Amazon Interview Question for Quality Assurance Engineers


Team: Kindle
Country: India
Interview Type: Written Test




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11
of 11 vote

if(s2==( s.substr(2)+s.substr(0,2)))
return true;
else
return false;

- Anonymous July 26, 2014 | Flag Reply
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0
of 0 votes

How could you come up with so succinct answer. It's awesome~

- ravio July 30, 2014 | Flag
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1
of 1 vote

Youre assuming the direction of rotation is always the same, which wasnt given. This fails if we have for example s="amazon" and s2="onamaz"

- Anonymous September 09, 2014 | Flag
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0
of 0 votes

This is the approach recommended in "Cracking the Coding Interview" by Gayle.

- Pavan Nayakanti April 03, 2015 | Flag
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1
of 1 vote

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main(){
    char s1[1000],s2[1000];
    int len1, len2,i,j,flag=1;
    scanf("%s",s1);
    scanf("%s",s2);
    len1=strlen(s1);
    len2=strlen(s2);
    for(i=2;i<len1;i++){
        if(s2[i-2]!=s1[i]){
            flag=0;
            break;
        }
    }
    if(flag){
        if(s1[0]!=s2[len2-2])
            flag=0;
        if(flag){
            if(s1[1]!=s2[len2-1])
                flag=0;
        }
    }
    if(flag)
        printf("True");
    else printf("False");
    return 0;
}

- aayush2k11 July 27, 2014 | Flag Reply
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1
of 1 vote

First, the length of S1 and S2 should equal.
It is ez to solve by duplicate s1 or s2 like s2 = s2+s2:

e.g S1="amazon" S2="azonam"
	Assuming we are duplicating S2:
		S2="azonamazonam";

Then, we can scan S2, find whether S2 contains S1.

public static boolean isRotated(String s1, String s2) {
	if (s1.length()!=s2.length())
		return false;
	int len = s1.length();

	// double String "s2";
	s2 = s2+s2;
	for (int i=0;i<len;i++) {
		if (s2.substring(i,i+len).equals(s1))
			return true;
	}

	return false;
}

The time complexity should be O(n), space complexity O(1);

}

- wyu277 August 17, 2014 | Flag Reply
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0
of 0 votes

Logic is works well .Adding addtional verification.

if (s1.length()!=s2.length())
return false;
int len = s1.length();

// double String "s2";
s1 = s1+s1;
for (int i=0;i<len;i++) {
if (s1.substring(i,i+len).equals(s2)) {
if (i==2) {
System.out.println("Left Roatation works");
return true;
} else {
i=i+len-1;
String ch = s1.substring(i+1,len*2);
if (ch.length() == 2) {
System.out.println("Right rotation works");
}
}
}
}

return false;
}

- sathyamoorthybe August 22, 2014 | Flag
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0
of 0 vote

int main(){string s="",s2="";
cin>>s>>s2;
for(int i=0;i<2;i++){string temp=s.substr(1)+s.substr(0,1);
s=temp;}
if(s==s2) cout<<"correct\n";
else
cout<<"wrong\n";
return 0;}

- Anonymous July 26, 2014 | Flag Reply
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0
of 0 vote

bool isSubstr2Places(string s1, string s2){
    return  (s2 == s.substr(2) + s.substr(0,2)) ||  (s2 == s.substr(0,2) + s.substr(2) );
}

- Anonymous July 26, 2014 | Flag Reply
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0
of 0 votes

Considers Both left rotation and right rotation by 2 places.

- kk July 27, 2014 | Flag
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0
of 0 vote

- Anonymous July 26, 2014 | Flag Reply
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0
of 0 vote

public boolean isSubString(String s1, String s2) {
  if(s1 == null || s2 == null) { return false; }
  return (s1 + s1).contains(s2);
}

- Max July 27, 2014 | Flag Reply
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0
of 0 vote

we can so this using hashtable also.......

- hellen July 27, 2014 | Flag Reply
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0
of 0 vote

public static boolean isRotatedByTwoChars(String s1, String s2)
    {
        String s3 = s1.substring(2, s1.length())+ s1.substring(0,2);
        String s4 = s1.substring(0, s1.length()-2) + s1.substring(s1.length()-2, s1.length());
        System.out.println(s3);
        System.out.println(s4);
        if (s2.equals(s3) || s2.equals(s4)) return true;
        else return false;
    }

- CODEBUG July 28, 2014 | Flag Reply
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0
of 0 vote

private static int myMod(int value, int mod){
		if( value >= 0 ){
			return value % mod;
		}
		
		return mod + value;
	}
	
	/**
	 * Rotate string to the right.
	 * time: O(N)
	 * space: O(N)
	 */
	public static String rotate(String str, int offset){
		
		if( str == null ){
			throw new IllegalArgumentException("Can't rotate NULL string");
		}
		
		if( offset == 0 ){
			return str;
		}
		
		char[] arr = str.toCharArray();
		BitSet handledPositions = new BitSet(str.length());
		
		for( int i = 0; i < arr.length; i++ ){
			
			int cur = i;
			int next = myMod((cur+offset), arr.length);
			char temp = arr[cur];
			
