Introduction to Dynamic Programming
Dynamic Programming is a technique used in problem-solving, where we divide a complex problem into smaller overlapping subproblems and store the solutions to those subproblems. By solving the subproblems only once and reusing their solutions, we can optimize the overall solution.
Why is Dynamic Programming important in problem-solving?
Dynamic Programming allows us to solve complex problems more efficiently by breaking them down into simpler subproblems. By storing the solutions to these subproblems, we eliminate redundant computations and improve the performance of our solutions.
For example, let's consider the problem of finding the nth Fibonacci number. The naive recursive approach would compute the same subproblem multiple times, resulting in exponential time complexity. However, by using dynamic programming and memoization, we can compute the Fibonacci numbers in linear time.
1class Main {
2 public static void main(String[] args) {
3 // replace with your Java logic here
4 System.out.println("Dynamic Programming is a technique used in problem-solving, where we divide a complex problem into smaller overlapping subproblems and store the solutions to those subproblems. By solving the subproblems only once and reusing their solutions, we can optimize the overall solution.");
5 }
6}xxxxxxxxxxclass Main { public static void main(String[] args) { // replace with your Java logic here System.out.println("Dynamic Programming is a technique used in problem-solving, where we divide a complex problem into smaller overlapping subproblems and store the solutions to those subproblems. By solving the subproblems only once and reusing their solutions, we can optimize the overall solution."); }}Are you sure you're getting this? Fill in the missing part by typing it in.
Dynamic Programming is a technique used in problem-solving, where we divide a complex problem into smaller overlapping subproblems and store the solutions to those subproblems. By solving the subproblems only once and reusing their solutions, we can optimize the overall solution.
Dynamic Programming allows us to solve __ problems more efficiently by breaking them down into simpler subproblems. By storing the solutions to these subproblems, we eliminate redundant computations and improve the performance of our solutions.
Write the missing line below.
Memoization
In the context of Dynamic Programming, memoization is a technique used to optimize the solutions to subproblems by storing their results so that we don't need to recompute them.
When solving problems using dynamic programming, we often encounter cases where the same subproblem is solved multiple times. This can result in redundant computations and can lead to inefficient solutions.
To address this issue, we can use memoization to store the results of solving subproblems and reuse these results when needed.
For example, let's consider the problem of computing the Fibonacci numbers. The Fibonacci sequence is defined as follows:
1F(0) = 0
2F(1) = 1
3F(n) = F(n-1) + F(n-2) for n > 1The naive recursive approach to compute the Fibonacci numbers has exponential time complexity since it solves the same subproblems multiple times. However, by using memoization, we can avoid redundant computations and solve the problem efficiently.
Here's an implementation of the Fibonacci function using memoization in Java:
1class Main {
2 public static void main(String[] args) {
3 // replace with your Java logic here
4 int result = fibonacci(6);
5 System.out.println("The 6th Fibonacci number is: " + result);
6 }
7
8 private static int fibonacci(int n) {
9 int[] memo = new int[n + 1];
10 return fibonacciHelper(n, memo);
11 }
12
13 private static int fibonacciHelper(int n, int[] memo) {
14 if (n <= 1) {
15 return n;
16 }
17
18 if (memo[n] != 0) {
19 return memo[n];
20 }
21
22 memo[n] = fibonacciHelper(n - 1, memo) + fibonacciHelper(n - 2, memo);
23 return memo[n];
24 }
25}In this implementation, we use an array memo to store the results of solving subproblems. Before solving a subproblem, we check if its result is already stored in memo and return it if it exists. Otherwise, we compute the result and store it in memo for future use.
By using memoization, we can significantly improve the performance of our dynamic programming solutions by avoiding redundant computations and reusing previously computed results.
xxxxxxxxxxclass Main { public static void main(String[] args) { // replace with your Java logic here int result = fibonacci(6); System.out.println("The 6th Fibonacci number is: " + result); } private static int fibonacci(int n) { int[] memo = new int[n + 1]; return fibonacciHelper(n, memo); } private static int fibonacciHelper(int n, int[] memo) { if (n <= 1) { return n; } if (memo[n] != 0) { return memo[n]; } memo[n] = fibonacciHelper(n - 1, memo) + fibonacciHelper(n - 2, memo); return memo[n]; }}Try this exercise. Fill in the missing part by typing it in.
