Amazon Interview Question for Computer Scientists


Country: India
Interview Type: Written Test




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of 0 vote

we need to remove parenthesis if complete string enclose with otherwise return same string.

- kamal suthar August 26, 2020 | Flag Reply
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public static String removeOuterBrace(String test){
if(! validate(test) )
return "INVALID";

return removeOuterBraceHelper(test);
}


private static String removeOuterBraceHelper(String test){

char[] elements = test.toCharArray();

if(elements[0] != '(')
return test;

if(elements[elements.length - 1] != ')')
return test;

int matched = 1;
for (int i = 1; i < elements.length; i++) {

if(elements[i] == '(') {
matched++;
}
else if(elements[i] == ')') {
matched --;

if(i == elements.length - 1 && matched == 0)
if(validate(test.substring(1, test.length()-1)))
return removeOuterBraceHelper(test.substring(1, test.length()-1));
}
}

return test;
}

private static boolean validate(String test) {
char[] elements = test.toCharArray();

int i = 0;
int matchCount = 0;
for (char c:
elements) {
if(c == '(') {
matchCount++;
i++;
}
else if(c == ')') {
matchCount--;
i--;
}
if(matchCount < 0)
return false;
}
return i == 0;
}
}

- nooreddin August 26, 2020 | Flag Reply
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0
of 0 votes

private static String removeOuterBraceHelper(String test){

char[] elements = test.toCharArray();

if(elements[0] != '(')
return test;

if(elements[elements.length - 1] != ')')
return test;

while(validate(test.substring(1, test.length()-1)) && test.startsWith("("))
test = test.substring(1, test.length()-1);

return test;
}

- nooreddin August 26, 2020 | Flag
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of 0 votes

private static String removeOuterBraceHelper(String test){

while(validate(test.substring(1, test.length()-1)) && test.startsWith("("))
test = test.substring(1, test.length()-1);

return test;
}

- nooreddin August 26, 2020 | Flag
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I have solved it using stack.
Time Complexity - O(n)
Space complexity - O(n)

#include<bits/stdc++.h>
using namespace std;

int main()
{
    
    string expression;
    cin>>expression;
    string s=expression;
    stack <int> exp_idx;
    bool is_invalid = 0;
    
    /*if while iterating i get a ')' and the top of the stack is '(' the bracket pair is redundant, so we place '$' signs in those places
    */
    for(int i=0;i<expression.length();i++)
    {
        if(expression[i]==')')
        {
            if(!exp_idx.empty() and expression[exp_idx.top()]=='(')
            {
                expression[exp_idx.top()]='$';
                expression[i]='$';
                exp_idx.pop();
            }
            else
            {
            
                while(!exp_idx.empty() and expression[exp_idx.top()]!='(')
                    exp_idx.pop();
                if(exp_idx.empty())
                {
                    i=expression.length();
                    is_invalid=1;
                }
                else
                {
                    exp_idx.pop();
                }
            }
            
        }
        else
            exp_idx.push(i);
        
    }
    
    //special check for the removal of first and last bracket
    
    int front=0,back=expression.length()-1;
    
    while(front<expression.length() and expression[front]=='$')
        front++;
    while(back>=0 and expression[back]=='$')
        back--;
        
    
    if(front<back and expression[front]=='(' and expression[back]==')')
    {
        int i=front;
        int bracket_balance=0;
        while(i<back)
        {
            if(expression[i]=='(')
                bracket_balance++;
            else if(expression[i]==')')
            {
                bracket_balance--;
                if(bracket_balance==0)
                    break;
            }
            i++;
        }
        if(i==back)
            expression[front]=expression[back]='$';
    }
    // special check ends
    
    if(is_invalid)
        cout<<"INVALID EXPRESSION";
    else
        for(auto it:expression)
            if(it!='$') cout<<it;
    
    return 0;    
}

- suvro_coder August 28, 2020 | Flag Reply
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0
of 0 vote

It is incredibly simple.
1. If the first char and last char are not "(" and ")" - then return the same string.
2. If they are, then,
2.1 get the substring "sub" from position 1 to len-2 (0 based index ).
2.2. Verify if this string has matched parenthesis
2.3 If it has then return the "sub".
2.4 Else return the original string.

- NoOne August 29, 2020 | Flag Reply
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0
of 0 vote

package com.company;

public class Main {

public static void main(String[] args) {
String s = "abc(c))";
//O(n)
System.out.println(parseStr(s));
}

static String parseStr(String s) {
//O(n)
if (!stringIsValid(s)) {
return "INVALID";
}

String result = null;
char[] chars = s.toCharArray();

//O(C)
for (int i = 0; i < chars.length; i++) {
int mirroredIndex = chars.length - i;
String validatedString = s.substring(i, mirroredIndex);
//O(n)
boolean valid = checkStringFullyInParethesis(validatedString);

if (!valid) {
result = s.substring(i, mirroredIndex);
break;
}
}

return result;
}

static boolean checkStringFullyInParethesis(String s) {
if (s.indexOf('(') == 0) {
int firstCloseParenthesisIndex = s.indexOf(')');

if (firstCloseParenthesisIndex > 0) {
for (var i = firstCloseParenthesisIndex + 1; i < s.length(); i++) {
char c = s.charAt(i);

if (c != ')' && c != '(') {
return false;
}
}

return true;
}
}

return false;
}

static boolean stringIsValid(String s) {
int countOpenParths = 0;
int countCloseParths = 0;

for (var i = 0; i < s.length(); i++) {
if (s.charAt(i) == '(') {
countOpenParths++;
} else if (s.charAt(i) == ')') {
countCloseParths++;
}
}

return countCloseParths == countOpenParths;
}
}

Complexity - O(n)
1) Verify if String is valid
2) In loop get substring sequentially for first and last element
2.1) Verify if substring is valid and entire fits in parenthesis
2.2) If yes - continue loop for second element, etc..
2.3) if no - exit with current substring

