# Minimum cost to make Longest Common Subsequence of length k

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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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 ith character from x.
3. Skip jth 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 ith character from x then i will be decreased by 1, no cost will be added and LCS will remain the same.
If you skip jth 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]```

.

## C++

 `#include ` `using` `namespace` `std; ` `const` `int` `N = 30; ` ` `  `// Return Minimum cost to make LCS of length k ` `int` `solve(``char` `X[], ``char` `Y[], ``int` `l, ``int` `r,  ` `                     ``int` `k, ``int` `dp[][N][N]) ` `{ ` `    ``// If k is 0. ` `    ``if` `(!k) ` `        ``return` `0; ` ` `  `    ``// If length become less than 0, return ` `    ``// big number. ` `    ``if` `(l < 0 | r < 0) ` `        ``return` `1e9; ` ` `  `    ``// If state already calculated. ` `    ``if` `(dp[l][r][k] != -1) ` `        ``return` `dp[l][r][k]; ` ` `  `    ``// Finding the cost ` `    ``int` `cost = (X[l] - ``'a'``) ^ (Y[r] - ``'a'``); ` ` `  `    ``// Finding minimum cost and saving the state value ` `    ``return` `dp[l][r][k] = min({cost + ` `                      ``solve(X, Y, l - 1, r - 1, k - 1, dp), ` `                             ``solve(X, Y, l - 1, r, k, dp),  ` `                             ``solve(X, Y, l, r - 1, k, dp)}); ` `} ` ` `  `// Driven Program ` `int` `main() ` `{ ` `    ``char` `X[] = ``"abble"``; ` `    ``char` `Y[] = ``"pie"``; ` `    ``int` `n = ``strlen``(X); ` `    ``int` `m = ``strlen``(Y); ` `    ``int` `k = 2; ` ` `  `    ``int` `dp[N][N][N]; ` `    ``memset``(dp, -1, ``sizeof` `dp); ` `    ``int` `ans = solve(X, Y, n - 1, m - 1, k, dp); ` ` `  `    ``cout << (ans == 1e9 ? -1 : ans) << endl; ` `    ``return` `0; ` `} `

## Python3

# Python3 program to calculate Minimum cost
# to make Longest Common Subsequence of length k
N = 30

# Return Minimum cost to make LCS of length k
def solve(X, Y, l, r, k, dp):

# If k is 0
if k == 0:
return 0

# If length become less than 0,
# return big number
if l < 0 or r < 0: return 1000000000 # If state already calculated if dp[l][r][k] != -1: return dp[l][r][k] # Finding cost cost = ((ord(X[l]) - ord('a')) ^ (ord(Y[r]) - ord('a'))) dp[l][r][k] = min([cost + solve(X, Y, l - 1, r - 1, k - 1, dp), solve(X, Y, l - 1, r, k, dp), solve(X, Y, l, r - 1, k, dp)]) return dp[l][r][k] # Driver Code if __name__ == "__main__": X = "abble" Y = "pie" n = len(X) m = len(Y) k = 2 dp = [[[-1] * N for __ in range(N)] for ___ in range(N)] ans = solve(X, Y, n - 1, m - 1, k, dp) print(-1 if ans == 1000000000 else ans) # This code is contributed # by vibhu4agarwal [tabbyending] Output:

```3
```