Memoization is a technique used to optimize the solutions to subproblems by ____ their results.
Write the missing line below.
Fibonacci Numbers
As a senior engineer interested in practice data structures and algorithms, you may be familiar with the concept of the Fibonacci sequence, which is a series of numbers where each number is the sum of the two preceding ones.
In the context of dynamic programming, we can use a technique called memoization to efficiently compute Fibonacci numbers and avoid redundant computations.
Let's consider the problem of computing the n-th Fibonacci number. The naive recursive approach to solve this problem has exponential time complexity since it solves the same subproblems multiple times. However, by using memoization, we can store the results of subproblems and reuse them when needed, resulting in a more efficient solution.
Here's an implementation of the Fibonacci function using memoization in Java:
1class Main {
2 public static void main(String[] args) {
3 // replace with your Java logic here
4 int result = fibonacci(6);
5 System.out.println("The 6th Fibonacci number is: " + result);
6 }
7
8 private static int fibonacci(int n) {
9 int[] memo = new int[n + 1];
10 return fibonacciHelper(n, memo);
11 }
12
13 private static int fibonacciHelper(int n, int[] memo) {
14 if (n <= 1) {
15 return n;
16 }
17
18 if (memo[n] != 0) {
19 return memo[n];
20 }
21
22 memo[n] = fibonacciHelper(n - 1, memo) + fibonacciHelper(n - 2, memo);
23 return memo[n];
24 }
25}xxxxxxxxxxclass Main { public static void main(String[] args) { // replace with your Java logic here int result = fibonacci(6); System.out.println("The 6th Fibonacci number is: " + result); } private static int fibonacci(int n) { int[] memo = new int[n + 1]; return fibonacciHelper(n, memo); } private static int fibonacciHelper(int n, int[] memo) { if (n <= 1) { return n; } if (memo[n] != 0) { return memo[n]; } memo[n] = fibonacciHelper(n - 1, memo) + fibonacciHelper(n - 2, memo); return memo[n]; }}Are you sure you're getting this? Fill in the missing part by typing it in.
The Fibonacci sequence is a series of numbers where each number is the __ of the two preceding ones.
Write the missing line below.
Longest Common Subsequence
As a senior engineer interested in practice data structures and algorithms, you may encounter a popular problem known as the Longest Common Subsequence (LCS) problem.
The LCS problem involves finding the length of the longest common subsequence between given strings str1 and str2.
A subsequence is a sequence that can be derived from another sequence by deleting some or no elements from the original sequence, without changing the order of the remaining elements.
For example, given the strings str1 = "AGGTAB" and str2 = "GXTXAYB", the longest common subsequence is GTAB, so the length of the LCS is 4.
To solve the LCS problem, we can use dynamic programming. We can define a 2D array dp of size (m + 1) x (n + 1), where m and n are the lengths of str1 and str2 respectively.
We can then use recursive function longestCommonSubsequence to calculate the length of the LCS as follows:
1// replace with your Java logic here
2String str1 = "AGGTAB";
3String str2 = "GXTXAYB";
4int m = str1.length();
5int n = str2.length();
6int[][] dp = new int[m + 1][n + 1];
7int len = longestCommonSubsequence(str1, str2, m, n, dp);
8System.out.println("The length of the longest common subsequence is: " + len);The longestCommonSubsequence function takes the two strings str1 and str2, along with their lengths m and n respectively, and the 2D array dp as parameters. It uses memoization to avoid redundant calculations of the LCS length.
In the longestCommonSubsequence function, we check the base cases where m or n is 0, which means one of the strings is empty. In this case, the length of the LCS is 0.
If the LCS length for the current characters str1[m - 1] and str2[n - 1] has already been calculated and stored in dp[m][n], we return that value.
Otherwise, we compare the last characters str1[m - 1] and str2[n - 1]. If they are the same, we increment the LCS length by 1 and recursively call longestCommonSubsequence for the remaining characters m - 1 and n - 1.
