kotlin coding test course, understanding dynamic programming

1. Overview of Dynamic Programming

Dynamic programming is a methodology for solving complex problems by breaking them down into simpler subproblems.
This technique is especially useful for solving optimization problems and addresses them repeatedly using well-defined rules.
Generally, dynamic programming must meet two conditions:

  • Optimal Substructure: An optimal solution to the problem must be composed of optimal solutions to its subproblems.
  • Overlapping Subproblems: There must be cases where subproblems are solved multiple times.

2. When to Use Dynamic Programming

Dynamic programming is primarily used in the following cases:

  • Calculating Fibonacci sequences
  • Longest Common Subsequence
  • Minimum Edit Distance
  • Knapsack Problem

3. Algorithm Problem: Knapsack Problem

Now, let’s take a deeper look at the knapsack problem using dynamic programming. This problem involves selecting items to achieve the maximum value within a given weight limit.

Problem Description

Given a weight limit for the bag, the problem is to find the maximum value achievable with a combination of items when their weights and values are known.
The input is in the following format:

  • n: Number of items
  • W: Maximum weight of the bag
  • weight[]: Array of weights of each item
  • value[]: Array of values of each item

Example Input

    n = 4
    W = 7
    weight[] = {1, 2, 3, 2}
    value[] = {20, 5, 10, 40}

Example Output

    Maximum Value: 60

4. Solving Process Using Dynamic Programming

To solve this problem, we will apply the basic idea of dynamic programming.
By selecting items one by one, we will calculate the maximum value considering the current weight that can be carried in the bag.

4.1 Initializing the DP Table

The DP table will be initialized as a two-dimensional array. Here, dp[i][j] stores the maximum value for a maximum weight j considering the first i items.
The initial value will be set to 0.

4.2 State Transition

We will define state transitions based on whether to include an item or not.
For each item i, the maximum value will be calculated under the following two conditions:

  • When item i is not included: dp[i][j] = dp[i-1][j]
  • When item i is included: dp[i][j] = dp[i-1][j-weight[i-1]] + value[i-1] (if j >= weight[i-1])

4.3 Final Implementation

We will now implement the above logic using Kotlin. 

    fun knapsack(n: Int, W: Int, weight: IntArray, value: IntArray): Int {
        val dp = Array(n + 1) { IntArray(W + 1) }

        for (i in 1..n) {
            for (w in 0..W) {
                if (weight[i - 1] <= w) {
                    dp[i][w] = maxOf(dp[i - 1][w],
                                    dp[i - 1][w - weight[i - 1]] + value[i - 1])
                } else {
                    dp[i][w] = dp[i - 1][w]
                }
            }
        }
        return dp[n][W]
    }

5. Complexity Analysis

The time complexity of this algorithm is O(nW), and the space complexity is also O(nW).
Here, n is the number of items, and W is the maximum weight of the bag.
However, there is a way to optimize space complexity.

5.1 Space Complexity Optimization Method

Since dp[i][j] only needs information from the previous row, it can be optimized using a one-dimensional array. 

    fun knapsackOptimized(n: Int, W: Int, weight: IntArray, value: IntArray): Int {
        val dp = IntArray(W + 1)

        for (i in 0 until n) {
            for (w in W downTo weight[i]) {
                dp[w] = maxOf(dp[w], dp[w - weight[i]] + value[i])
            }
        }
        return dp[W]
    }

6. Conclusion

Dynamic programming is a powerful tool for efficiently solving various problems.
We learned the concepts and implementation methods of dynamic programming through the knapsack problem.
I hope you continue to practice and learn through different problems in the future.

I hope this article helps you prepare for coding tests using Kotlin. Keep learning and practicing to find the optimal solution!