Doubt in the solution

#1

please check my code and what is wrong in my approach???

#2
#3

@Adhyayan,

Your code is not working for input:
abbve
gbdea
Correct answer : be

You can follow this approach while traversing your dp array:

  1. Construct dp[m+1][n+1] using the count LCS dynamic programming solution.

  2. The value dp[m][n] contains length of LCS. Create a character array lcs[] of length equal to the length of lcs plus 1 (one extra to store \0).

  3. Traverse the 2D array starting from dp[m][n]. Do following for every cell dp[i][j]

  • If characters (in X and Y) corresponding to dp[i][j] are same (Or X[i-1] == Y[j-1]), then include this character as part of LCS.
  • Else compare values of dp[i-1][j] and dp[i][j-1] and go in direction of greater value.
#4

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