If the last characters are different, we take the maximum of the LCS lengths obtained by removing the last character of str1 (keeping str2 intact) and removing the last character of str2 (keeping str1 intact).
Finally, we store the calculated LCS length in dp[m][n] and return that value.
By using dynamic programming and memoization, we can efficiently solve the longest common subsequence problem. The time complexity of the dynamic programming solution is O(mn), where m and n are the lengths of str1 and str2 respectively.
xxxxxxxxxx}class Main { public static void main(String[] args) { // replace with your Java logic here String str1 = "AGGTAB"; String str2 = "GXTXAYB"; int m = str1.length(); int n = str2.length(); int[][] dp = new int[m + 1][n + 1]; int len = longestCommonSubsequence(str1, str2, m, n, dp); System.out.println("The length of the longest common subsequence is: " + len); } private static int longestCommonSubsequence(String str1, String str2, int m, int n, int[][] dp) { if (m == 0 || n == 0) { return 0; } if (dp[m][n] != 0) { return dp[m][n]; } if (str1.charAt(m - 1) == str2.charAt(n - 1)) { dp[m][n] = 1 + longestCommonSubsequence(str1, str2, m - 1, n - 1, dp); } else { dp[m][n] = Math.max(longestCommonSubsequence(str1, str2, m - 1, n, dp), longestCommonSubsequence(str1, str2, m, n - 1, dp)); } return dp[m][n];Try this exercise. Fill in the missing part by typing it in.
In the longest common subsequence problem, we can use dynamic programming to calculate the ___ of the longest common subsequence between two strings.
The ___ method is defined as follows:
1int longestCommonSubsequence(String str1, String str2) {
2 int m = str1.length();
3 int n = str2.length();
4 int[][] dp = new int[m + 1][n + 1];
5
6 for (int i = 1; i <= m; i++) {
7 for (int j = 1; j <= n; j++) {
8 if (str1.charAt(i - 1) == str2.charAt(j - 1)) {
9 dp[i][j] = dp[i - 1][j - 1] + 1;
10 } else {
11 dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
12 }
13 }
14 }
15
16 return dp[m][n];
17}Write the missing line below.
Knapsack Problem
The Knapsack Problem is a classic optimization problem in computer science and mathematics. Given a set of items, each with a weight and a value, and a maximum weight capacity for a knapsack, the goal is to determine the most valuable combination of items to include in the knapsack without exceeding its capacity.
In other words, we want to maximize the total value of the items in the knapsack while keeping the total weight below the maximum capacity.
The 0/1 in the name of the problem refers to the restriction that each item can only be included once (0) or excluded (1) from the knapsack. We either take the item or we don't, we cannot take a fraction of it.
To solve the 0/1 knapsack problem, we can use dynamic programming.
Let's define a 2D array dp of size (n + 1) x (capacity + 1), where
nis the number of itemscapacityis the maximum weight capacity of the knapsack
The value of dp[i][j] represents the maximum value that can be obtained using the first i items and a knapsack of capacity j.
We can use the following recursive formula to fill the dp array:
1if (weights[i - 1] <= j) {
2 dp[i][j] = max(values[i - 1] + dp[i - 1][j - weights[i - 1]], dp[i - 1][j]);
3} else {
4 dp[i][j] = dp[i - 1][j];
5}xxxxxxxxxxclass Main { public static void main(String[] args) { // replace with your Java logic here int[] values = {60, 100, 120}; int[] weights = {10, 20, 30}; int capacity = 50; int maxProfit = knapSack(capacity, weights, values, weights.length); System.out.println("The maximum profit is: " + maxProfit); } static int knapSack(int capacity, int[] weights, int[] values, int n) { int[][] dp = new int[n + 1][capacity + 1]; for (int i = 0; i <= n; i++) { for (int j = 0; j <= capacity; j++) { if (i == 0 || j == 0) { dp[i][j] = 0; } else if (weights[i - 1] <= j) { dp[i][j] = Math.max(values[i - 1] + dp[i - 1][j - weights[i - 1]], dp[i - 1][j]); } else { dp[i][j] = dp[i - 1][j]; } } } return dp[n][capacity]; }}Are you sure you're getting this? Click the correct answer from the options.
