Amazon Interview Question for Software Engineer Interns


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3
of 3 vote

We can use a data structure stack to solve this problem. My Java code:
Time O(n)
Space O(n)

private boolean isValidParenthesis(char[] s) {
    // assume we have a map contains all type of parenthesis
    Map<Character, Character> parenthesisType = new HashMap<Character, Character>();
    parenthesisType.put('{', '}');
    parenthesisType.put('(', ')');
    parenthesisType.put('[', ']');

    // initial a stack
    Stack<Character> stack = new Stack<Character>();

    for (int i = 0; i < s.length; i++) {
        // if s[i] is a valid open parenthesis, push to stack
        if (parenthesisType.containsKey(s[i])) {
            stack.push(s[i]);
        }
        // if not, this may be a close parenthesis, so
        // check the top of stack, if it's same type
        // with s[i], remove this top open parenthesis
        else if (!stack.empty() && parenthesisType.get(stack.peek()) == s[i]) {
            stack.pop();
        }
        // if this parenthesis not exist or stack is empty, return false
        else {
            return false;
        }
    }

    // if our stack is empty, this mean s is valid parenthesis
    return stack.isEmpty();
}

- techinterviewquestion.com January 31, 2016 | Flag Reply
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0
of 0 votes

Will your program not return on the first character if it is not a braces?

- nmathur February 05, 2016 | Flag
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0
of 0 votes

Hi nmathur, my program will return false immediately when it meets an invalid brace because this mean string s will be an invalid parenthesis sequence too.

- techinterviewquestion.com February 06, 2016 | Flag
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0
of 0 vote

c# implementation.
O(n) time.
O(1) space.
Return values: true - balanced, false - not balanced.

public static bool ValidateBrackets1( string str ) {
    const uint n = 0;
    // all possible brackets coupled
    var dic = new Dictionary<string, uint> { {"{}", n}, {"()", n}, {"[]", n}, {"<>", n} };
    // all possible closing brackets
    var dic0 = new Dictionary<char, uint> { { '}', n }, { ')', n }, { ']', n }, { '>', n } };

    foreach ( char c in str ) {
        foreach ( var item in dic ) {
            if ( item.Key[ 0 ] == c ) {
                dic[ item.Key ]++;
                dic0[ item.Key[ 1 ]  ] = dic0.Values.Max() + 1;
                break;
            }
            if ( item.Key[ 1 ] == c ) {
                if ( dic0[ item.Key[ 1 ] ] != dic0.Values.Max() ) { return false; }
                try {
                    checked { dic[ item.Key ]--; }
                    dic0[ item.Key[ 1 ] ] = (dic[ item.Key ] == 0) 
                        ? 0 
                        : dic0.Values.Where( x => x > 0 ).Min() - 1;
                } catch ( OverflowException ) {
                    return false;
                }
                break;
            }
        }
    }
    return dic.All( item => item.Value == 0 );
}

- zr.roman January 31, 2016 | Flag Reply
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0
of 0 votes

does this check for in order as well?
for example {[}]...

- djtrance February 01, 2016 | Flag
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0
of 0 votes

you are right, thank you.
This check is not performed.
So, looks like using stack with O(n) space is inevitable.

- zr.roman February 01, 2016 | Flag
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0
of 0 votes

well, actually, O(n) space is not inevitable.
I improved this solution such that it is still O(n) time and O(1) space.

- zr.roman February 01, 2016 | Flag
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0
of 0 votes

Can you explain how it happens in O(1) space?

- Anonymous February 06, 2016 | Flag
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0
of 0 votes

yep.
It's quite obvious. The space is needed only for 2 dictionaries, both of constant size. Constant size means O(1).
That's it.

