**Delete From the Ends and Reinsert to Target - Google Top Interview Questions**

### Problem Statement :

You are given two strings s and t that are anagrams of each other. Consider an operation where we remove either the first or the last character in s and insert it back anywhere in the string. Return the minimum number of operations required to convert s into t. Constraints n ≤ 1,000 where n is the length of s Example 1 Input s = "edacb" t = "abcde" Output 3 Explanation The three operations are: We can remove "b" and insert it after "a" to get "edabc" We can remove "e" and insert it after "c" to get "dabce" We can remove "d" and insert it after "c" to get "abcde"

### Solution :

` ````
Solution in C++ :
int LCS(int i, int j, int n, string& s, string& t, vector<vector<vector<int> > >& dp,
bool canskip) {
if (i >= n || j >= n) {
return 0;
}
int op1 = 0, op2 = 0, op3 = 0;
if (dp[i][j][canskip] != -1) {
return dp[i][j][canskip];
}
if (s[i] == t[j]) {
op1 = 1 + LCS(i + 1, j + 1, n, s, t, dp, 0);
}
if (canskip) {
op2 = LCS(i + 1, j, n, s, t, dp, 1);
}
op3 = LCS(i, j + 1, n, s, t, dp, canskip);
return dp[i][j][canskip] = max(op1, max(op2, op3));
}
int solve(string s, string t) {
int n = s.length();
vector<vector<vector<int> > > dp(n, vector<vector<int> >(n, vector<int>(2, -1)));
return (n - LCS(0, 0, n, s, t, dp, 1));
}
```

` ````
Solution in Python :
class Solution:
def solve(self, s, t):
lhs = 1
rhs = len(s)
while lhs < rhs:
mid = (lhs + rhs + 1) // 2
def is_subsequence(start):
idx = 0
for j in range(len(t)):
if s[idx + start] == t[j]:
idx += 1
if idx == mid:
return True
return False
has_subsequence = any(is_subsequence(i) for i in range(len(s) - mid + 1))
if has_subsequence:
lhs = mid
else:
rhs = mid - 1
return len(s) - lhs
```

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