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A sub-sequence of a given sequence is just the given sequence with some elements (possibly none or all) left out. We are given two sequences $X[m]$ and $Y[n]$ of lengths $m$ and $n$, respectively with indexes of $X$ and $Y$ starting from $0$.

We wish to find the length of the longest common sub-sequence (LCS) of $X[m]$ and $Y[n]$ as $l(m, n)$, where an incomplete recursive definition for the function $I(i, j)$ to compute the length of the LCS of $X[m]$ and $Y[n]$ is given below:

l(i,j)  = 0, if either i = 0 or j = 0
        = expr1, if i,j > 0 and X[i-1] = Y[j-1]
        = expr2, if i,j > 0 and X[i-1] ≠ Y[j-1]

Which one of the following options is correct?

  1. $\text{expr1} = l\left(i-1, j\right) +1$

  2. $\text{expr1} = l\left(i, j-1\right)$

  3. $\text{expr2} = \max\left(l\left(i-1, j\right), l\left(i,j-1\right)\right)$

  4. $\text{expr2} = \max\left(l\left(i-1, j-1\right), l\left(i,j\right)\right)$

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2 Answers

Best answer
37 votes
37 votes

Answer is C. When the currently compared elements doesn't match, we have two possibilities for the LCS, one including X[i] but not Y[j] and other including Y[j] but not X[i].

/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int lcs( char *X, char *Y, int m, int n )
{
   if (m == 0 || n == 0)
     return 0;
   if (X[m-1] == Y[n-1])
     return 1 + lcs(X, Y, m-1, n-1);
   else
     return max(lcs(X, Y, m, n-1), lcs(X, Y, m-1, n));
}
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24 votes
24 votes

Answer is C. 

When the currently compared elements doesn't match, we have two possibilities for the LCS, one including $X\left [ i \right ]$ but not $Y\left [ j \right ]$  and other including $Y\left [ j \right ]$ but not $X\left [ i \right ]$.

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