text: /* Copyright (c) 2019-2021 Grant Hadlich
   *
   * Permission is hereby granted, free of charge, to any person obtaining a copy
   * of this software and associated documentation files (the Software), to deal
   * in the Software without restriction, including without limitation the rights
   * to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
   * copies of the Software, and to permit persons to whom the Software is
   * furnished to do so, subject to the following conditions:
   *
   * The above copyright notice and this permission notice shall be included in
   * all copies or substantial portions of the Software.
   *
   * THE SOFTWARE IS PROVIDED AS IS, WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
   * IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
   * FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
   * AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
   * LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
   * OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
   * THE SOFTWARE.
   */
  class Solution {
  public:
      string shortestCommonSupersequence(string a, string b) {
      vector<vector<int>> dp(a.size() + 1, vector<int> (b.size() + 1, 0));
      
      for(int i = 1; i <= a.size(); i++){
          for(int j = 1; j <= b.size(); j++){
              if(a[i - 1] == b[j - 1]){
                  dp[i][j] = dp[i - 1][j - 1] + 1;
              }else {
                  dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
              }
          }
      }
      
      // Now need to traverse back from end till 0
      string ans = "";
      
      int i = a.size(), j = b.size();
      while(i > 0 || j > 0){
          if(i > 0 && dp[i][j] == dp[i - 1][j]){
              i--;
              ans = a[i] + ans;
          }else if(j > 0 && dp[i][j] == dp[i][j - 1]){
              j--;
              ans = b[j] + ans;
          }else {
              i--;
              j--;
              ans = a[i] + ans;
          }
      }
      
      return ans;
      }
  };