2-D Dynamic Programming Data-Structures and Algorithms

01. Unique Paths

There is a robot on an m x n grid. The robot is initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

Given the two integers m and n, return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The test cases are generated so that the answer will be less than or equal to 2 * 109.

Example 1:

Input: m = 3, n = 7
Output: 28

Example 2:

Input: m = 3, n = 2
Output: 3
Explanation: From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Down -> Down
2. Down -> Down -> Right
3. Down -> Right -> Down

Explanation

To solve this problem, I've used dynamic programming (DP). The idea is to break down the problem into smaller subproblems and solve them. We create a DP array where each element dp[i][j] represents the number of unique paths to reach the cell (i, j) from the top-left corner.

You're right. The initialization should indeed refer to the last row, given that the solution is implemented in a bottom-up way. Let's correct that and provide the detailed explanation accordingly.

Detailed Steps

Initialization:
- Initialize a 1D DP array row of size n with all elements set to 1. This array represents the number of ways to reach each cell in the last row, which is always 1 because there is only one way to move right to reach any cell in the last row.
Filling the DP Table:
- Iterate over the rows of the grid starting from the second last row up to the first row.
- For each row, create a new DP array newRow of size n initialized to 1.
- Iterate over the columns of the current row from right to left. For each cell (i, j), update newRow[j] to be the sum of newRow[j+1] (number of ways to reach the cell to the right) and row[j] (number of ways to reach the cell below).
- Update row to be newRow after processing each row.
Result:
- The answer is found in row[0], which represents the number of unique paths to reach the bottom-right corner from the top-left corner.

Example Walkthrough

Let's walk through the example m = 3, n = 7:

Initialization:
- row = [1, 1, 1, 1, 1, 1, 1]
Iteration:
- For the first row iteration (i = 1):
  - newRow = [1, 1, 1, 1, 1, 1, 1]
  - For j = 5 to 0:
    - newRow[5] = newRow[6] + row[5] = 1 + 1 = 2
    - newRow[4] = newRow[5] + row[4] = 2 + 1 = 3
    - newRow[3] = newRow[4] + row[3] = 3 + 1 = 4
    - newRow[2] = newRow[3] + row[2] = 4 + 1 = 5
    - newRow[1] = newRow[2] + row[1] = 5 + 1 = 6
    - newRow[0] = newRow[1] + row[0] = 6 + 1 = 7
  - Update row to newRow
- For the second row iteration (i = 0):
  - newRow = [1, 1, 1, 1, 1, 1, 1]
  - For j = 5 to 0:
    - newRow[5] = newRow[6] + row[5] = 1 + 2 = 3
    - newRow[4] = newRow[5] + row[4] = 3 + 3 = 6
    - newRow[3] = newRow[4] + row[3] = 6 + 4 = 10
    - newRow[2] = newRow[3] + row[2] = 10 + 5 = 15
    - newRow[1] = newRow[2] + row[1] = 15 + 6 = 21
    - newRow[0] = newRow[1] + row[0] = 21 + 7 = 28
  - Update row to newRow
Result:
- The final row is [28, 21, 15, 10, 6, 3, 1]
- row[0] is 28, which is the number of unique paths from the top-left to the bottom-right corner.

Efficiency Analysis

Time Complexity: $O(m \cdot n)$, where m is the number of rows and n is the number of columns. We iterate through each cell of the grid once.
Space Complexity: $O(n)$, where n is the number of columns. We use a 1D DP array of size n to store the number of unique paths.

02. Longest Common Subsequence

Leetcode Problem URL

Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0.

A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

For example, "ace" is a subsequence of "abcde".

A common subsequence of two strings is a subsequence that is common to both strings.

Example 1:

   Example matrix
   [[0, 0, 0, 0],
   [0, 0, 0, 0],
   [0, 0, 0, 0],
   [0, 0, 0, 0],
   [0, 0, 0, 0],
   [0, 0, 0, 0]]

Input: text1 = "abcde", text2 = "ace"
Output: 3
Explanation: The longest common subsequence is "ace" and its length is 3.

Example 2:

Input: text1 = "abc", text2 = "abc"
Output: 3
Explanation: The longest common subsequence is "abc" and its length is 3.

Example 3:

Input: text1 = "abc", text2 = "def"
Output: 0
Explanation: There is no such common subsequence, so the result is 0.

Explanation

This problem can be solved using dynamic programming. The idea is to create a 2D table dp where dp[i][j] represents the length of the longest common subsequence of the substrings text1[0:i] and text2[0:j].