What type of problem is the Knapsack Problem?
Click the option that best answers the question.
- Sorting problem
- Greedy problem
- Optimization problem
- Search problem
Matrix Chain Multiplication
In the context of dynamic programming, the Matrix Chain Multiplication problem involves finding the most efficient way to multiply a sequence of matrices.
Given a chain of matrices, where the i-th matrix has dimensions d[i-1] x d[i], the goal is to determine the minimum number of multiplications required to calculate the product of the entire chain.
To solve this problem, we can use a dynamic programming approach. We can define a 2D array dp of size (n x n), where n is the number of matrices in the chain.
The value of dp[i][j] represents the minimum number of multiplications required to calculate the product of matrices from the i-th matrix to the j-th matrix.
We can use the following recursive formula to fill the dp array:
1if (i == j) {
2 dp[i][j] = 0;
3} else {
4 dp[i][j] = Integer.MAX_VALUE;
5
6 for (int k = i; k < j; k++) {
7 int cost = dp[i][k] + dp[k+1][j] + d[i-1] * d[k] * d[j];
8
9 if (cost < dp[i][j]) {
10 dp[i][j] = cost;
11 }
12 }
13}xxxxxxxxxx}import java.util.Arrays;public class MatrixChainMultiplication { public static int matrixChainMultiplication(int[] dimensions) { int n = dimensions.length; int[][] dp = new int[n][n]; for (int d = 1; d < n - 1; d++) { for (int i = 1; i < n - d; i++) { int j = i + d; dp[i][j] = Integer.MAX_VALUE; for (int k = i; k < j; k++) { int cost = dp[i][k] + dp[k + 1][j] + dimensions[i - 1] * dimensions[k] * dimensions[j]; if (cost < dp[i][j]) { dp[i][j] = cost; } } } } return dp[1][n - 1]; } public static void main(String[] args) { int[] dimensions = {10, 5, 20, 2, 12}; int minCost = matrixChainMultiplication(dimensions); System.out.println("Minimum cost for matrix chain multiplication: " + minCost);Build your intuition. Fill in the missing part by typing it in.
The value of dp[i][j] represents the minimum number of ___ required to calculate the product of matrices from the i-th matrix to the j-th matrix.
Write the missing line below.
Coin Change Problem
The Coin Change Problem is a classic problem in computer science where we need to find the minimum number of coins required to make a given amount of money.
For example, suppose we have the following coins:
- 1 cent
- 2 cents
- 5 cents
And we want to make a total of 11 cents. To find the minimum number of coins required, we can use a dynamic programming approach.
We can create an array dp of size amount + 1 and initialize all the values to amount + 1. This will act as our memoization table, storing the minimum number of coins required for each amount from 0 to amount.
We then iterate through the array of coins and for each coin, iterate through all the amounts from coin to amount. For each amount i, we update dp[i] with the minimum of its current value and dp[i - coin] + 1 (where dp[i - coin] represents the minimum number of coins required to make the remaining amount after subtracting the coin).
Finally, the value at dp[amount] represents the minimum number of coins required to make the given amount. If the value is greater than the amount, it means it is not possible to make the amount using the given coins.
Here's the Java code to solve the Coin Change Problem using dynamic programming:
1import java.util.Arrays;
2
3class CoinChangeProblem {
4 public static int coinChange(int[] coins, int amount) {
5 int[] dp = new int[amount + 1];
6 Arrays.fill(dp, amount + 1);
7 dp[0] = 0;
8
9 for (int coin : coins) {
10 for (int i = coin; i <= amount; i++) {
11 dp[i] = Math.min(dp[i], dp[i - coin] + 1);
12 }
13 }
14
15 return dp[amount] > amount ? -1 : dp[amount];
16 }
17
18 public static void main(String[] args) {
19 int[] coins = {1, 2, 5};
20 int amount = 11;
21 int result = coinChange(coins, amount);
22 System.out.println("Minimum number of coins required: " + result);
23 }
24}xxxxxxxxxximport java.util.Arrays;class CoinChangeProblem { public static int coinChange(int[] coins, int amount) { int[] dp = new int[amount + 1]; Arrays.fill(dp, amount + 1); dp[0] = 0; for (int coin : coins) { for (int i = coin; i <= amount; i++) { dp[i] = Math.min(dp[i], dp[i - coin] + 1); } } return dp[amount] > amount ? -1 : dp[amount]; } public static void main(String[] args) { int[] coins = {1, 2, 5}; int amount = 11; int result = coinChange(coins, amount); System.out.println("Minimum number of coins required: " + result); }}Build your intuition. Is this statement true or false?
