Given two string **X**, **Y** and an integer **k**. Now the task is to convert string X with minimum cost such that the Longest Common Subsequence of X and Y after conversion is of length k. The cost of conversion is calculated as XOR of old character value and new character value. Character value of ‘a’ is 0, ‘b’ is 1 and so on.

**Examples:**

Input : X = "abble", Y = "pie", k = 2 Output : 25 If you changed 'a' to 'z', it will cost 0 XOR 25.

The problem can be solved by slight change in Dynamic Programming problem of Longest Increasing Subsequence. Instead of two states, we maintain three states.

Note, that if k > min(n, m) then it’s impossible to attain LCS of atleast k length, else it’s always possible.

Let dp[i][j][p] stores the minimum cost to achieve LCS of length p in x[0…i] and y[0….j].

With base step as dp[i][j][0] = 0 because we can achieve LCS of 0 legth without any cost and for i < 0 or j 0 in such case).

Else there are 3 cases:

1. Convert x[i] to y[j].

2. Skip i^{th} character from x.

3. Skip j^{th} character from y.

If we convert x[i] to y[j], then cost = f(x[i]) XOR f(y[j]) will be added and LCS will decrease by 1. f(x) will return the character value of x.

Note that the minimum cost to convert a charcater ‘a’ to any character ‘c’ is always f(a) XOR f(c) because f(a) XOR f(c) <= (f(a) XOR f(b) + f(b) XOR f(c)) for all a, b, c.

If you skip i^{th} character from x then i will be decreased by 1, no cost will be added and LCS will remain the same.

If you skip j^{th} character from x then j will be decreased by 1, no cost will be added and LCS will remain the same.

Therefore,

dp[i][j][k] = min(cost + dp[i - 1][j - 1][k - 1], dp[i - 1][j][k], dp[i][j - 1][k]) The minimum cost to make the length of their LCS atleast k is dp[n - 1][m - 1][k]

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