DP Table Initialization:
- We initialize a 2D list dp with dimensions (len(text1) + 1) x (len(text2) + 1) filled with 0s. This table will help us store the lengths of LCS for substrings of text1 and text2.
- The extra row and column are to handle the case when either string is empty, in which case the LCS length is 0.
Filling the DP Table:
- We start filling the table from the bottom-right corner.
- For each pair of indices i and j, if the characters text1[i] and text2[j] match, then dp[i][j] is 1 + dp[i + 1][j + 1], because the current character can be included in the LCS.
- If the characters don't match, we take the maximum value between dp[i + 1][j] and dp[i][j + 1], which represents skipping one of the characters in either text1 or text2.
Result:
- After filling the table, the top-left corner dp[0][0] contains the length of the longest common subsequence of the two strings.

Example Walkthrough

Let's go through the example "abcde" and "ace" in detail to understand how we can find the length of the longest common subsequence (LCS) step by step.

Iteration 1:
- Start with i = 4 and j = 2 (Comparing text1[4] = "e" with text2[2] = "e")
- The characters match ("e" == "e").
- Update dp[4][2] = 1 + dp[5][3] = 1 + 0 = 1.
- The updated dp table:
```
dp = [[0, 0, 0, 0],
      [0, 0, 0, 0],
      [0, 0, 0, 0],
      [0, 0, 0, 0],
      [0, 0, 1, 0],
      [0, 0, 0, 0]]
```
Iteration 2:
- We move to i = 4 and j = 1 (comparing text1[4] = "e" with text2[1] = "c").
- The characters do not match ("e" != "c"), so we cannot include this character in our common subsequence.
- To decide the value of dp[4][1], we look at the right cell dp[4][2] and the cell below dp[5][1].
  - dp[4][2] gives us the LCS length if we skip the current character in text2.
  - dp[5][1] gives us the LCS length if we skip the current character in text1.
- We take the maximum of these two values: dp[4][1] = max(dp[4][2], dp[5][1]) = max(1, 0) = 1.
- This means that up to this point, the maximum LCS length considering these two characters remains 1.
- The updated dp table is:
```
dp = [[0, 0, 0, 0],
      [0, 0, 0, 0],
      [0, 0, 0, 0],
      [0, 0, 0, 0],
      [0, 1, 1, 0],
      [0, 0, 0, 0]]
```
Iteration 3:
- Move to i = 4 and j = 0 (Comparing text1[4] = "e" with text2[0] = "a")
- The characters do not match ("e" != "a").
- Update dp[4][0] = max(dp[4][1], dp[5][0]) = max(1, 0) = 1.
- The updated dp table:
```
dp = [[0, 0, 0, 0],
      [0, 0, 0, 0],
      [0, 0, 0, 0],
      [0, 0, 0, 0],
      [1, 1, 1, 0],
      [0, 0, 0, 0]]
```
Iteration 4:
- Move to i = 3 and j = 2 (Comparing text1[3] = "d" with text2[2] = "e")
- The characters do not match ("d" != "e").
- Update dp[3][2] = max(dp[3][3], dp[4][2]) = max(0, 1) = 1.
- The updated dp table:
```
dp = [[0, 0, 0, 0],
      [0, 0, 0, 0],
      [0, 0, 0, 0],
      [0, 0, 1, 0],
      [1, 1, 1, 0],
      [0, 0, 0, 0]]
```
Iteration 5:
- Move to i = 3 and j = 1 (Comparing text1[3] = "d" with text2[1] = "c")
- The characters do not match ("d" != "c").
- Update dp[3][1] = max(dp[3][2], dp[4][1]) = max(1, 1) = 1.
- The updated dp table:
```
dp = [[0, 0, 0, 0],
      [0, 0, 0, 0],
      [0, 0, 0, 0],
      [0, 1, 1, 0],
      [1, 1, 1, 0],
      [0, 0, 0, 0]]
```
Iteration 6:
- Move to i = 3 and j = 0 (Comparing text1[3] = "d" with text2[0] = "a")
- The characters do not match ("d" != "a").
- Update dp[3][0] = max(dp[3][1], dp[4][0]) = max(1, 1) = 1.
- The updated dp table:
```
dp = [[0, 0, 0, 0],
      [0, 0, 0, 0],
      [0, 0, 0, 0],
      [1, 1, 1, 0],
      [1, 1, 1, 0],
      [0, 0, 0, 0]]
```
Iteration 7:
- Move to i = 2 and j = 2 (Comparing text1[2] = "c" with text2[2] = "e")
- The characters do not match ("c" != "e").
- Update dp[2][2] = max(dp[2][3], dp[3][2]) = max(0, 1) = 1.
- The updated dp table:
```
dp = [[0, 0, 0, 0],
      [0, 0, 0, 0],
      [0, 0, 1, 0],
      [1, 1, 1, 0],
