Amazon Interview Question for Quality Assurance Engineers


Team: QAE 1
Country: India
Interview Type: In-Person




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Use a hash table.. loop through the char array and for each char insert the character as key and count as the value...

- VVS February 13, 2012 | Flag Reply
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map<char, int> countChar(char* str) {
    for(int i=0; str[i]; i++) {
        cnt[str[i]]++;
    } 
}

- !@$#@! February 13, 2012 | Flag Reply
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map<char, int> countChar(char* str) {
map<char, int> map;
for(int i=0; str[i]; i++) {
cnt[str[i]]++;
}
}

- #$#@% February 13, 2012 | Flag
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0
of 0 vote

THe easiest way is to use integer array with length 256 and update the value as we enounter each character.
int [] array = new int[256]
char charArray[] = str.toCharArray();
foreach(char c in charArray)
{
array[c]++;
}

- VVS February 13, 2012 | Flag Reply
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This is the classic solution from the times when a character was 8 bits. These days you should consider Unicode, too.

- Selmeczy, Péter February 13, 2012 | Flag
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For unicode array len would becm 2^16 since unicode char hv max len of 16 bits ....(hence it is good to check whether inp string is ascii or unicode)

- saurabh February 13, 2012 | Flag
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THe easiest way is to use integer array with length 256 and update the value as we enounter each character.
int [] array = new int[256]
char charArray[] = str.toCharArray();
foreach(char c in charArray)
{
array[c]++;
}

- VVS February 13, 2012 | Flag Reply
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0
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package com.amazon.techinterview;

import java.util.Hashtable;
import java.util.Iterator;
import java.util.Map;
import java.util.Set;
import java.util.Map.Entry;

public class NoOfChars {
public static void main (String args[]){
Hashtable<Character, Integer> myHash = new Hashtable<Character, Integer>(26);
String myString = new String("gbvsyuxfasyusdfaxsasghdvsgcydegfrdywegvfwehyicv");
int length = myString.length();
for (int i=0; i<length; i++){
char ch = myString.charAt(i);
if (!myHash.containsKey(ch)){
myHash.put(ch,1);
}
else {
int presentCount = myHash.get(ch);
myHash.remove(ch);
myHash.put(ch, presentCount + 1);
}
}
Set<Entry<Character, Integer>> mySet = myHash.entrySet();
Iterator it = mySet.iterator();

while (it.hasNext()){
Map.Entry entry = (Map.Entry) it.next();
System.out.println(entry.getKey() + " : " + entry.getValue());
}
}
}

- Nilanjan Sil February 13, 2012 | Flag Reply
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int numberOfOccuranceOfPattern(char * str, char * pattern)
{
	int numOcc = 0;
	int i = 0;
	int j = 0;
	while(str[i] != '\0')
	{
		if(pattern[j] == str[i])
		{
			while(pattern[j] != '\0')
			{
				if(pattern[j] != str[i])
				{
					j = 0;
					break;
				}				

				j++;
				i++;

				if((pattern[j] == '\0') && ((str[i] == ' ') || (str[i] == '\0')))
				{
					numOcc++;
					j = 0;
				}
			}
		}
		else
		{
			i++;
		}
	}
	return numOcc;
}

- biswaranjan February 13, 2012 | Flag Reply
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import java.util.*;

public class B {
public static void main(String[] args) throws Exception{

Map<Character, Integer> map = new TreeMap<Character, Integer>();
String input = "Varun Saxena";

for (int i = 0; i < input.length(); i++) {

char in = input.charAt(i);
if(in == ' ' )continue;
if(map.get(in) != null){
int count = map.get(in);

map.put(in, ++count);

}else {
map.put(input.charAt(i), 1);
}
}

System.out.print(map);


}
}

- saxenavarun61 February 16, 2012 | Flag Reply
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def charCount(s):
  m = {}
  for c in s:
    if c in m:
      m[c] += 1
    else: m[c] = 1
  for k in m.keys():
    print str(k)+":"+str(m[k])

