]
7	Max left half max	[0..1]	2	-1	1	2	[2, -1]
8	Divide left-right half	[2..3]	-	-	-	-	[3, 4]
9	Base case single element	[2..2]	3	-	-	3	[3]
10	Base case single element	[3..3]	4	-	-	4	[4]
11	Max crossing sum [2..3]	[2..3]	3	4	7	7	[3, 4]
12	Max left half max	[0..3]	2	7	8	8	[2, -1, 3, 4]
13	Divide right half	[4..7]	-	-	-	-	[-2, 1, 5, -3]
14	Divide right-left half	[4..5]	-	-	-	-	[-2, 1]
15	Base case single element	[4..4]	-2	-	-	-2	[-2]
16	Base case single element	[5..5]	1	-	-	1	[1]
17	Max crossing sum [4..5]	[4..5]	-2	1	-1	1	[-2, 1]
18	Divide right-right half	[6..7]	-	-	-	-	[5, -3]
19	Base case single element	[6..6]	5	-	-	5	[5]
20	Base case single element	[7..7]	-3	-	-	-3	[-3]
21	Max crossing sum [6..7]	[6..7]	5	-3	2	5	[5, -3]
22	Max right half max	[4..7]	1	5	6	6	[-2, 1, 5, -3]
23	Max crossing sum full [0..7]	[0..7]	8	6	12	12	[2, -1, 3, 4, -2, 1, 5, -3]
24	Final max subarray sum	[0..7]	8	6	12	12	[2, -1, 3, 4, -2, 1, 5, -3]

💡 All subproblems solved, final max subarray sum is 12 for subarray [2, -1, 3, 4, -2, 1, 5]

Variable Tracker

Variable	Start	After Step 6	After Step 11	After Step 12	After Step 17	After Step 21	After Step 22	After Step 23	Final
maxLeftSum	-	2	7	8	1	5	6	8	8
maxRightSum	-	-1	4	7	1	-3	5	6	6
crossSum	-	1	7	8	-1	2	6	12	12
maxSubarraySum	-	2	7	8	1	5	6	12	12

Key Moments - 3 Insights

Why do we calculate max crossing sum separately?

Because the maximum subarray might cross the middle point, so we must check the sum crossing left and right halves (see steps 6, 11, 17, 21, 23 in execution_table).

Why do we return the maximum of left, right, and crossing sums?

Because the max subarray can be fully in the left half, fully in the right half, or crossing the middle. We compare all three to find the overall max (see step 23).

What happens at base cases with single elements?

At base cases (steps 4,5,9,10,15,16,19,20), the max subarray is the element itself, so we return that element's value.

Visual Quiz - 3 Questions

Look at the execution_table at step 23, what is the max crossing sum for the full array?

A 6

B 12

C 8

D 7

💡 Hint

Check the 'Cross Sum' column at step 23 in execution_table.

At which step does the max subarray sum first become 7?

A Step 6

B Step 17

C Step 11

D Step 21

💡 Hint

Look at the 'Max Subarray Sum' column in execution_table and find when it first reaches 7.

If the array had all negative numbers, how would the base case steps change?

A Base cases would return the least negative element as max subarray.

B Base cases would return zero.

C Base cases would return the sum of all elements.

D Base cases would return positive infinity.

💡 Hint

Recall base cases return the single element value, so max subarray is the largest single element (see steps 4,5,9,10,15,16,19,20).

Concept Snapshot

Divide and Conquer approach to find max subarray:
- Divide array into halves
- Recursively find max subarray in left and right halves
- Find max crossing subarray crossing middle
- Return max of left, right, crossing sums
- Base case: single element subarray
- Time complexity: O(n log n)

Full Transcript

This visualization shows the divide and conquer method to find the maximum subarray sum. The array is split into halves recursively until single elements are reached. At each step, the maximum subarray sum is found for the left half, right half, and the subarray crossing the middle. The maximum of these three is returned. The execution table tracks these steps with subarray ranges and sums. The variable tracker shows how max sums change. Key moments clarify why crossing sums are needed and how base cases work. The quiz tests understanding of these steps and results.

