* * Runtime : 12ms, faster than 98.40% * Memory : 10.7MB, faster than 85.57% * prerequisite : least common subsequence in string */
| 5 | * prerequisite : least common subsequence in string |
| 6 | */ |
| 7 | class Solution { |
| 8 | public: |
| 9 | string shortestCommonSupersequence(string str1, string str2) { |
| 10 | |
| 11 | int i = str1.length(); |
| 12 | int j = str2.length(); |
| 13 | |
| 14 | |
| 15 | |
| 16 | int t[i+1][j+1]; |
| 17 | |
| 18 | // create matrix of size str1 length as row |
| 19 | // and str2 length as column |
| 20 | |
| 21 | // base condition |
| 22 | for(int k = 0;k<=i;k++) |
| 23 | { |
| 24 | t[k][0] = 0; |
| 25 | } |
| 26 | |
| 27 | // base condition |
| 28 | for(int k = 0;k<=j;k++) |
| 29 | { |
| 30 | t[0][k] = 0; |
| 31 | } |
| 32 | |
| 33 | // to form different solution from all possible length string |
| 34 | for(int k = 1;k<=i;k++) |
| 35 | { |
| 36 | for(int l = 1;l<=j;l++) |
| 37 | { |
| 38 | if(str1[k-1] == str2[l-1]) |
| 39 | { |
| 40 | t[k][l] = 1+t[k-1][l-1]; |
| 41 | } |
| 42 | else{ |
| 43 | t[k][l] = max(t[k-1][l],t[k][l-1]); |
| 44 | } |
| 45 | } |
| 46 | } |
| 47 | |
| 48 | string res=""; |
| 49 | |
| 50 | while(i>0 && j>0) |
| 51 | { |
| 52 | // if char are same in str1 and str2, add it one time |
| 53 | if(str1[i-1]==str2[j-1]) |
| 54 | { |
| 55 | res.push_back(str1[i-1]); |
| 56 | i--;j--; |
| 57 | } |
| 58 | else{ |
| 59 | |
| 60 | // comparing matrix value to form string |
| 61 | // add both characters of string to answer |
| 62 | if(t[i][j-1] > t[i-1][j]) |
| 63 | { |
| 64 | res.push_back(str2[j-1]); |
nothing calls this directly
no outgoing calls
no test coverage detected