Find Maximum Subarray Divide and Conquer in DSA Java - Build from Scratch

Find Maximum Subarray using Divide and Conquer

📖 Scenario: You are analyzing daily temperature changes to find the longest warm spell. The temperature changes are stored in an array of integers, where positive numbers mean warmer days and negative numbers mean colder days. You want to find the contiguous period with the highest total temperature increase.

🎯 Goal: Build a Java program that uses the divide and conquer method to find the maximum sum of a contiguous subarray within the given array of temperature changes.

📋 What You'll Learn

Create an integer array called tempChanges with the exact values: {-2, 1, -3, 4, -1, 2, 1, -5, 4}

Create an integer variable called n that stores the length of tempChanges

Implement a method called maxCrossingSum that finds the maximum sum of a subarray crossing the middle point

Implement a method called maxSubArraySum that uses divide and conquer to find the maximum subarray sum

Print the maximum subarray sum found by maxSubArraySum

💡 Why This Matters

🌍 Real World

Finding maximum subarrays is useful in financial analysis to find the best time to buy and sell stocks, or in signal processing to detect strong signals.

💼 Career

Understanding divide and conquer algorithms and recursion is essential for software engineering roles that involve algorithm design and optimization.

1

Create the temperature changes array

Create an integer array called tempChanges with these exact values: {-2, 1, -3, 4, -1, 2, 1, -5, 4}

DSA Java

# Create the integer array tempChanges with the given values
# Your code here

Hint

Use int[] tempChanges = { ... }; with the exact numbers inside the braces.

2

Create the length variable

Create an integer variable called n that stores the length of the tempChanges array

DSA Java

public class MaxSubarray {
    public static void main(String[] args) {
        int[] tempChanges = {-2, 1, -3, 4, -1, 2, 1, -5, 4};
        // Create an integer variable n to store the length of tempChanges
        // Your code here
    }
}

Hint

Use int n = tempChanges.length; to get the array size.

3

Implement the divide and conquer methods

Implement the methods maxCrossingSum and maxSubArraySum inside the MaxSubarray class. maxCrossingSum takes parameters int[] arr, int left, int mid, int right and returns the maximum sum of a subarray crossing mid. maxSubArraySum takes parameters int[] arr, int left, int right and returns the maximum subarray sum using divide and conquer.

DSA Java

public class MaxSubarray {
    public static void main(String[] args) {
        int[] tempChanges = {-2, 1, -3, 4, -1, 2, 1, -5, 4};
        int n = tempChanges.length;
        // Implement maxCrossingSum and maxSubArraySum methods below
        // Your code here
    }
}

Hint

Use two loops in maxCrossingSum to find max sums on left and right of mid. Use recursion in maxSubArraySum to divide the array.

4

Print the maximum subarray sum

Add a print statement inside the main method to display the maximum subarray sum by calling maxSubArraySum(tempChanges, 0, n - 1)

DSA Java

public class MaxSubarray {
    public static void main(String[] args) {
        int[] tempChanges = {-2, 1, -3, 4, -1, 2, 1, -5, 4};
        int n = tempChanges.length;
        // Print the maximum subarray sum using maxSubArraySum
        // Your code here
    }

    static int maxCrossingSum(int[] arr, int left, int mid, int right) {
        int sum = 0;
        int leftSum = Integer.MIN_VALUE;
        for (int i = mid; i >= left; i--) {
            sum += arr[i];
            if (sum > leftSum) {
                leftSum = sum;
            }
        }

        sum = 0;
        int rightSum = Integer.MIN_VALUE;
        for (int i = mid + 1; i  rightSum) {
                rightSum = sum;
            }
        }

        return leftSum + rightSum;
    }

    static int maxSubArraySum(int[] arr, int left, int right) {
        if (left == right) {
            return arr[left];
        }

        int mid = (left + right) / 2;

        int leftMax = maxSubArraySum(arr, left, mid);
        int rightMax = maxSubArraySum(arr, mid + 1, right);
        int crossMax = maxCrossingSum(arr, left, mid, right);

        return Math.max(Math.max(leftMax, rightMax), crossMax);
    }
}

Hint

Use System.out.println(maxSubArraySum(tempChanges, 0, n - 1)); to print the result.

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