- zr.roman February 06, 2016 | Flag
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0
of 0 vote

variant #2.
c#.
Time O(n).
Space O(1).

public static bool ValidateBrackets2( string str ) {
    const uint n = 0;
    // all possible pairs of brackets in forward order
    var dic1 = new Dictionary<char, char> { {'{', '}'}, {'(', ')'}, {'[', ']'}, {'<', '>'} };
    // all possible pairs of brackets in reverse order
    var dic2 = new Dictionary<char, char> { {'}', '{'}, {')', '('}, {']', '['}, {'>', '<'} };
    // all possible pairs of brackets coupled
    var dic3 = new Dictionary<string, uint> { {"{}", n}, {"()", n}, {"[]", n}, {"<>", n} };
    // all possible closing brackets
    var dic0 = new Dictionary<char, uint> { { '}', n }, { ')', n }, { ']', n }, { '>', n } };

    foreach ( char ch in str ) {
        if ( dic1.ContainsKey( ch ) ) {
            dic3[ ch.ToString() + dic1[ ch ] ]++;
            dic0[ dic1[ ch ]  ] = dic0.Values.Max() + 1;
            continue;
        }
        if ( dic2.ContainsKey( ch ) ) {
            if ( dic0[ ch ] != dic0.Values.Max() ) { return false; }
            try {
                checked { dic3[ dic2[ ch ] + ch.ToString() ]--; }
                dic0[ ch ] = (dic3[ dic2[ ch ] + ch.ToString() ] == 0) 
                        ? 0 
                        : dic0.Values.Where( x => x > 0 ).Min() - 1;
            } catch ( OverflowException ) {
                return false;
            }
        }
    }
    return dic3.All( item => item.Value == 0 );
}

- zr.roman January 31, 2016 | Flag Reply
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0
of 0 vote

Do we have also have to keep order into consideration?

Example ')(' would be unbalanced as the order is not right of opening and closing braces?

- Rajdeep January 31, 2016 | Flag Reply
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0
of 0 vote

python
time O(m)
space O(1)

import stack

def isBalanced(symString):
    stcl = stack.Stack()
    balanced = True
    i = 0
    while i < len(symString) and balanced:
        if symString[i] in "{[(":
            stck.push(symString[i])
        else:
	    if stck.isEmpty():
	        balanced =False
	    else:
		top = stck.pop()
		if not matches(top, symString[i]):
		    balanced = False
	i += 1
    # if after all, symbols are still balanced
    if balanced and stck.isEmpty():
	return True
    return False

def matches(open, close):
    pOpen = "{[("
    pClose = "}])"
    return pOpen.index(open) == pClose.index(close)

- davron.sh.karimov February 01, 2016 | Flag Reply
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3
of 3 votes

Would it be possible to explain why this is O(1) space? wouldn't you be pushing potentially n/2 elements on the stack?

- djtrance February 01, 2016 | Flag
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0
of 0 vote

C++ solution with example string

#include <iostream>
#include <string>
#include <stack>
using namespace std;

bool isOpen(char c)
{
	return c == '(' || c == '{' || c == '[';
}

bool isClose(char c)
{
	return c == ')' || c == '}' || c == ']';
}

bool isMatch(char c1, char c2)
{
	return (((c1 == '(') && (c2 == ')')) ||
	     	((c1 == '[') && (c2 == ']')) ||
	     	((c1 == '{') && (c2 == '}')) );
}

bool isBalanced(const string& str)
{
	stack<char> charStack;
	for (unsigned int i = 0; i < str.size(); ++i ) {
		if (isOpen(str[i])) {
			charStack.push(str[i]);
		} else if (isClose(str[i])) {
			if (!charStack.empty() && isMatch(charStack.top(), str[i])) {
				charStack.pop();
			} else {
				return 0;
			}
		} 
	}
	return charStack.empty();
}

int main()
{
	string str = "([[(){}]])({})";
	
	cout << str << " : " << isBalanced(str) << endl;

}

- utkubulkan February 03, 2016 | Flag Reply
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0
of 0 vote

