Flatten dictionary - April 28, 2018 - 10:00 AM mock interview - one more practice on mock interview

FlattenDictionary_April28_2018.cs

using System;
using System.Collections.Generic;

class Solution
{
  public static Dictionary<string, string> FlattenDictionary(Dictionary<string, object> dict)
  {
    if(dict == null )
      return new Dictionary<string, string>(); 
    
    return FlattenDictionaryHelper(dict, "");
  }
  
  private static Dictionary<string, string> FlattenDictionaryHelper(Dictionary<string, object> dict, string prefix)
  {
    var flatten = new Dictionary<string, string>(); 
    
    foreach(var pair in dict)
    {
      var key      = pair.Key; 
      var valueStr = pair.Value; 
      
       var newKey = key; 
       if(prefix.Length > 0)
            newKey = prefix + "." + key; 
       if(key.Length == 0)
            newKey = prefix; 
      
      if(valueStr.GetType() == typeof(Dictionary<string,object>) )
      {
        // solve subproblem 
        var newKeyObject = (Dictionary<string, object>) valueStr;                
        
        var childFlatten = FlattenDictionaryHelper(newKeyObject, newKey);   // key ? 
        
        foreach(var childPair in childFlatten)
        {
          flatten.Add(childPair.Key, childPair.Value);
        }       
      }
      else 
      {                        
        flatten.Add(newKey, valueStr.ToString()); 
      }      
    }
    
    return flatten; 
  }

  static void Main(string[] args)
  {

  }
}

/*
keywords:

dictionary
key
value - primitive type / dictionary
   primitive type: integer, boolean, double, string, excluding boolean, double
   
ask: flattenDictionary 
  concatenate by dot 
  Key2.a
  Key2.b
  
  parent dictionary key as prefix, then prefix + '.' + key
  
  recursive/ depth first search 
  
  iterate every key/value pair, if value is primtive type then put back to flatten as is; 
  otherwise it is dictioinary type, and then solve subproblem with string/ string flatten dict - sub dict
    add current key as prefix by going over subdict
   
 */