Find Maximum Subarray Divide and Conquer in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Java

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

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A-

B+

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition.

Using multiplication which is incorrect here.

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

2fill in blank

medium

Complete the code to find the maximum of three integers.

DSA Java

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

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AmidMax

BmaxMid

CmaxCross

DcrossMax

Attempts:

3 left

💡 Hint

Common Mistakes

Using variables that do not exist.

Confusing leftMax and rightMax with crossMax.

The crossMax variable holds the maximum sum crossing the middle.

3fill in blank

hard

Fix the error in the loop that calculates max crossing sum on the left side.

DSA Java

for (int i = mid; i >= [1]; i--) {

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Alow

Bmid

Chigh

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid as the lower bound causes an empty loop.

Using high causes index out of range.

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

4fill in blank

hard

Fill both blanks to correctly calculate the max crossing sum on the right side.

DSA Java

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

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

A<=

B++

C<

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Using < excludes the last element.

Using -- decrements j causing infinite loop.

The loop runs from mid+1 up to high inclusive, incrementing j each time.

5fill in blank

hard

Fill all three blanks to complete the recursive max subarray function call and return statement.

DSA Java

int leftMax = findMaxSubarray(arr, low, [1]);
int rightMax = findMaxSubarray(arr, [2], high);
int crossMax = maxCrossingSum(arr, low, mid, high);
return [3](leftMax, Math.max(rightMax, crossMax));

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

Amid

Bmid + 1

CMath.max

DMath.min

Attempts:

3 left

💡 Hint

Common Mistakes

Using Math.min instead of Math.max.

Incorrect indices causing stack overflow.

The left half ends at mid, right half starts at mid+1, and return the max of three sums.

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 max sum subarray in one pass

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

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

D. Calculate all possible subarrays and pick the one with the maximum sum

Solution

1. Step 1: Understand divide and conquer strategy

   The array is divided 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.

3. Final Answer:

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

4. Quick Check:

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

Hint: Divide array, solve halves, combine crossing max [OK]

Common Mistakes:

• Thinking sorting helps find max subarray sum
• Confusing divide and conquer with sliding window
• Ignoring subarrays crossing the middle

2. Which of the following is the correct base case for the recursive function in the divide and conquer approach to find the maximum subarray?

easy

A. When the array is empty, return 0

B. When the array has only one element, return that element

C. When the array has two elements, return their sum

D. When the array has more than one element, return the max of left and right

Solution

1. Step 1: Identify base case for recursion

   The recursion stops when the subarray has only one element.

2. Step 2: Return the single element as max subarray sum

   Since one element is the smallest subarray, return it directly.

3. Final Answer:

   Return single element when array size is one -> Option B

4. Quick Check:

   Base case = single element return [OK]

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

Common Mistakes:

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

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

int[] arr = {-2, 1, -3, 4, -1, 2, 1, -5, 4};
System.out.println(maxSubArray(arr, 0, arr.length - 1));

// Assume maxSubArray is correctly implemented divide and conquer method

medium

A. 4

B. 7

C. 5

D. 6

Solution

1. Step 1: Identify max subarray in given array

   The max subarray is [4, -1, 2, 1] with sum 4 + (-1) + 2 + 1 = 6.

2. Step 2: Check if any larger sum exists

   Adding the next element -5 or 4 does not increase sum beyond 6, but including the last 4 gives total 5 for subarray [4, -1, 2, 1, -5, 4]?

3. Step 3: Correct max subarray is [4, -1, 2, 1] with sum 6

   Actually, the max sum is 6, not 5.

4. Final Answer:

   6 -> Option D

5. Quick Check:

   Max subarray sum = 6 [OK]

Hint: Find max sum subarray by checking sums around middle [OK]

Common Mistakes:

• Choosing sum including negative elements that reduce total
• Miscounting elements in max subarray
• Confusing max sum with max element

4. Identify the error in this divide and conquer code snippet for max subarray sum:

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

medium

A. No error, code correctly finds max crossing sum

B. Should initialize sum to Integer.MIN_VALUE instead of 0

C. Return statement should be max of left_sum and right_sum, not sum

D. Loop ranges are incorrect; should start from left+1 and mid-1

Solution

1. Step 1: Analyze loops for left and right sums

   Loops correctly accumulate sums from mid to left and mid+1 to right.

2. Step 2: Check return statement

   Returns sum of max left and right sums crossing mid, which is correct for crossing subarray.

3. Final Answer:

   No error, code correctly finds max crossing sum -> Option A

4. Quick Check:

   Crossing sum = left max + right max [OK]

Hint: Crossing sum = max left + max right sums around mid [OK]

Common Mistakes:

• Initializing sum to wrong value causing incorrect sums
• Returning max instead of sum of left and right max
• Incorrect loop boundaries skipping elements

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

hard

A. -1

B. -10

C. 0

D. -4

Solution

1. Step 1: Understand max subarray with all negative numbers

   When all numbers are negative, max subarray is the single largest (least negative) element.

2. Step 2: Identify largest element in array

   Largest element is -1, so max subarray sum is -1.

3. Final Answer:

   -1 -> Option A

4. Quick Check:

   Max subarray sum with negatives = max element [OK]

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

Common Mistakes:

• Summing all negatives instead of picking max element
• Returning zero incorrectly assuming empty subarray
• Confusing max subarray sum with min element