这个问题可以通过动态规划来解决。首先,我们需要定义dp[i][j]表示s的前i个字符和t的前j个字符相等所需的最少操作次数。然后,我们可以根据s[i]和t[j]是否相等来更新dp[i][j]。如果s[i]和t[j]相等,那么dp[i][j] = dp[i-1][j-1]。如果s[i]和t[j]不相等,那么dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1。最后,我们需要从dp[n][n]开始,根据dp数组倒推出操作序列。具体实现可以参考以下代码:
```python
def min_operations(s, t):
n = len(s)
dp = [[0] * (n + 1) for _ in range(n + 1)]
for i in range(1, n + 1):
for j in range(1, n + 1):
if s[i - 1] == t[j - 1]:
dp[i][j] = dp[i - 1][j - 1]
else:
dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1
i, j, operations = n, n, 0
while i > 0 and j > 0:
if s[i - 1] == t[j - 1]:
i -= 1
j -= 1
elif dp[i - 1][j - 1] < dp[i - 1][j]:
i -= 1
operations += 1
else:
j -= 1
operations += 1
print(operations)
i, j, operations = n, n, 0
while i > 0 and j > 0:
if s[i - 1] == t[j - 1]:
i -=