The Coin Change Problem can be solved using a greedy algorithm.
Press true if you believe the statement is correct, or false otherwise.
Minimum Path Sum
In dynamic programming, the Minimum Path Sum problem involves finding the minimum cost to reach the bottom-right corner of a grid from the top-left corner. Each cell in the grid represents the cost of moving to that cell. We can only move down or right.
Let's consider a simple example. Suppose we have the following grid:
1grid = [
2 [1, 3, 1],
3 [1, 5, 1],
4 [4, 2, 1]
5]To find the minimum path sum, we can use a dynamic programming approach. We can create a 2D array dp of the same size as the grid, where dp[i][j] represents the minimum cost to reach cell (i, j). We initialize dp[0][0] to the value of the top-left cell in the grid.
Then, we iterate through the grid starting from the second row, updating the values of dp[i][j] by taking the minimum between dp[i-1][j] and dp[i][j-1] and adding the cost of the current cell. This represents the minimum cost to reach the current cell from either the cell above or the cell to the left.
Finally, the value at the bottom-right corner of dp represents the minimum path sum.
Here's the Java code to solve the Minimum Path Sum problem using dynamic programming:
1import java.util.Arrays;
2
3class MinimumPathSum {
4 public static int minPathSum(int[][] grid) {
5 int m = grid.length;
6 int n = grid[0].length;
7 int[][] dp = new int[m][n];
8
9 dp[0][0] = grid[0][0];
10
11 for (int i = 1; i < m; i++) {
12 dp[i][0] = dp[i-1][0] + grid[i][0];
13 }
14
15 for (int j = 1; j < n; j++) {
16 dp[0][j] = dp[0][j-1] + grid[0][j];
17 }
18
19 for (int i = 1; i < m; i++) {
20 for (int j = 1; j < n; j++) {
21 dp[i][j] = Math.min(dp[i-1][j], dp[i][j-1]) + grid[i][j];
22 }
23 }
24
25 return dp[m-1][n-1];
26 }
27
28 public static void main(String[] args) {
29 int[][] grid = {
30 {1, 3, 1},
31 {1, 5, 1},
32 {4, 2, 1}
33 };
34
35 int minSum = minPathSum(grid);
36 System.out.println("Minimum Path Sum: " + minSum);
37 }
38}Let's test your knowledge. Is this statement true or false?
Dynamic programming involves finding the maximum cost to reach the bottom-right corner of a grid from the top-left corner.
Press true if you believe the statement is correct, or false otherwise.
Edit Distance
In dynamic programming, the Edit Distance is a measure of the similarity between two strings. It gives us the minimum number of operations (insertions, deletions, and substitutions) required to transform one string into another.
For example, let's consider two strings: "developer" and "designer". The edit distance between these two strings is 3. We can transform the string "developer" into "designer" by substituting 'v' with 's', 'e' with 'i', and 'r' with 'g'.
To calculate the edit distance between two strings, we can use a dynamic programming approach. We can create a 2D array dp of size (m+1) x (n+1), where m and n are the lengths of the two strings.
We initialize the first row of dp with values from 0 to m and the first column with values from 0 to n. This represents the edit distance between an empty string and a substring of the first string, and vice versa.
Then, we iterate through the characters of the two strings. If the characters are the same, we copy the value from dp[i-1][j-1] to dp[i][j]. Otherwise, we take the minimum of the three adjacent cells (dp[i-1][j-1], dp[i][j-1], and dp[i-1][j]) and add 1, representing the cost of the respective operation.
