Amazon Interview Question for SDE-2s


Country: United States
Interview Type: Phone Interview




Comment hidden because of low score. Click to expand.
2
of 2 vote

def check_bracket(str):
    stack = list()
    for char in str:
        if char == '<' or char == '(' or char == '{' or\
           char == '[':
            stack.append(char)
        else:
            bracket = stack.pop()
            if char == '>' and bracket != '<' or\
               char == '}' and bracket != '{' or\
               char == ')' and bracket != '(' or\
               char == ']' and bracket != '[':
                return False
    return True

- Kool October 31, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

Easy and adaptable solution:

private final static HashMap<Char, Char> parenthesis = new HashMap<>();
static{
	parenthesis.add(“(“, “)”);
	parenthesis.add(“<“, “>”);
	parenthesis.add(“{“, “}”);
	parenthesis.add(“[“, “]”);
}


public boolean isBalanced(String str){


	Stack<Character> stack = new Stack<>();
	
	for(Char c : str.toCharArray()){
		if(parenthesis.contains(c)){
			stack.push(c);
		}else{
			if(parenthesis.(stack.pop()) != c)
				return false;
		}
	}


	if(stack.isEmpty())
		return true;
	return false;


}

- libertythecoder November 01, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 2 vote

// ZoomBA
paren_arr = [ { _'<' : 1 ,  _'>' : -1 , 'c' : 0 },
              { _'(' : 1 ,  _')' : -1 , 'c' : 0 },
              { _'{' : 1 ,  _'}' : -1 , 'c' : 0 },
              { _'[' : 1 ,  _']' : -1 , 'c' : 0 } ]
// now the indexing 
parens = { _'<' : 0 , _'>' : 0 , 
           _'(' : 1 , _')' : 1 ,
           _'{' : 2 , _'}' : 2 ,
           _'[' : 3 , _']' : 3 }

def match_paren(string){
   fold ( string.value , true ) -> {
     matcher = paren_arr[ parens[ $.item ] ]
     matcher.c += matcher[ $.item ]
     break ( matcher.c < 0 ){ false }
     true
   } && ! ( exists ( paren_arr ) :: {  $.item.c != 0 } )
}
println ( match_paren (  "<({()})[]>"  ) ) 
println ( match_paren ( "<({([)})[]>" ) )

This is fully declarative. The logic is actually implemented in the maps, while the engine simply runs it through, also fully extensible.

- NoOne October 30, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void BalanceStrings()
{
	var patternList = new List<string>() { "<({()})[]>" , "<({([)})[]>" };
	char[] allowedChars = { '(', '[', '{', ')', ']', '}', '<', '>' };
	var updatedPattern = new StringBuilder();

	foreach (var pattern in patternList)
	{
		var patternToCharacters = pattern.ToCharArray();

		foreach (var character in patternToCharacters)
		{
			var isMatch = allowedChars.Any(x => x == character);
			if (isMatch)
			{
				updatedPattern.Append(character);
			}
		}

		Console.WriteLine(pattern.Equals(updatedPattern.ToString()) ? "balanced" : "Not balanced");
	}
}

- gunjan.prmr October 31, 2016 | Flag Reply
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0
of 0 vote

import java.util.*;
class BalancedString {
  public static void main(String[] args) {
    System.out.println(isBalanced("<({()})[]>"));
    System.out.println(isBalanced("<({([)})[]>"));
  }
  public static boolean isBalanced(String s) {
  	Stack<String> stack = new Stack<>();
    HashMap<String, String> map = new HashMap<>();
    map.put("(", ")");
    map.put("{", "}");
    map.put("[", "]");
    map.put("<", ">");
    for (int i=0; i<s.length(); i++) {
    	String current = s.charAt(i) + "";
    	if (!stack.isEmpty() && map.get(stack.peek()) != null && map.get(stack.peek()).equals(current)) {
	    	stack.pop();
    	} else {
    		stack.push(current);
    	}
    }
    return stack.isEmpty();
  }

}

- dnsh November 01, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Tested in C# and also can be extended easily

static string openingChars = "<{[(";
        static string closingChars = ">}])";

        static bool IsBalanced(IEnumerable<char> characters)
        {
            int[] counters = new int[openingChars.Length];

            int i = 0;

            foreach (var c in characters)
            {
                i = openingChars.IndexOf(c);

                if (i != -1)
                {
                    counters[i]++;
                    continue;
                }

                i = closingChars.IndexOf(c);

                if (i != -1)
                {
                    if (counters[i] == 0) return false;
                    else counters[i]--;
                }
            }

            for (i = 0; i < counters.Length; i++)
            {
                if (counters[i] != 0)
                {
                    return false;
                }
            }

