\n Dynamic programming is a methodology for solving complex problems by breaking them down into simpler subproblems.
\n This technique is especially useful for solving optimization problems and addresses them repeatedly using well-defined rules.
\n Generally, dynamic programming must meet two conditions:\n <\/p>\n
\n Dynamic programming is primarily used in the following cases:\n <\/p>\n
\n Now, let\u2019s 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.\n <\/p>\n
\n 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.
\n The input is in the following format:\n <\/p>\n
\n n = 4\n W = 7\n weight[] = {1, 2, 3, 2}\n value[] = {20, 5, 10, 40}\n <\/pre>\nExample Output<\/h3>\n\n Maximum Value: 60\n <\/pre>\n4. Solving Process Using Dynamic Programming<\/h2>\n
\n To solve this problem, we will apply the basic idea of dynamic programming.
\n By selecting items one by one, we will calculate the maximum value considering the current weight that can be carried in the bag.\n <\/p>\n
4.1 Initializing the DP Table<\/h3>\n
\n The DP table will be initialized as a two-dimensional array. Here, dp[i][j]<\/code> stores the maximum value for a maximum weight j<\/code> considering the first i<\/i> items.
\n The initial value will be set to 0.\n <\/p>\n4.2 State Transition<\/h3>\n
\n We will define state transitions based on whether to include an item or not.
\n For each item i<\/code>, the maximum value will be calculated under the following two conditions:\n <\/p>\n\n- When item
i<\/code> is not included: dp[i][j] = dp[i-1][j]<\/code><\/li>\n- When item
i<\/code> is included: dp[i][j] = dp[i-1][j-weight[i-1]] + value[i-1]<\/code> (if j >= weight[i-1]<\/code>)<\/li>\n<\/ul>\n4.3 Final Implementation<\/h3>\n
\n We will now implement the above logic using Kotlin.\n <\/p>\n
\n fun knapsack(n: Int, W: Int, weight: IntArray, value: IntArray): Int {\n val dp = Array(n + 1) { IntArray(W + 1) }\n\n for (i in 1..n) {\n for (w in 0..W) {\n if (weight[i - 1] <= w) {\n dp[i][w] = maxOf(dp[i - 1][w],\n dp[i - 1][w - weight[i - 1]] + value[i - 1])\n } else {\n dp[i][w] = dp[i - 1][w]\n }\n }\n }\n return dp[n][W]\n }\n <\/pre>\n5. Complexity Analysis<\/h2>\n
\n The time complexity of this algorithm is O(nW)<\/code>, and the space complexity is also O(nW)<\/code>.
\n Here, n<\/code> is the number of items, and W<\/code> is the maximum weight of the bag.
\n However, there is a way to optimize space complexity.\n <\/p>\n5.1 Space Complexity Optimization Method<\/h3>\n