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.
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 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.
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]
# 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:
# 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:
- Length of longest common subsequence containing vowels
- Minimum cost to make a string free of a subsequence
- Maximum length subsequence such that adjacent elements in the subsequence have a common factor
- Longest Common Subsequence | DP-4
- Printing Longest Common Subsequence
- LCS (Longest Common Subsequence) of three strings
- C++ Program for Longest Common Subsequence
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- Longest Common Increasing Subsequence (LCS + LIS)
- Longest common subsequence with permutations allowed
- Edit distance and LCS (Longest Common Subsequence)
- Java Program for Longest Common Subsequence
- Python Program for Longest Common Subsequence
- Longest subsequence such that adjacent elements have at least one common digit
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Improved By : vibhu4agarwal