1. What is Dynamic Programming?
Dynamic Programming (DP) is an algorithmic technique that involves breaking a problem down into smaller subproblems and solving them. It typically uses a recursive approach but is characterized by storing the results of subproblems to reduce redundant calculations.
DP is mainly effective for optimization problems and is divided into two main concepts: ‘memoization’ and ‘tabulation’.
2. Problem Description
2.1 Problem: Longest Common Subsequence (LCS)
Given two strings A and B, the problem is to find the length of the longest common subsequence between A and B. A subsequence is a set of characters that can be selected from A and B while maintaining their order but possibly leaving out some characters.
2.2 Example Input
A: "ABCBDAB"
B: "BDCAB"
2.3 Example Output
"The length of the LCS is 4." // "BCAB" or "BDAB"
3. Problem Solving Process
3.1 Problem Analysis
To solve the problem, we use the following approach:
- Define the lengths of strings A and B as n and m, respectively.
- Create a 2D array dp to store the lengths of LCS up to the i-th character of A and the j-th character of B such that dp[i][j] holds this value.
- If the characters match, set dp[i][j] = dp[i-1][j-1] + 1; otherwise, set dp[i][j] = max(dp[i-1][j], dp[i][j-1]).
3.2 State Transition Relation
The state transition in dynamic programming can be defined as follows:
dp[i][j] =
{
dp[i-1][j-1] + 1, if A[i-1] == B[j-1]
max(dp[i-1][j], dp[i][j-1]), if A[i-1] != B[j-1]
}
3.3 Java Implementation Code
Now, let’s implement LCS in Java.
public class LCS {
public static int longestCommonSubsequence(String A, String B) {
int n = A.length();
int m = B.length();
int[][] dp = new int[n + 1][m + 1];
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (A.charAt(i - 1) == B.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1] + 1;
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return dp[n][m];
}
public static void main(String[] args) {
String A = "ABCBDAB";
String B = "BDCAB";
int length = longestCommonSubsequence(A, B);
System.out.println("The length of the LCS is " + length + ".");
}
}
4. Overall Time Complexity and Space Complexity
The time complexity of this algorithm is O(n * m), and the space complexity is O(n * m). However, it can be reduced to O(min(n, m)) through space optimization.
5. Additional Solutions and Examples
We will look at several other problems or variations that can be solved by extending this problem. For example, after finding the longest common subsequence of the given strings, we can also add a function to print that subsequence.
6. Final Summary
Dynamic programming is a very useful technique for breaking down complex problems into subproblems. If you have understood the basic concept of DP through the LCS problem, it is recommended to practice by solving your own problems.
7. Next Course Announcement
In the next course, we will cover other examples of dynamic programming and learn about various optimization techniques. If you are interested, we encourage your participation!