			/** not handled yet */
			while( ! handledPositions.get(next) ){
				
				char ch = temp;
				temp = arr[next];
				arr[next] = ch;
				cur = next;
				next = myMod((next+offset), arr.length);
				handledPositions.set(cur);
			}
		}
		
		return new String(arr);
		
	}
	
	/**
	 * Check if 'rotated' string is rotation by 2 positions to the left/right of a 'base' string.
	 * time: O(N)
	 * space: O(N)
	 */
	public static boolean isRotateByTwo(String base, String rotated){
		
		if( base == null || rotated == null ){
			throw new IllegalArgumentException("NULL 'base' and/or 'rotated' stirng(s) passed");
		}
		
		if( base.length() != rotated.length() ){
			return false;
		}
			
		/** 
		 * 1. shift right by 2 and check;
		 * 2. shift left by 2 and check.
		 */
		return rotate(base, 2).equals(rotated) || rotate(base, -2).equals(rotated);

	}

- m@}{ July 28, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int _tmain(int argc, _TCHAR* argv[])
{
char* a = "quality";
char* b = "lityqua";
bool flag = true;
int count = 0;
char* t = a + 2;
char* t2 = b;
while (*t != '\0')
{
if (*t == *t2)
{
++t;
++t2;
}
else
{
flag = false;
break;
}
}

if (flag)
{
t = a;
while (*t2 != '\0')
{
if (*t2 == *t)
{
++t2;
++t;
}
else
{
flag = false;
}
}
}

if (flag) cout << "Yes" << endl;
else cout << "No" << endl;
_getch();
return 0;
}

- Siddharth More July 28, 2014 | Flag Reply
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0
of 0 vote

public boolean rotate(String s1, String s2){
if(s1 == null || s2 == null || s1.length() != s2.length()) return false;
if(s1.length() <= 2) return s2.equals(s1);
return s2.equals(s1.substring(2) + s1.substring(0,2))
|| s2.equals(s1.substring(s1.length() - 2) + s1.substring(0, s1.length() - 2));
}

- averill July 28, 2014 | Flag Reply
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0
of 0 vote

def isSwappable?(str1, str2)
  return false if str1 == nil || str2 == nil || str1.length != str2.length
  extra_in_string1 = []
  extra_in_string2 = []
  (1..str1.length).each do |i|
    if str1[i] != str2[i]
      extra_in_string1 << str1[i]
      extra_in_string2 << str2[i]
    end
  end
  extra_in_string1.length == 2 && extra_in_string1.sort == extra_in_string2.sort
end

Test cases:
isSwappable?(nil, nil) # false
isSwappable?(nil, "a") # false
isSwappable?("a", nil) # false
isSwappable?("a", "asd") # false
isSwappable?("asx", "asd") # false
isSwappable?("abc", "acb") # false
isSwappable?("amazon", "azamon") # true
isSwappable?("amazox", "azamon") # false

- Vijay Raghavan Aravamudhan July 28, 2014 | Flag Reply
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0
of 0 vote

for python single line code will work:

print s1[0:2]==s2[-2:]

- Akshay Singh Jetawat July 29, 2014 | Flag Reply
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0
of 0 vote

for(int i=0;i<S1.length();i++){
if(i<2){
S4=S4+S1.charAt(i);
}else{
S5=S5+S1.charAt(i);
}
}

- Sathishwaran Selvaraj July 30, 2014 | Flag Reply
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0
of 0 vote

for(int i=0;i<S1.length();i++){
if(i<2){
S4=S4+S1.charAt(i);
}else{
S5=S5+S1.charAt(i);
}
}
if(S2.equals(S5+S4){
return true
else
return false

- Sathish July 30, 2014 | Flag Reply
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0
of 0 vote

public static boolean twoplaces(String s1, String s2){
return s2.equals((s1+s1).substring(2,2+s1.length()));
}

- tesh July 31, 2014 | Flag Reply
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0
of 0 vote

private static boolean checkString(String input,String test) {
	input=input+input.substring(0, 2);
return	test.equals(input.substring(2));
	}

- harish August 04, 2014 | Flag Reply
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0
of 0 vote

public static bool IsRotation(string s1, string s2, int x)
{
    string s1New = string.Concat(s1, s1);
    int index = s1New.IndexOf(s2);
    return x == index;
}

- Anonymous August 09, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

First, the length of S1 and S2 should equal.
It is ez to solve by duplicate s1 or s2 like s2 = s2+s2:

e.g S1="amazon" S2="azonam"
	Assuming we are duplicating S2:
		S2="azonamazonam";

Then, we can scan S2, find whether S2 contains S1.

public static boolean isRotated(String s1, String s2) {
	if (s1.length()!=s2.length())
		return false;
	int len = s1.length();

