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Prashant Mishra

Posted on • Originally published at leetcode.com

# Shortest Common Super-sequence Leetcode (Same as Lcs string)

Given two strings str1 and str2, return the shortest string that has both str1 and str2 as subsequences. If there are multiple valid strings, return any of them.

A string s is a subsequence of string t if deleting some number of characters from t (possibly 0) results in the string s.

Example 1:

``````Input: str1 = "abac", str2 = "cab"
Output: "cabac"
Explanation:
str1 = "abac" is a subsequence of "cabac" because we can delete the first "c".
str2 = "cab" is a subsequence of "cabac" because we can delete the last "ac".
The answer provided is the shortest such string that satisfies these properties.
``````

Solution:
Time complexity is same as that of longest common subsequence
i.e. O(m*n) , space complexity is O(m*n) for using dp array.

``````class Solution {
String str = "";
public String shortestCommonSupersequence(String str1, String str2) {
int dp[][] = new int[str1.length()+1][str2.length()+1];

for(int i=0;i<=str1.length();i++){
dp[i] =0;
}
for( int i =0;i<=str2.length();i++){
dp[i] = 0;
}

for( int i =1;i<=str1.length();i++){
for(int j =1;j<=str2.length();j++){
if(str1.charAt(i-1)==str2.charAt(j-1)){
dp[i][j] = 1 + dp[i-1][j-1];
}
else dp[i][j] = Integer.max(dp[i][j-1],dp[i-1][j]);
}
}
int p = str1.length(), q = str2.length();

while(p>0 && q>0){
if(str1.charAt(p-1) == str2.charAt(q-1)){
str = str1.charAt(p-1)+ str;
p--;
q--;

}
else if(dp[p][q-1] > dp[p-1][q]){
str = str2.charAt(q-1) + str;
q--;
}
else {
str = str1.charAt(p-1) + str;
p--;
}
}

while(p>0){
str  = str1.charAt(p-1)+ str;
p--;
}
while(q>0){
str = str2.charAt(q-1) +  str;
q--;
}
return str;
}
}
`````` Hey! We would love your help!