Swiggy Interview Question for SDE-3s


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def get_node(value, nodes){
  if ( value !@ nodes ){
    node = { 'v' : value, 'c' : list() , 'p' : null }
    nodes[value] = node 
  }
  nodes[value] // return 
}

def create_graph( list_of_pairs ){
  nodes = dict()
  for ( pair : list_of_pairs ){
    #(parent,child) = pair 
    p_node = get_node( parent, nodes )
    panic ( size( p_node.c ) > 1 , "It supposed to be having only 2 child!" )
    c_node = get_node( child , nodes )
    assert ( empty( c_node.p ) , "Duplicate - or - multiple parent for a child!" )
    c_node.p = p_node.v // parent the child 
    p_node.c += c_node.v // child the parent 
  }
  nodes // return 
}

def find_root_for_valid_bin_tree(graph){
  // check if graph valid binary tree? 
  possible_roots = select( graph ) where { 
    n = $.o.value 
    empty(n.p ) 
  }
  assert( size(possible_roots) == 1, "Is a graph, with multiple roots" )
  // check for cycles ? easy... do dfs and check ... 
  root = possible_roots[0]
  stack = list()
  stack.push( root.key )
  visited = set()
  while (!empty(stack) ){
    nv = stack.pop()
    panic( nv @ visited , "There are cycles!" )
    visited += nv 
    for ( c : graph[nv].c ){ stack.push( c ) }
  }
  root.key // return root 
}

def str_bin_tree( root_value, graph ){
  root = graph[root_value]
  left_str_rep = str_bin_tree( root.c[0] , graph  ) ?? "" 
  right_str_rep = str_bin_tree( root.c[1] , graph ) ?? "" 
  str( "(%s%s%s)", root_value, left_str_rep, right_str_rep )
}

def do_swiggy(list_of_pairs){
  g = create_graph(list_of_pairs)
  r = find_root_for_valid_bin_tree(g)
  println( str_bin_tree( r , g ) )
}

pair_list = [ ['A','B'], ['B','C'], ['A','D'], ['C','E']  ]
do_swiggy( pair_list )

- NoOne June 01, 2019 | Flag Reply
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public static void main(String[] args) throws Exception {
        System.out.print(swiggySDE3(Arrays.asList("(A,B)", "(B,C)", "(A,D)", "(C,E)", "(A,F)")));
    }

    public static String swiggySDE3(List<String> str) {
        Map<String, List<String>> map = new HashMap<>();
        for (String t : str) {
            String[] pair = t.substring(1, t.length() - 1).split(",");
            final List<String> orDefault = map.getOrDefault(String.valueOf(pair[0]), new ArrayList<>());
            orDefault.add(pair[1]);
            map.put(pair[0], orDefault);
        }
        StringBuffer stringBuffer = new StringBuffer();
        swiggySDE3UTIL("A", map, stringBuffer);
        return stringBuffer.toString().substring(0, stringBuffer.length()-1);
    }

    private static void swiggySDE3UTIL(String start, Map<String, List<String>> map, StringBuffer stringBuffer) {
        final List<String> values = map.get(start);
        if (values == null || values.isEmpty()) {
            stringBuffer.append(start + ")");
            return;
        }
        stringBuffer.append(start+"(");
        values.forEach(t->swiggySDE3UTIL(t,map,stringBuffer));
        stringBuffer.append(")");
    }

- koustav.adorable June 06, 2019 | Flag Reply
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Simply, maintain two HashMap. Parent Map and child map. In parent map keep a child count. In child map keep a parent info. When inserting new tuple check.

1. parent map already has more than 2 child count.
2. child already has same parent as in tuple
3. child already has a parent.

4. After all tuples are entered run through the parent table and count entries with 0 child if more than 1 flag error.

- Mukul June 14, 2019 | Flag Reply


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