- amshali February 22, 2012 | Flag Reply
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of 0 vote

import java.util.Hashtable;
public class CountCharOcurrence {
	public static void main(String[] args) {
		String word = "PROGRAM";
		char[] charArray = word.toCharArray();
		Hashtable<Character, Integer> hashtable = new Hashtable<Character, Integer>();
		for (char c : charArray) {
			System.out.println(c);
			if (hashtable.get(c) != null) {
				hashtable.put(c, hashtable.get(c) + 1);
			} else {
				hashtable.put(c, 1);
			}
		}
		System.out.println(hashtable.entrySet());
	}
}

This is my approach.

- sudhi.joshi April 22, 2012 | Flag Reply
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0
of 0 vote

please see my approach.

String name="rnagabandi";
HashMap<Character, Integer> hashMap = new HashMap<Character, Integer>();

int temp=0;
hashMap.put(name.charAt(0),1);

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

if(hashMap.get(name.charAt(i))!=null){
temp=hashMap.get(name.charAt(i));
hashMap.put(name.charAt(i),temp+1);
}else{
hashMap.put(name.charAt(i),1);
}

temp=0;
}

System.out.println(hashMap);

- raghu.nagabandi July 16, 2012 | Flag Reply
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of 0 vote

char string [] = "Test";
int Count = 1;
char Temp [] = string;
int i,j;


while ( temp[i] = '\0');
while(string[j] = '\0'){
if (temp[i] = string[j]){
count++;
}
j++;
}
cout<<temp[i]<<"occurance"<<count;
i++;
count = 1;
j = i;
}

- Mahes Karthic August 20, 2012 | Flag Reply
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of 0 vote

public static void main(String args[]){
String sat = "test";
sat =sat.toUpperCase();
int num = 0;

char[] sat1=sat.toCharArray();

for(int j=0;j<sat1.length;j++){
int count = 0;
for(int i=0;i<sat1.length;i++){
// System.out.print(sat1[i] + " ");

if(sat1[num]==sat1[i]){
count = count +1;

}

}


System.out.println(sat1[num] + " " + count);




num=num+1;
}


}

- sathishwaran January 06, 2013 | Flag Reply
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of 0 vote

# include <stdio.h>
# include <conio.h>

void main ()
{
int i,j,arr[256],size;
char *text;

printf("Enter the text : ");
scanf("%s",text);
i = 0;

for (i=0;i<255;i++)
{
arr[i]=0;

}
i=0;

while(text[i] != '\0')
{
	arr[(int)text[i]]++;
	i++;
}

for(i=0;i<255;i++)
{
 if( arr[i]>0)
 {
	printf("%c : %d\n",(char)i,arr[i]);
 }
}

printf("\n%s",text);


getch ();
clrscr();
}

- abhi.20dec September 26, 2013 | Flag Reply
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0
of 0 vote

Ruby Code

#finding number of occurance of the characters

data = String.new("good")
cnt = Array.new

l= data.length
for i in 0...l
  cnt[i]= data.count(data[i])  #counting characters
 end 
for k in 0...l
   puts  "Char #{data[k]} displayed  #{cnt[k]} times"
end

- Nitin Nazare November 20, 2013 | Flag Reply
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0
of 0 vote

public class CountCharOccurance
{

public static void main(String[] args)
{

String value = "Test@Tutor";

int[] count = new int[256];

value = value.toLowerCase();

for (int i = 0; i< value.length(); i++)
{

count[(int)value.charAt(i)]++;
}

List<Character> printed = new ArrayList<Character>();

for (int i = 0; i< value.length(); i++)
{

if (printed.contains(value.charAt(i)))
continue;

printed.add(value.charAt(i));

System.out.println("Occurrence of char [" + value.charAt(i) + "] - " + count[(int)value.charAt(i)]);
}
}

}

- Raghav July 29, 2014 | Flag Reply


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