Practice

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

easy

A. Use a sliding window to find the maximum sum subarray in one pass

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

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

D. Calculate sum of all subarrays and pick the maximum

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 subarrays crossing the middle, then take the maximum of these.

3. Final Answer:

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

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 subarrays crossing the middle
• Trying to sum all subarrays directly

2. Which of the following is the correct base case condition in the divide and conquer function for maximum subarray?

easy

A. if (low == high) return 0;

B. if (low > high) return 0;

C. if (low < high) return arr[low] + arr[high];

D. if (low == high) return arr[low];

Solution

1. Step 1: Identify base case meaning

   When the subarray has only one element (low equals high), the max subarray sum is that element itself.

2. Step 2: Match base case condition

   The condition if (low == high) return arr[low]; correctly returns the single element's value.

3. Final Answer:

   if (low == high) return arr[low]; -> Option D

4. Quick Check:

   Single element subarray returns element value [OK]

Hint: Base case: one element means return that element [OK]

Common Mistakes:

• Returning 0 when low == high
• Using incorrect comparison operators
• Returning sum of two elements instead of one

3. What is the output of the following C code snippet using divide and conquer to find max subarray sum?

int arr[] = {-2, 1, -3, 4, -1, 2, 1, -5, 4};
int result = maxSubArray(arr, 0, 8);
printf("%d", result);

Assume maxSubArray is correctly implemented divide and conquer function.

medium

A. 6

B. 7

C. 5

D. 4

Solution

1. Step 1: Identify max subarray in given array

   The maximum subarray is [4, -1, 2, 1] with sum 4 + (-1) + 2 + 1 = 6, but check if including more elements helps.

2. Step 2: Check if extending subarray increases sum

   Including the element before 4 is -3 (reduces sum), after 1 is -5 (reduces sum), so max sum is 6.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Max subarray sum = 6 [OK]

Hint: Find max sum of continuous elements [OK]

Common Mistakes:

• Counting non-continuous elements
• Miscomputing sums with negative numbers
• Assuming max element alone is answer

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

int maxCrossingSum(int arr[], int low, int mid, int high) {
    int left_sum = 0, sum = 0;
    for (int i = mid; i >= low; i--) {
        sum += arr[i];
        if (sum > left_sum)
            left_sum = sum;
    }
    int right_sum = 0;
    sum = 0;
    for (int i = mid + 1; i <= high; i++) {
        sum += arr[i];
        if (sum > right_sum)
            right_sum = sum;
    }
    return left_sum + right_sum;
}

medium

A. left_sum and right_sum should be initialized to INT_MIN, not 0

B. The loops should start from low and high, not mid

C. sum variable should not be reset between loops

D. Return statement should subtract left_sum from right_sum

Solution

1. Step 1: Understand initialization of sums

   Initializing left_sum and right_sum to 0 causes errors if all numbers are negative, as max sum could be less than 0.

2. Step 2: Correct initialization

   They should be initialized to the smallest integer (INT_MIN) to correctly track max sums even if negative.

3. Final Answer:

   left_sum and right_sum should be initialized to INT_MIN, not 0 -> Option A

4. Quick Check:

   Initialize max sums to INT_MIN to handle negatives [OK]

Hint: Use INT_MIN for max sums to handle negative arrays [OK]

Common Mistakes:

• Initializing max sums to zero causes wrong results
• Not resetting sum variable properly
• Incorrect loop ranges

5. Given an array arr = [-1, -2, -3, -4], what will the divide and conquer max subarray function return and why?

hard

A. 0 because empty subarray sum is considered

B. -1 because the max subarray is the single largest element

C. -10 because sum of all elements is max

D. -4 because last element is max

Solution

1. Step 1: Analyze array with all negative numbers

   All elements are negative, so the max subarray is the one with the least negative value (largest number).

2. Step 2: Identify max element

   The largest element is -1, so max subarray sum is -1 (single element subarray).

3. Final Answer:

   -1 because the max subarray is the single largest element -> Option B

4. Quick Check:

   Max subarray in negatives = max single element [OK]

Hint: Max subarray in all negatives is max single element [OK]

Common Mistakes:

• Assuming sum of all negatives is max
• Considering empty subarray sum as zero incorrectly
• Picking last element without comparison