Finally, the value at dp[m][n] represents the edit distance between the two strings.
Here's the Java code to calculate the edit distance:
1<<code>>xxxxxxxxxx}class Main { public static void main(String[] args) { String str1 = "developer"; String str2 = "designer"; int distance = editDistance(str1, str2); System.out.println("Edit Distance between " + str1 + " and " + str2 + ": " + distance); } public static int editDistance(String word1, String word2) { int m = word1.length(); int n = word2.length(); int[][] dp = new int[m + 1][n + 1]; for (int i = 0; i <= m; i++) { dp[i][0] = i; } for (int j = 0; j <= n; j++) { dp[0][j] = j; } for (int i = 1; i <= m; i++) { for (int j = 1; j <= n; j++) { if (word1.charAt(i - 1) == word2.charAt(j - 1)) { dp[i][j] = dp[i - 1][j - 1]; } else { dp[i][j] = 1 + Math.min(dp[i - 1][j - 1], Math.min(dp[i][j - 1], dp[i - 1][j])); } }Try this exercise. Is this statement true or false?
The edit distance between two strings measures the number of operations required to transform one string into another. The operations can include insertions, deletions, and substitutions of characters.
Press true if you believe the statement is correct, or false otherwise.
Subset Sum Problem
The Subset Sum Problem is a classic problem in computer science and is often used to introduce the concept of dynamic programming. Given a set of positive integers and a target sum, the goal is to determine whether there is a subset of the given set that adds up to the target sum.
For example, consider the set of integers [3, 1, 5, 9, 12] and the target sum 8. We can form the subset [3, 5] which adds up to 8.
To solve the Subset Sum Problem using dynamic programming, we can use a 2D array dp of size (n+1) x (targetSum+1), where n is the number of elements in the set. The value dp[i][j] represents whether there is a subset of the first i elements that adds up to the sum j.
We initialize the first row of dp with false values, except for dp[0][0] which is set to true. This represents the case where the sum is 0 and no elements are selected.
Then, we iterate through the elements and the possible sums. If an element arr[i-1] is greater than the current sum j, we copy the value from the previous row dp[i-1][j] to dp[i][j]. Otherwise, we take the logical OR of two conditions:
dp[i-1][j]: The current element is not included in the subset.dp[i-1][j-arr[i-1]]: The current element is included in the subset. We subtract the current element from the sumjand check if there is a subset of the previous elements that adds up to this reduced sum.
Finally, the value at dp[n][targetSum] represents whether there is a subset of the entire set that adds up to the target sum.
Here's the Java code to solve the Subset Sum Problem:
1<<code>
2 class Main {
3 /* Returns true if there is a subset of
4 elements in arr[] that has given sum */
5 static boolean isSubsetSum(int[] arr, int n, int sum)
6 {
7 boolean[][] dp = new boolean[n + 1][sum + 1];
8
9 // If sum is 0, there is always a subset
10 // with 0 elements
11 for (int i = 0; i <= n; i++)
12 dp[i][0] = true;
13
14 // If sum is not 0 and set is empty,
15 // then answer is false
16 for (int i = 1; i <= sum; i++)
17 dp[0][i] = false;
18
19 // Fill the subset table in bottom-up manner
20 for (int i = 1; i <= n; i++) {
21 for (int j = 1; j <= sum; j++) {
22 if (arr[i - 1] > j)
23 dp[i][j] = dp[i - 1][j];
24 else
25 dp[i][j] = dp[i - 1][j] || dp[i - 1][j - arr[i - 1]];
26 }
27 }
28
29 return dp[n][sum];
30 }
31
32 public static void main(String[] args)
33 {
34 int[] arr = { 3, 1, 5, 9, 12 };
35 int sum = 8;
36 int n = arr.length;
37 if (isSubsetSum(arr, n, sum))
38 System.out.println("There is a subset of the given set with the given sum");
39 else
40 System.out.println("No subset of the given set has the given sum");
41 }
42 }Build your intuition. Is this statement true or false?
The Subset Sum Problem can be solved using recursive backtracking.
Press true if you believe the statement is correct, or false otherwise.
Generating complete for this lesson!