Solution in Java
Time : O(n)
Space : O(n)

private static void checkBracketBalancing(String input){
        char[] inputArr=input.toCharArray();
        Map<Character,Character> charPairs=new HashMap<Character,Character>();        
        charPairs.put(')', '(');
        charPairs.put('}', '{');
        charPairs.put(']', '[');
        Stack<Character> charst=new Stack<Character>();
        for(Character c : inputArr){
            if(charPairs.get(c)!=null){
                if(!charst.isEmpty() && charPairs.get(c).equals(charst.peek())){                    
                    charst.pop();
                }else{
                    System.out.println("Unbalanced");
                    return;
                }
            }else{
                charst.push(c);
            }
        }
        if(charst.isEmpty()){
            System.out.println("Balanced");
        }
    }

- t@_@n February 04, 2016 | Flag Reply
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0
of 0 vote

#include <stdio.h>

int arr[][2] = {
	{ '{','}'},
	{ '[',']'},
	{ '(',')'} 
	/* Keep on adding new set of brackets*/
}; 

void getvalue(char ch, int *f, int *t){
	if (ch == 0) return;
	for (int i = *f + 1; i < sizeof(arr) / sizeof(arr[0]); i++){
		if (ch == arr[i][0]){
			*f = i;
			*t = 0;
			break;
		}
		if (ch == arr[i][1]){
			*f = i;
			*t = 1;
			break;
		}
	}
}
bool fun(char *str){
	if (*str == 0) return true;
	int f = -1;
	int t = 0;
	getvalue(*str, &f, &t);
	if (f == -1) return true;
	else if (t == 0) return (true && fun(str + 1));
	else {
		int ff = -1;
		int tt = 0;
		getvalue(*(str - 1), &ff, &tt);
		if (ff == f && tt == 0) return true;
		else return false;
	}
}

int main(){
	printf("%d",fun("([{exp[exp]})"));
	return 0;
}

- praveen pandit March 09, 2016 | Flag Reply
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0
of 0 vote

it wasn't well tested..
please comment on below.

int arr[][2] = {
	{ '{','}'},
	{ '[',']'},
	{ '(',')'} 
	/* Keep on adding new set of brackets*/
}; 

void getvalue(char ch, int *f, int *t){
	if (ch == 0) return;
	for (int i = *f + 1; i < sizeof(arr) / sizeof(arr[0]); i++){
		if (ch == arr[i][0]){
			*f = i;
			*t = 0;
			break;
		}
		if (ch == arr[i][1]){
			*f = i;
			*t = 1;
			break;
		}
	}
}
bool fun1(char *str, stack<char> *s){
	if (*str == 0){
		if (s->empty()) return true;
		else return false;
	}
	int f = 0;
	int t = 0;
	getvalue(*str, &f, &t);
	if (t == 0) s->push(*str);
	else{
		if (s->empty()) return false;
		else{
			char ch = s->top();
			int ff = 0;
			int tt = 0;
			getvalue(ch, &ff, &tt);
			if (ff == f && tt == 0) {
				s->pop();
				return fun1(str + 1, s);
			}
			else return false;
		}
	}
	return (true && fun1(str + 1, s));
}

void fun2(char *str){
	int *arr = (int*)calloc(sizeof(int)*128,1);
}
int main(){
	stack<char> s;
	printf("%d",fun1("(({}[(){}]))",&s));
	return 0;
}

- praveen pandit March 09, 2016 | Flag Reply
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-1
of 1 vote

boolean validParenthsis(String str) {


Stack<Character> stk  = new Stack<>();

for (int i = 0 ; i < str.length ; i++) {
	char currChar = str.charAt(i);
	if (currChar == '{' || currChar == '[' || currChar == '(' ){
		stk.push(currChar);
	} else {
		if (stk.isEmpty() || stk.pop() != currChar) {
			return false;
		}
	}
}

}

return stk.isEmpty();
}

- er.vishalmehta83 January 31, 2016 | Flag Reply
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0
of 0 votes

You forgot to pop the stack.

- ASDFASDFASDFASDFASDF February 01, 2016 | Flag
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0
of 0 votes

You forgot to pop the stack.

- asdQwspypfmb February 01, 2016 | Flag
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0
of 0 votes

that, and the fact that '}' is not equal to '{'

- djtrance February 01, 2016 | Flag


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