	// double String "s2";
	s2 = s2+s2;
	for (int i=0;i<len;i++) {
		if (s2.substring(i,i+len).equals(s1))
			return true;
	}

	return false;
}

The time complexity should be O(n), space complexity O(1);

}

- wyu277 August 17, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

{{
S1="amazon"
S2="azonam"

def RotateString():
if (S1[2:]+ S1[:2]) == S2:
return True
else:
return False

print RotateString()
}}}

- Sunil August 22, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

S1="amazon" 
S2="azonam"


def RotateString():
    if (S1[2:]+ S1[:2]) == S2:
        return True
    else:
        return False
    
print RotateString()

- Sunil August 22, 2014 | Flag Reply
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0
of 0 vote

bool AreRotatedStrings(string const &s1, string const &s2, size_t n)
{
	string str(s1);
	rotate(str.begin(), str.begin() + n, str.end());
	return (str == s2);
}

- Teh Kok How September 03, 2014 | Flag Reply
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0
of 0 vote

public static boolean reversetwospaces(String str1,String str2){

if(str1.length()!=str2.length()){
return false;
}
str2=str2.substring(str2.length()-2, str2.length())+str2.substring(0, str2.length()-2);;
if(str2.equals(str1)){
return true;
}
else{
return false;
}
}

- sudip January 14, 2015 | Flag Reply
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0
of 0 vote

public class StringRotation
{
public static boolean isRotated(String s1, String s2,int rotationCount)
{
if(s1.length()!=s2.length())
{
return false;
}
String s=s1.substring(rotationCount)+s1.substring(0,rotationCount);
if(s.equals(s2))return true;
else
return false;
}
public static void main(String[] args)
{
System.out.println(isRotated("amazon","azonam",2));
}
}

- JAI March 12, 2015 | Flag Reply
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0
of 0 vote

public class StringRotation
{
public static boolean isRotated(String s1, String s2,int rotationCount)
{
if(s1.length()!=s2.length())
{
return false;
}
String s=s1.substring(rotationCount)+s1.substring(0,rotationCount);
if(s.equals(s2))return true;
else
return false;
}
public static void main(String[] args)
{
System.out.println(isRotated("amazon","azonam",2));
}
}

- JAI March 12, 2015 | Flag Reply
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0
of 0 vote

public class RoatingDemo {

	public static void main(String[] args) {
		TestRoatation("amazon","azonam");
		TestRoatation("quality","lityqua");
		
	}

	
	public static void TestRoatation(String actualData, String rotationData)
	{
		
		String rotationlogicalData="";
		int roation=2;
		for(int i=roation;i<actualData.length();i++)
		{
			rotationlogicalData=rotationlogicalData+actualData.charAt(i);
		}
		
		for(int i=0;i<roation;i++)
		{
			rotationlogicalData=rotationlogicalData+actualData.charAt(i);
		}
		System.out.println(rotationlogicalData);
		
		
		if(rotationlogicalData.equals(rotationData))
		{
			System.out.println("pass");
		}
		else
		{
			System.out.println("fail");
		}
	}
}

- Anonymous November 03, 2016 | Flag Reply
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0
of 0 vote

ublic class RoatingDemo {

	public static void main(String[] args) {
		TestRoatation("amazon","azonam");
		TestRoatation("quality","lityqua");
		
	}

	
	public static void TestRoatation(String actualData, String rotationData)
	{
		
		String rotationlogicalData="";
		int roation=2;
		for(int i=roation;i<actualData.length();i++)
		{
			rotationlogicalData=rotationlogicalData+actualData.charAt(i);
		}
		
		for(int i=0;i<roation;i++)
		{
			rotationlogicalData=rotationlogicalData+actualData.charAt(i);
		}
		System.out.println(rotationlogicalData);
		
		
		if(rotationlogicalData.equals(rotationData))
		{
			System.out.println("pass");
		}
		else
		{
			System.out.println("fail");
		}
	}
}

- Anonymous November 03, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void main(String[] args) {
		String s1 = "amazon", s2 = "azonam";
		boolean matchFound = false;
		
		for(int i=0; i<s2.length(); i++) {
			for(int j=0; j<s2.length(); j++) {
				if(i!=j) {
					String strTemp = "" + s2.charAt(i) + s2.charAt(j) + collectRestOfWords(s2, i, j);
					if(strTemp.equals(s1)) {
						System.out.println("Word " + strTemp + " matched!");
						matchFound = true;
						break;
					}
				}
			}
			if(matchFound) {
				return;
			}
		}
		System.out.println("No match found");
	}

- GK May 03, 2019 | Flag Reply
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0
of 0 votes

Adding the missing method collectRestOfWords

public static String collectRestOfWords(String input, int pos1, int pos2) {
		String strTemp = "";
		for(int i=0; i<input.length(); i++) {
			if(i!= pos1 && i!=pos2) {
				strTemp += "" + input.charAt(i);
			}
		}
		return strTemp;
	}

- GK May 03, 2019 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

What if he asks for n places instead of 2?

- Shubham January 16, 2020 | Flag Reply


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