      [1, 1, 1, 0],
      [0, 0, 0, 0]]
```
Iteration 8:
- Move to i = 2 and j = 1 (Comparing text1[2] = "c" with text2[1] = "c")
- The characters match ("c" == "c").
- Update dp[2][1] = 1 + dp[3][2] = 1 + 1 = 2.
- The updated dp table:
```
dp = [[0, 0, 0, 0],
      [0, 0, 0, 0],
      [0, 2, 1, 0],
      [1, 1, 1, 0],
      [1, 1, 1, 0],
      [0, 0, 0, 0]]
```
Iteration 9:
- Move to i = 2 and j = 0 (Comparing text1[2] = "c" with text2[0] = "a")
- The characters do not match ("c" != "a").
- Update dp[2][0] = max(dp[2][1], dp[3][0]) = max(2, 1) = 2.
- The updated dp table:
```
dp = [[0, 0, 0, 0],
      [0, 0, 0, 0],
      [2, 2, 1, 0],
      [1, 1, 1, 0],
      [1, 1, 1, 0],
      [0, 0, 0, 0]]
```
Iteration 10:
- Move to i = 1 and j = 2 (Comparing text1[1] = "b" with text2[2] = "e")
- The characters do not match ("b" != "e").
- Update dp[1][2] = max(dp[1][3], dp[2][2]) = max(0, 1) = 1.
- The updated dp table:
```
dp = [[0, 0, 0, 0],
      [0, 0, 1, 0],
      [2, 2, 1, 0],
      [1, 1, 1, 0],
      [1, 1, 1, 0],
      [0, 0, 0, 0]]
```
Iteration 11:
- Move to i = 1 and j = 1 (Comparing text1[1] = "b" with text2[1] = "c")
- The characters do not match ("b" != "c").
- Update dp[1][1] = max(dp[1][2], dp[2][1]) = max(1, 2) = 2.
- The updated dp table:
```
dp = [[0, 0, 0, 0],
      [0, 2, 1, 0],
      [2, 2, 1, 0],
      [1, 1, 1, 0],
      [1, 1, 1, 0],
      [0, 0, 0, 0]]
```
Iteration 12:
- Move to i = 1 and j = 0 (Comparing text1[1] = "b" with text2[0] = "a")
- The characters do not match ("b" != "a").
- Update dp[1][0] = max(dp[1][1], dp[2][0]) = max(2, 2) = 2.
- The updated dp table:
```
dp = [[0, 0, 0, 0],
      [2, 2, 1, 0],
      [2, 2, 1, 0],
      [1, 1, 1, 0],
      [1, 1, 1, 0],
      [0, 0, 0, 0]]
```
Iteration 13:
- Move to i = 0 and j = 2 (Comparing text1[0] = "a" with text2[2] = "e")
- The characters do not match ("a" != "e").
- Update dp[0][2] = max(dp[0][3], dp[1][2]) = max(0, 1) = 1.
- The updated dp table:
```
dp = [[0, 0, 1, 0],
      [2, 2, 1, 0],
      [2, 2, 1, 0],
      [1, 1, 1, 0],
      [1, 1, 1, 0],
      [0, 0, 0, 0]]
```
Iteration 14:
- Move to i = 0 and j = 1 (Comparing text1[0] = "a" with text2[1] = "c")
- The characters do not match ("a" != "c").
- Update dp[0][1] = max(dp[0][2], dp[1][1]) = max(1, 2) = 2.
- The updated dp table:
```
dp = [[0, 2, 1, 0],
      [2, 2, 1, 0],
      [2, 2, 1, 0],
      [1, 1, 1, 0],
      [1, 1, 1, 0],
      [0, 0, 0, 0]]
```
Iteration 15:
- Move to i = 0 and j = 0 (Comparing text1[0] = "a" with text2[0] = "a")
- The characters match ("a" == "a").
- Update dp[0][0] = 1 + dp[1][1] = 1 + 2 = 3.
- The final dp table:
```
dp = [[3, 2, 1, 0],
      [2, 2, 1, 0],
      [2, 2, 1, 0],
      [1, 1, 1, 0],
      [1, 1, 1, 0],
      [0, 0, 0, 0]]
```

Result:

The value at dp[0][0] is 3, which is the length of the longest common subsequence between text1 = "abcde" and text2 = "ace".

Efficiency Analysis

Time Complexity: $O(m \times n)$, where m is the length of text1 and n is the length of text2.
- We need to fill a 2D table of size m x n.
Space Complexity: $O(m \times n)$.
- We use a 2D table to store the lengths of the longest common subsequence for each pair of indices.

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README.md

README.md

2-D Dynamic Programming Data-Structures and Algorithms

01. Unique Paths

Explanation

Detailed Steps

Example Walkthrough

Efficiency Analysis

02. Longest Common Subsequence

Explanation

Example Walkthrough

Result:

Efficiency Analysis

Files

README.md

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README.md

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2-D Dynamic Programming Data-Structures and Algorithms

01. Unique Paths

Explanation

Detailed Steps

Example Walkthrough

Efficiency Analysis

02. Longest Common Subsequence

Explanation

Example Walkthrough

Result:

Efficiency Analysis