How to Solve Array Manipulation Problems Using the Difference Array Technique

How to Solve Array Manipulation Problems Using the Difference Array Technique

Array manipulation problems are common in technical interviews, and one popular example is finding the maximum value in an array after applying multiple range updates efficiently.

As a seasoned software engineer specialising in algorithms and data structures, I understand how overwhelming these problems can feel at first.

This article simplifies the solution step-by-step so that even beginners can grasp it.

We'll explore the difference array technique, an efficient way to handle range updates in O(n+q) time. Where n is the size of the array, and q is the number of operations.

By the end of this article, you'll not only understand the approach but also be able to confidently implement it in any coding environment.

What Is the Problem We're Solving?

Imagine you have an array of zeros, and someone asks you to:

• Add a number k to every element between two indices, a and b (inclusive).
• Do this multiple times with different values of a, b, and k.
• After all operations, find the largest number in the array.

The challenge is that the array could be huge, and directly updating every element in the range for each operation would be slow. 

For example:

You have an array of size n=5: [0, 0, 0, 0, 0]

You perform these operations:

• Add 100 to indices 1–2.
• Add 100 to indices 2–5.
• Add 100 to indices 3–4.

After performing these operations, you'll find that the largest value in the array is 200.

Queries are interpreted as follows:
   

 a b k
 1 5 3
 4 8 7
 6 9 1

Add the values between the indices and inclusive:
 

index->  1 2 3  4  5 6 7 8 9 10

        [0,0,0, 0, 0,0,0,0,0, 0]
        [3,3,3, 3, 3,0,0,0,0, 0]
        [3,3,3,10,10,7,7,7,0, 0]
        [3,3,3,10,10,8,8,8,1, 0]

The largest value is after all operations are performed.

# Sample Input

5 3
1 2 100
2 5 100
3 4 100

# Sample Output

200

Explanation

• After the first update, the list is 100 100 0 0 0.
• After the second update list is 100 200 100 100 100.
• After the third update list is 100 200 200 200 100.
• The maximum value is 200.

The Difference Array

We don't update every element in the range [a,b].

Alternatively:

• Mark the start of the range with k and the end of the range with −k.
• Then, calculate the actual values of the array using a prefix sum.
• This way, you only modify two elements for each operation, no matter how large the range is.

Step-by-Step Explanation

Let's break the problem into three parts:

Set Up the Difference Array

• Create an array of zeros with one extra element to handle boundaries.
• For each operation, update the start and end of the range.

Compute the Prefix Sum

• Traverse the difference array and calculate the actual values of the original array using a running sum.

Find the Maximum

• As you compute the prefix sum, track the largest value.

Code Implementation

Here's how the difference array solution looks in Python:
 

def arrayManipulation(n, queries):

    # Step 1: Initialize the difference array with zeros
    diff_array = [0] * (n + 1)

    # Step 2: Apply each query as a range update
    for a, b, k in queries:
        diff_array[a] += k  # Start the range with k
        if b + 1 <= n:
            diff_array[b + 1] -= k  # End the range with -k

    # Step 3: Compute the prefix sum and track the maximum
    max_value = 0
    current_value = 0
    for value in diff_array:
        current_value += value
        max_value = max(max_value, current_value)

    return max_value

# Testing

if __name__ == '__main__':
     fptr = open(os.environ['OUTPUT_PATH'], 'w')

     first_multiple_input = input().rstrip().split()

     n = int(first_multiple_input[0])

     m = int(first_multiple_input[1])

     queries = []

     for _ in range(m):
          queries.append(list(map(int, input().rstrip().split())))

     result = arrayManipulation(n, queries)

     fptr.write(str(result) + '\n')

     fptr.close()

Let's apply this code step by step for clarity.

Input: n = 5
queries = [
    [1, 2, 100],  # Add 100 to indices 1–2
    [2, 5, 100],  # Add 100 to indices 2–5
    [3, 4, 100]   # Add 100 to indices 3–4
]

Step 1: Initialize the Difference Array

diff_array = [0, 0, 0, 0, 0, 0]  # One extra space for boundaries

Step 2: Apply Queries

# Query [1, 2, 100]
diff_array = [0, 100, 0, -100, 0, 0]

# Query [2, 5, 100]
diff_array = [0, 100, 100, -100, 0, 0, -100]

# Query [3, 4, 100]
diff_array = [0, 100, 100, 0, 0, -100, -100]

Step 3: Compute the Prefix Sum

• As you calculate the prefix sum, track the maximum value:
• Prefix sum: [0, 100, 200, 200, 200, 100]
• Maximum value: 200
• Output: The maximum value is 200.

What's Next for You?

Congratulations! You've just learned how to solve a complex array manipulation problem efficiently.

The difference array technique is a powerful tool, especially for range update problems in competitive programming.

Try implementing this in other languages or applying it to similar problems. Practice will make this approach second nature.

Still have questions? Dive deeper into array algorithms or explore other optimization techniques to build your problem-solving toolkit!