- artem.shustov August 31, 2020 | Flag Reply
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0
of 0 vote

import java.util.*;

/*
Examples:
(((abc))) --> abc
(ab(c)) --> ab(c)
(abc09%(c)) --> abc09%(c)
ab(c) --> ab(c)
(ab)c --> (ab)c
abc(c)) → INVALID
(abc)(def) --> (abc)(def)
(abc)typ(def) --> (abc)typ(def)
((abc)(def)) --> (abc)(def)
*/

class RemoveSmallBrackets {
	public static void main(String[] args) {
		String[] inputs = new String[]{
		"(((abc)))",
		"(ab(c))",
		"(ab09%(c))",
		"ab(c)",
		"abc(c))",
		"(abc)(def)",
		"(abc)typ(def)",
		"((abc)typ(def))"	
		};
		char leftChar = '(';
		char rightChar = ')';
		
		for(String s : inputs){
			char[] ch = s.toCharArray();
			System.out.println(String.valueOf(ch)+"->"+String.valueOf(filter(ch,leftChar,rightChar)));	
		}
	}
	
	private static char[] filter(char[] input, char x, char y){
		char[] result = null;
		if(count(input,x)!=count(input,y))
			return new char[]{'I','N','V','A','L','I','D'};
		
		if(input.length>0 && input[0]==x && input[input.length-1]==y){
			result =  filter(Arrays.copyOfRange(input, 1, input.length-1),x,y);
			return (isValidFormat(result,x,y))?result:input;
		}
		
		return input;
	}
	
	private  static boolean isValidFormat(char[] input, char x, char y){
		int fx = firstIndexOf(input,x);
		int fy = firstIndexOf(input,y);
		
		if((fx==0 && fy==0) || (fx<fy))
			return true;
		else
			return false;
	}
	
	private static int count(char[] input, char toBeSearched){
		int count = 0;
		for(char c: input)
			if(c==toBeSearched)
				count++;
		
		return count;
	}
	
	private static int firstIndexOf(char[] input, char x){
		int result = 0;
		for(char c: input){
			if(c==x)
				return result;
			result++;
		}
		return 0;	
				
	}
}

- KritarthD September 03, 2020 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

import java.util.*;

/*
Examples:
(((abc))) --> abc
(ab(c)) --> ab(c)
(abc09%(c)) --> abc09%(c)
ab(c) --> ab(c)
(ab)c --> (ab)c
abc(c)) → INVALID
(abc)(def) --> (abc)(def)
(abc)typ(def) --> (abc)typ(def)
((abc)(def)) --> (abc)(def)
*/

class RemoveSmallBrackets {
	public static void main(String[] args) {
		String[] inputs = new String[]{
		"(((abc)))",
		"(ab(c))",
		"(ab09%(c))",
		"ab(c)",
		"abc(c))",
		"(abc)(def)",
		"(abc)typ(def)",
		"((abc)typ(def))"	
		};
		char leftChar = '(';
		char rightChar = ')';
		
		for(String s : inputs){
			char[] ch = s.toCharArray();
			System.out.println(String.valueOf(ch)+"->"+String.valueOf(filter(ch,leftChar,rightChar)));	
		}
	}
	
	private static char[] filter(char[] input, char x, char y){
		char[] result = null;
		if(count(input,x)!=count(input,y))
			return new char[]{'I','N','V','A','L','I','D'};
		
		if(input.length>0 && input[0]==x && input[input.length-1]==y){
			result =  filter(Arrays.copyOfRange(input, 1, input.length-1),x,y);
			return (isValidFormat(result,x,y))?result:input;
		}
		
		return input;
	}
	
	private  static boolean isValidFormat(char[] input, char x, char y){
		int fx = firstIndexOf(input,x);
		int fy = firstIndexOf(input,y);
		
		if((fx==0 && fy==0) || (fx<fy))
			return true;
		else
			return false;
	}
	
	private static int count(char[] input, char toBeSearched){
		int count = 0;
		for(char c: input)
			if(c==toBeSearched)
				count++;
		
		return count;
	}
	
	private static int firstIndexOf(char[] input, char x){
		int result = 0;
		for(char c: input){
			if(c==x)
				return result;
			result++;
		}
		return 0;	
				
	}
}

- KritarthD September 03, 2020 | Flag Reply
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0
of 0 vote

private static string RemoveParans(string statement)
        {
            if (statement.Length == 1)
                if (statement[0] != '(' && statement[0] != ')') return statement; else return "INVALID";

            int opened = 0;
            int closed = 0;
            var removeCandidate = new List<int>();

            for (int i = 0; i < statement.Length; i++)
            {
                if (closed > opened) return "INVALID";

                if (statement[i] == '(')
                {
                    opened++;
                    removeCandidate.Add(i);
                }
                if (statement[i] == ')')
                {
                    closed++;
                    if (statement[i - 1] != '(' && statement[i - 1] != ')') removeCandidate.RemoveAt(removeCandidate.Count - 1);
                    else removeCandidate.Add(i);
                }
            }

            string result = string.Empty;
            for (int i = 0; i < statement.Length; i++)
            {
                if (!removeCandidate.Contains(i)) result += statement[i];
            }
            if (result.Length > 2 && result.Substring(1, result.Length - 2).All(c => c != '(' && c != ')')) result = result.Substring(1, result.Length - 2);

            return result;
        }

- Hassan Kachal September 12, 2020 | Flag Reply


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