Find Maximum Subarray Divide and Conquer in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Maximum Subarray Sum Using Divide and Conquer

What is the output of the following Java code that finds the maximum subarray sum using divide and conquer?

DSA Java

public class MaxSubarray {
    public static int maxCrossingSum(int[] arr, int left, int mid, int right) {
        int sum = 0;
        int 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;
    }

    public static int maxSubArraySum(int[] arr, int left, int right) {
        if (left == right) return arr[left];
        int mid = (left + right) / 2;
        int left_max = maxSubArraySum(arr, left, mid);
        int right_max = maxSubArraySum(arr, mid + 1, right);
        int cross_max = maxCrossingSum(arr, left, mid, right);
        return Math.max(Math.max(left_max, right_max), cross_max);
    }

    public static void main(String[] args) {
        int[] arr = {-2, 1, -3, 4, -1, 2, 1, -5, 4};
        System.out.println(maxSubArraySum(arr, 0, arr.length - 1));
    }
}

A 6

B 7

C 5

D 4

💡 Hint

Think about the subarray with the largest sum in the array.

The maximum subarray is [4, -1, 2, 1] which sums to 6. The divide and conquer approach correctly finds this sum.

🧠 Conceptual

intermediate

Key Idea Behind Divide and Conquer for Maximum Subarray

Which statement best describes the key idea behind the divide and conquer approach to find the maximum subarray?

A Divide the array into two halves, find max subarray in left, right, and crossing middle, then take the maximum.

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

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

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

💡 Hint

Divide and conquer splits the problem into smaller parts and combines results.

The divide and conquer method splits the array, solves for left and right halves, and also considers subarrays crossing the middle, then returns the maximum of these.

🔧 Debug

advanced

Identify the Bug in maxCrossingSum Implementation

What is the bug in the following maxCrossingSum method that causes incorrect maximum crossing sum calculation?

DSA Java

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

A The sum variable should not be reset between the two loops.

B The loops should run from left to mid and mid+1 to right instead of backwards.

C The method should return the maximum of left_sum and right_sum, not their sum.

D left_sum and right_sum should be initialized to Integer.MIN_VALUE instead of 0.

💡 Hint

Consider arrays with all negative numbers.

Initializing left_sum and right_sum to 0 causes incorrect results when all numbers are negative because 0 is greater than any negative sum. Using Integer.MIN_VALUE ensures correct max sums.

❓ Predict Output

advanced

Output of Maximum Subarray Sum for All Negative Array

What is the output of the following code when the input array contains all negative numbers?

DSA Java

public class MaxSubarray {
    public static int maxCrossingSum(int[] arr, int left, int mid, int right) {
        int sum = 0;
        int 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;
    }

    public static int maxSubArraySum(int[] arr, int left, int right) {
        if (left == right) return arr[left];
        int mid = (left + right) / 2;
        int left_max = maxSubArraySum(arr, left, mid);
        int right_max = maxSubArraySum(arr, mid + 1, right);
        int cross_max = maxCrossingSum(arr, left, mid, right);
        return Math.max(Math.max(left_max, right_max), cross_max);
    }

    public static void main(String[] args) {
        int[] arr = {-8, -3, -6, -2, -5, -4};
        System.out.println(maxSubArraySum(arr, 0, arr.length - 1));
    }
}

A 0

B -8

C -2

D -28

💡 Hint

The maximum subarray in all negative numbers is the least negative number.

The maximum subarray is the single element -2, which is the largest (least negative) number in the array.

🚀 Application

expert

Time Complexity of Divide and Conquer Maximum Subarray Algorithm

What is the time complexity of the divide and conquer algorithm for finding the maximum subarray sum for an array of size n?

A O(n^2)

B O(n log n)

C O(n)

D O(log n)

💡 Hint

Consider the recurrence relation T(n) = 2T(n/2) + O(n).

The algorithm divides the array into two halves recursively (2T(n/2)) and combines results in linear time O(n), resulting in O(n log n) time complexity.

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