Implement feline_fixes, which is a diff function that returns the minimum number of edit operations needed to transform the start word into the goal word. There are three kinds of edit operations: Add a letter to start, Remove a letter from start, Substitute a letter in start for another. Each edit operation contributes 1 to the difference between two words.
时间: 2024-02-25 15:58:08 浏览: 132
Sure, here's the implementation of the `feline_fixes` function in Python:
```python
def feline_fixes(start, goal):
m, n = len(start), len(goal)
dp = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(m + 1):
dp[i][0] = i
for j in range(n + 1):
dp[0][j] = j
for i in range(1, m + 1):
for j in range(1, n + 1):
if start[i - 1] == goal[j - 1]:
dp[i][j] = dp[i - 1][j - 1]
else:
dp[i][j] = 1 + min(dp[i - 1][j - 1], dp[i][j - 1], dp[i - 1][j])
return dp[m][n]
```
This function uses dynamic programming to solve the problem. The `dp` matrix is initialized with 0s, and the first row and column are filled with the distances between the empty string and the prefixes of the start and goal words. Then, the matrix is filled in row-major order using the following recurrence:
- If the i-th character of start is equal to the j-th character of goal, then the distance between the prefixes of length i and j is the same as the distance between the prefixes of length i-1 and j-1.
- Otherwise, we can transform the prefix of start into the prefix of goal using one of three operations: substitution (if we replace the i-th character of start with the j-th character of goal), deletion (if we delete the i-th character of start), or insertion (if we insert the j-th character of goal after the i-th character of start). We take the minimum of the costs of these three operations plus 1 (to account for the current mismatch) to get the distance between the prefixes of length i and j.
Finally, the function returns the value in the bottom-right corner of the matrix, which represents the distance between the entire start and goal words.
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