Pick from Both Sides! A Comprehensive Guide to Solving the Interview Bit Problem in Java

Solving array problems in coding interviews can be challenging, especially when the problem demands an optimal solution in both time and space complexity. One such intriguing problem is “Pick from both sides!” available on Interview Bit. In this blog, we’ll dive deep into understanding this problem, discuss the optimal approach using Java, and demonstrate a detailed dry run to help you grasp the solution thoroughly.

Understanding the Problem Statement

The “Pick from both sides!” problem is a typical array problem that requires you to maximize the sum of a specific number of elements picked from both ends of the array. Here’s a brief summary of the problem:

• You are given an integer array A of size N and an integer B.
• You need to select exactly B elements from the array.
• You can pick elements from either the beginning or the end of the array.
• The goal is to maximize the sum of the selected elements.

Also Read: Reverse An Array | Program to Reverse an Array

Example:

If A = [5, -2, 3, 1, 2] and B = 3, you can pick elements as follows:

• Pick 5, -2, and 3 from the beginning.
• Or pick 3, 1, and 2 from the end.
• Or pick 5, 1, and 2 from both sides.

The optimal solution in this case would be to pick 5, 1, and 2, which results in the maximum sum of 8.

Approach to Solve the Problem

To solve this problem, a brute-force approach would involve checking all possible combinations of picking elements from both ends. However, this approach is inefficient, especially when the array size is large. Instead, we can utilize a more optimal solution using a sliding window technique.

Step-by-Step Breakdown of the Approach:

1. Calculate Initial Sum:
   First, calculate the sum of the first B elements from the start of the array. This sum will serve as our initial maximum sum.
2. Sliding Window:
   Then, gradually start picking elements from the end of the array and simultaneously dropping elements from the beginning. With each new combination, calculate the sum and update the maximum sum if the new sum is greater.
3. Optimization:
   The sliding window technique allows us to consider all combinations of picking B elements without explicitly generating each combination, thus optimizing both time and space complexity.

Also Read: Subarray Sum Equals K | Leet Code Subarray Sum Equals K

Java Implementation of the Optimal Solution

Let’s dive into the code to implement the approach discussed above.

public class PickFromBothSides {

    public static int maxSum(int[] A, int B) {
        int n = A.length;

        // Calculate the initial sum of the first B elements
        int currentSum = 0;
        for (int i = 0; i < B; i++) {
            currentSum += A[i];
        }

        int maxSum = currentSum;

        // Use a sliding window to consider sums by removing elements from the start
        // and adding elements from the end
        for (int i = 0; i < B; i++) {
            currentSum = currentSum - A[B - 1 - i] + A[n - 1 - i];
            maxSum = Math.max(maxSum, currentSum);
        }

        return maxSum;
    }

    public static void main(String[] args) {
        int[] A = {5, -2, 3, 1, 2};
        int B = 3;

        System.out.println("Maximum sum: " + maxSum(A, B)); // Output should be 8
    }
}

Also Read: Subarray with Given Sum | Sliding Window Technique

Dry Run of the Code

Let’s break down the execution of the code with the given example:

Input:

• A = [5, -2, 3, 1, 2]
• B = 3

Step 1: Initial Sum Calculation

• Sum of the first B elements: 5 + (-2) + 3 = 6
• Set maxSum = 6

Step 2: Sliding Window Technique

• Iteration 1:
• Remove A[2] (3) and add A[4] (2).
• New sum: 6 - 3 + 2 = 5
• maxSum remains 6 as 5 is less than 6.
• Iteration 2:
• Remove A[1] (-2) and add A[3] (1).
• New sum: 5 - (-2) + 1 = 8
• maxSum is updated to 8.
• Iteration 3:
• Remove A[0] (5) and add A[2] (3).
• New sum: 8 - 5 + 3 = 6
• maxSum remains 8.

Final Output:

• The maximum sum obtained is 8, which corresponds to picking elements [5, 1, 2].

Also Read: Noble Integer | Noble integer in an Array | InterviewBit Solution

Optimization and Complexity Analysis

This approach significantly reduces the number of calculations needed compared to a brute-force method. By limiting the sum calculation to just the necessary combinations through a sliding window, we achieve an O(B) time complexity, making it scalable even for larger arrays.

Time Complexity: O(B)
Space Complexity: O(1)

Common Mistakes and Tips

• Edge Cases: Consider edge cases where B is 0 or equal to the size of the array. Also, handle cases where all elements are negative, as the solution might involve picking no elements at all.
• Array Bounds: Ensure that your code handles array bounds correctly when picking elements from both ends.
• Negative Numbers: The presence of negative numbers can affect the optimal sum, so carefully consider their impact during the sum calculation.

Read More: Longest Palindromic Substring | LeetCode Problem Solution | Interview bit Solution

Conclusion

The “Pick from both sides!” problem on Interview Bit is a great exercise in optimizing array operations. By using a sliding window approach, we reduce the complexity and enhance the efficiency of the solution. Understanding and mastering this technique can be a significant asset during technical interviews. Ensure to practice variations of this problem to solidify your understanding and adaptability.

FAQs

What if B is greater than the array size?
If B is greater than the array size, the problem becomes invalid since we can’t pick more elements than available. This scenario should be handled as an edge case with appropriate error handling.

Can this approach be used for circular arrays?
Yes, with slight modifications, this approach can be adapted to handle circular arrays by considering the array’s wrap-around behavior.

What if all elements in the array are negative?
If all elements are negative, the optimal solution might be to pick no elements or the least negative number. The logic should be adjusted accordingly to handle such cases.

Is there a way to optimize the space complexity further?
The current solution already operates in O(1) space complexity, which is optimal. Further optimization in space complexity isn’t necessary.

Can this method be used for other similar problems?
Yes, the sliding window technique is versatile and can be applied to various problems involving subarray sums or similar constraints.