**Shortest Common Supersequence - Google Top Interview Questions**

### Problem Statement :

Given strings a and b, return the length of the shortest string that has both a and b as subsequences. Example 1 Input a = "bell" b = "yellow" Output 7 Explanation One possible solution is "ybellow".

### Solution :

` ````
Solution in C++ :
int solve(string a, string b) {
int M = a.size(), N = b.size();
vector<int> dp1(N + 1), dp2(N + 1);
for (int i = 1; i <= M; i++) {
for (int j = 1; j <= N; j++)
dp1[j] =
a[i - 1] == b[j - 1] ? dp2[j - 1] + 1 : max(max(dp2[j], dp1[j - 1]), dp2[j - 1]);
dp2.swap(dp1);
}
return M + N - dp2[N];
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(String a, String b) {
int ans = 0;
int[][] dp = new int[a.length() + 1][b.length() + 1];
for (int i = 1; i < dp.length; i++) {
for (int j = 1; j < dp[0].length; j++) {
if (a.charAt(i - 1) == b.charAt(j - 1))
dp[i][j] = 1 + dp[i - 1][j - 1];
else {
dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
}
}
}
int temp = dp[a.length()][b.length()];
ans = temp + (a.length() - temp) + (b.length() - temp);
return ans;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, a, b):
return len(a) + len(b) - self.LCS(a, b)
def LCS(self, a, b):
m = len(a)
n = len(b)
# declaring the array for storing the dp values
L = [[None] * (n + 1) for i in range(m + 1)]
"""Following steps build L[m+1][n+1] in bottom up fashion
Note: L[i][j] contains length of LCS of X[0..i-1]
and Y[0..j-1]"""
for i in range(m + 1):
for j in range(n + 1):
if i == 0 or j == 0:
L[i][j] = ""
elif a[i - 1] == b[j - 1]:
L[i][j] = L[i - 1][j - 1] + a[i - 1]
else:
if len(L[i - 1][j]) >= len(L[i][j - 1]):
L[i][j] = L[i - 1][j]
elif len(L[i - 1][j]) < len(L[i][j - 1]):
L[i][j] = L[i][j - 1]
# L[m][n] contains the length of LCS of X[0..n-1] & Y[0..m-1]
return len(L[m][n])
```

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