            return true;
        }

- xyz November 04, 2016 | Flag Reply
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0
of 0 vote

import java.util.ArrayList;
import java.util.Scanner;

public class BalancedString {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		String str = in.nextLine();
		ArrayList<Character> stack = new ArrayList<>();
		stack.add(str.charAt(0));
		for(int i=1; i<str.length(); i++){
			if((str.charAt(i) == ')' && stack.get(0) == '(') || 
					(str.charAt(i) == '}' && stack.get(0) == '{') ||
					(str.charAt(i) == ']' && stack.get(0) == '[') ||
					(str.charAt(i) == '>' && stack.get(0) == '<'))
				stack.remove(0);
			else
				stack.add(0, str.charAt(i));
		}
		if(stack.isEmpty())
			System.out.println("Balanced");
		else
			System.out.println("not");
		in.close();
	}

}

- Anonymous November 15, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Is this string considered balanced?
<({([)})]>

- Anonymous November 23, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

@Dmytri
i don't think this is the right solution - see these cases imo is incorrect:

cowbell pts/1 try>./a.out {[}]
String is balanced

cowbell pts/1 try>./a.out {{]]
String is balanced

- rravir November 27, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

class CheckCode{
    public static boolean check(String s){
        Stack ll = new Stack();
        HashMap<Character,Character> map = new HashMap<Character,Character>();
        map.put('{', '}');
        map.put('(', ')');
        map.put('[', ']');
        char c;
        char k;
        if (s == null) return false;
        
        for (int i = 0;i<s.length();i++){
            c = s.charAt(i);
            if (map.get(c) != null) {
                ll.push(c);
            }
            if (map.containsValue(c)){
                if (ll.peek() != null) {
                    k = (char) ll.pop();
                    if (map.get(k) != c) return false;
                }
                else return false;
            }
        }
        
        return ll.peek() == null;
    }
}

- edanir December 05, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include<iostream>
#include<stack>
#include<string>
using namespace std;
// Function to check whether two characters are opening
// and closing of same type.
bool ArePair(char opening,char closing)
{
if(opening == '(' && closing == ')') return true;
else if(opening == '{' && closing == '}') return true;
else if(opening == '[' && closing == ']') return true;
return false;
}
bool AreParanthesesBalanced(string exp)
{
stack<char> S;
for(int i =0;i<exp.length();i++)
{
if(exp[i] == '(' || exp[i] == '{' || exp[i] == '[')
S.push(exp[i]);
else if(exp[i] == ')' || exp[i] == '}' || exp[i] == ']')
{
if(S.empty() || !ArePair(S.top(),exp[i]))
return false;
else
S.pop();
}
}
return S.empty() ? true:false;
}

int main()
{
/*Code to test the function AreParanthesesBalanced*/
string expression;
cout<<"Enter an expression: "; // input expression from STDIN/Console
cin>>expression;
if(AreParanthesesBalanced(expression))
cout<<"Balanced\n";
else
cout<<"Not Balanced\n";
}

- Salim Amrani December 20, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def CheckIfBalance(mystring):
    stack = list()
    for eachchar in mystring:
        if eachchar == '(' or eachchar == '{' or eachchar == '[' or eachchar == '<':
            stack.append(eachchar)
        elif eachchar == ')' and len(stack) > 0 and stack.pop() == '(':
            continue
        elif eachchar == '}' and len(stack) > 0 and stack.pop() == '{':
            continue
        elif eachchar == ']' and len(stack) > 0 and stack.pop() == '[':
            continue
        elif eachchar == '>' and len(stack) > 0 and stack.pop() == '<':
            continue
        else:
            return "Not Balanced"
    if len(stack) == 0:
        return "Balanced"
    else:
        return "Not Balanced"

print CheckIfBalance("<({()})[]>") #Returns Balanced
print CheckIfBalance("<({([)})[]>") #Returns Not Balanced
print CheckIfBalance("))((") #Returns Not Balanced
print CheckIfBalance("((()") #Returns Not Balanced

- ankitpatel2100 February 18, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

Using counter to determine whether string is balanced.

#include <stdio.h>	// Input-Output requests

int main(int argc, char** argv) {
	char *paren = argv[1];
	int counter = 0; // Counters for these
	
	while (*paren != '\0') {
		switch(*paren) {
			case '{':
			case '(':
			case '[':
				counter++;
				break;
			case '}':
			case ')':
			case ']':
				counter--;
				break;
		}
		paren++;	
	}

	if (counter == 0)
		printf("String is balanced\n");
	else
		printf("String is not balanced.\n");		
		
	return 0;	// Successfully quit.
}

- Dmytri November 01, 2016 | Flag Reply


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