Find Maximum Subarray Divide and Conquer in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the middle index of the array segment.

DSA C#

int mid = (low + [1]) / 2;

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

Alength

Blow

Cmid

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Using low twice instead of high.

Using length instead of high.

Using mid variable before assignment.

The middle index is calculated by adding low and high indices and dividing by 2.

2fill in blank

medium

Complete the code to return the maximum of three values.

DSA C#

return Math.Max(leftMax, Math.Max(rightMax, [1]));

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

AmidMax

BcrossMax

CmaxSum

DtotalMax

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable not defined in the function.

Using maxSum which is not declared.

Using midMax which is not part of the algorithm.

The maximum subarray sum is the max of left, right, and crossing sums.

3fill in blank

hard

Fix the error in the loop that calculates the left crossing sum.

DSA C#

for (int i = mid; i >= [1]; i--) { sum += arr[i]; if (sum > leftSum) leftSum = sum; }

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

Alow

Bhigh

Cmid

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using high as lower bound in left loop.

Using mid as lower bound causing infinite loop.

Using 0 which may be out of current segment.

The loop must go from mid down to low index to cover left half.

4fill in blank

hard

Fill both blanks to calculate the right crossing sum correctly.

DSA C#

int sum = 0; int rightSum = int.MinValue; for (int j = mid + 1; j [1] [2]; j++) { sum += arr[j]; if (sum > rightSum) rightSum = sum; }

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

A<=

B<

Chigh

Dlow

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' excludes the high index.

Using low as upper bound is incorrect.

Using '>' causes loop not to run.

The loop runs from mid+1 up to and including high index. Using '<=' and 'high' is correct.

5fill in blank

hard

Fill all three blanks to complete the recursive maximum subarray function.

DSA C#

if (low == high) return arr[[1]]; int mid = (low + high) / 2; int leftMax = MaxSubArray(arr, low, [2]); int rightMax = MaxSubArray(arr, [3], high);

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

Alow

Bmid

Cmid + 1

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Using high in base case index.

Using mid+1 in left recursive call.

Using mid in right recursive call.

Base case returns arr[low]. Left recursive call is from low to mid. Right recursive call is from mid+1 to high.

Practice

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

easy

A. Calculate sum of all subarrays and pick the maximum

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

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

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

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 subproblems

   Find max subarray in left half, right half, and crossing the middle, then take the maximum of these three.

3. Final Answer:

   Split array, find max left, right, crossing, then combine -> Option D

4. Quick Check:

   Divide and conquer = split and combine max subarrays [OK]

Hint: Divide array, conquer halves, combine max sums [OK]

Common Mistakes:

• Thinking sorting helps find max subarray sum
• Using brute force sum of all subarrays
• Confusing divide and conquer with sliding window

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

easy

A. If the array length is 0, return 0

B. If the array length is 1, return the single element

C. If the array length is 2, return the sum of both elements

D. If the array length is 1, return 0

Solution

1. Step 1: Identify base case for recursion

   When the array has only one element, that element itself is the max subarray sum.

2. Step 2: Return the single element value

   Return the value of that single element as the maximum sum.

3. Final Answer:

   Return single element when array length is 1 -> Option B

4. Quick Check:

   Base case = single element value [OK]

Hint: Base case: single element returns itself [OK]

Common Mistakes:

• Returning 0 for single element array
• Returning sum of two elements as base case
• Not defining a base case causing infinite recursion

3. What will be 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 maxSum = FindMaxSubarray(arr, 0, arr.Length - 1);
Console.WriteLine(maxSum);

int FindMaxSubarray(int[] a, int left, int right) {
    if (left == right) return a[left];
    int mid = (left + right) / 2;
    int leftMax = FindMaxSubarray(a, left, mid);
    int rightMax = FindMaxSubarray(a, mid + 1, right);
    int crossMax = FindMaxCrossingSubarray(a, left, mid, right);
    return Math.Max(Math.Max(leftMax, rightMax), crossMax);
}

int FindMaxCrossingSubarray(int[] a, int left, int mid, int right) {
    int leftSum = int.MinValue, sum = 0;
    for (int i = mid; i >= left; i--) {
        sum += a[i];
        if (sum > leftSum) leftSum = sum;
    }
    int rightSum = int.MinValue;
    sum = 0;
    for (int j = mid + 1; j <= right; j++) {
        sum += a[j];
        if (sum > rightSum) rightSum = sum;
    }
    return leftSum + rightSum;
}

medium

A. 6

B. 7

C. 5

D. 4

Solution

1. Step 1: Identify the maximum subarray sum

   The array is [-2, 1, -3, 4, -1, 2, 1, -5, 4]. The subarray [4, -1, 2, 1] (indices 3-6) sums to 6, which is the maximum.

2. Step 2: Trace the code execution

   Split at mid=4: leftMax (0-4) = 4, rightMax (5-8) = 4, crossMax = 6 ([3-6]). Returns Math.Max(4, 4, 6) = 6.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   max(leftMax=4, rightMax=4, crossMax=6) = 6 [OK]

Hint: Trace subarray sums carefully to find max [OK]

Common Mistakes:

• Miscomputing crossing sum
• Confusing max subarray with max element
• Ignoring negative numbers impact

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

int FindMaxCrossingSubarray(int[] a, int left, int mid, int right) {
    int leftSum = 0, sum = 0;
    for (int i = mid; i >= left; i--) {
        sum += a[i];
        if (sum > leftSum) leftSum = sum;
    }
    int rightSum = 0;
    sum = 0;
    for (int j = mid + 1; j <= right; j++) {
        sum += a[j];
        if (sum > rightSum) rightSum = sum;
    }
    return leftSum + rightSum;
}

medium

A. leftSum and rightSum should be initialized to int.MinValue, not 0

B. The loops should run from left to mid and mid+1 to right respectively

C. sum variable should not be reset between loops

D. The function should return max of leftSum and rightSum, not their sum

Solution

1. Step 1: Check initialization of leftSum and rightSum

   They are initialized to 0, which fails if all numbers are negative because max sum could be less than 0.

2. Step 2: Correct initialization to int.MinValue

   Initialize leftSum and rightSum to int.MinValue to correctly track max sums even if negative.

3. Final Answer:

   Initialize leftSum and rightSum to int.MinValue -> Option A

4. Quick Check:

   Init max sums to int.MinValue to handle negatives [OK]

Hint: Init max sums to int.MinValue to handle negatives [OK]

Common Mistakes:

• Initializing max sums to 0 causing wrong results
• Resetting sum incorrectly between loops
• Returning sum instead of max in crossing function

5. Given an array int[] arr = { -1, -2, -3, -4 }, what will the divide and conquer algorithm return as the maximum subarray sum?

hard

A. -4

B. 0

C. -1

D. -10

Solution

1. Step 1: Analyze array with all negative numbers

   All elements are negative, so max subarray is the single largest element.

2. Step 2: Identify largest element

   The largest element is -1, which is the max subarray sum.

3. Final Answer:

   -1 -> Option C

4. Quick Check:

   Max subarray in negatives = max element [OK]

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

Common Mistakes:

• Returning 0 instead of max negative element
• Summing all negatives incorrectly
• Assuming empty subarray sum is allowed