## Amazon Interview Question for Front-end Software Engineers

Country: India
Interview Type: Written Test

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

Implementing a hexdec counter.
Assumption: Max value is FFFFFFFFF The counter will reset when it reaches this max value.

``````char IncrementChar(char ch)
{
switch (ch)
{
case '0':
case '1':
case '2':
case '3':
case '4':
case '5':
case '6':
case '7':
case '8':
return (char) ((int) ch + 1);
break;
case '9':
return 'A';
break;
case 'A':
case 'B':
case 'C':
case 'D':
case 'E':
return (char) ((int) ch + 1);
break;
case 'F':
return '0';
break;
default:
return ' ';
break;
}
}

char* IncrementString (char *str)
{
if( !strcmp(str,"FFFFFFFFF"))
return "000000000"; // Reset the counter after it reaches max

for (int i=8; i>=0;i--)
{
str[i] = IncrementChar(str[i]);
if ( str[i] != '0')
return str;
}
return str;
}

int _tmain(int argc, _TCHAR* argv[])
{

char num[10] = "000000000";
// Print value till FFF
while(strcmp(num,"000000FFF") != 0)
printf("%s\n" , IncrementString(num));
return 0;
}``````

Comment hidden because of low score. Click to expand.
1
of 1 vote

Is it hexadecimal ?. If it isnt then the max value of counter be ZZ9ZZ9999

Comment hidden because of low score. Click to expand.
0
of 0 vote

The given string contains Hex values hence convert it to Decimal values fist then increment the value again convert it back to Hex. You will get the answer. Simple.

Comment hidden because of low score. Click to expand.
0
of 0 vote

create a map from single "digit" n to n+1
increase the tail digit, if overflow, recursively increase the previous digit.
The solution can handle unlimited length strings.
js code:

``````var map = {
0: "1",
1: "2",
...
C: "D",
D: "0"
}
function increase(str) {
var array = [];
for(var i=str.length-1; i >= 0;i--){
array.push(str.charAt(i));
}
array = increaseAt(array, 0);

return  array.reverse().join("");
}

function increaseAt(array, index) {
var d = array[index];
var next = map[d];
array[index] = next;

if(next === "0") {// overflow
var nextIndex = index+1;
if(nextIndex  >= array.length) {
array.push("1");
} else {
array = increaseAt(array, nextIndex);
}
}
return array;
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

Very ambigious question . Please provide with proper requirements

Comment hidden because of low score. Click to expand.
0
of 0 vote

public static String couterString(String s, int length){
char[] schar = s.toCharArray();
char c = schar[length-1];
if(c == '9'){
c = 'A';
schar[length -1] = c;
return String.valueOf(schar,0,length);
}else if(c == 'F'){
c= '0';
return couterString(s, length-1) + String.valueOf(c);
}else{
c++;
schar[length -1] = c;
return String.valueOf(schar,0,length);
}
}

Comment hidden because of low score. Click to expand.
0
of 0 vote

assuming that character values will be unchanged, i.e. max value will be AB9CD9999

``````public class StringCounter {

public static void main(String[]args)
{
String input = "AB0CD00000", tmp ="", result="";

int number = 0;

for(int i=0;i<input.length();i++)
{
if(Character.isDigit(input.charAt(i)))
{
tmp+=Character.toString(input.charAt(i));
}
}

tmp ="1"+tmp;

number = Integer.parseInt(tmp);

if(number+1<2*Math.pow(10, tmp.length()-1))
number++;

tmp = Integer.toString(number);

int j= 1;

for(int i=0;i<input.length();i++)
{
if(Character.isDigit(input.charAt(i)))
{
result+= tmp.charAt(j);
j++;
}
else
{
result+= input.charAt(i);
}
}

System.out.println(result);
}
}``````

Comment hidden because of low score. Click to expand.
1
of 1 vote

Nice Answer Dude ..But i mentioned about the ALGORITHM,u forgot that..anyway gooddddddddddd

Comment hidden because of low score. Click to expand.
0

Nice dude but i want to increment the character also for examples if input AB0CD9999 then the output is AB0CE0000 please help me

Comment hidden because of low score. Click to expand.
0
of 0 vote

public partial class Form1 : Form
{
ArrayList arr = null;
string sv = string.Empty;
public Form1()
{
InitializeComponent();
sv = "AB0CD1010";
int iv = sv.Length;
arr = new ArrayList();
foreach (char ch in sv)
{
}
}

private void button1_Click(object sender, EventArgs e)
{
int inum = 0;
bool bOnce = false;
for(int i=0;i<=arr.Count-1;i++)
{
bool bNum = int.TryParse(arr[i].ToString(), out inum);
if (bNum)
{
int iFirst=0;
bool bFirst = int.TryParse(arr[i - 1].ToString(), out iFirst);
if (bFirst && !bOnce)
{
int ival = Convert.ToInt32(arr[arr.Count - 1].ToString());
int Inc = ival + 1;
arr[arr.Count - 1] = Inc.ToString();
bOnce = true;
}
}
textBox1.Text += arr[i].ToString();
}
textBox1.Text += "\r\n";
}
}

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````char[] incrementString(char str[])
{
int index = lengthOf(str)-1;
str[index]++;
while((str[index]==58 || str[index]==91) && index>=0) //58 is ASCII_9 + 1, 91 is ASCII_Z + 1
{
if(str[index] == 58) //If 1 is added to 9
{
str[index] = 65; // Set it to A
return str;
}
if(str[index] == 91) //If 1 is added to Z
{
str[index] = 48; // Set it to 0
if(index == 0)
{
char incStr[lengthOf(str)+1];
incStr = concatenateCharToArray('1', str); //Its trivial to write this function. It shall be omitted. For example if str is ABC234 All it does is retrun 1ABC234
return incStr;
}
else
str[--index]++;
}

}
}``````

Comment hidden because of low score. Click to expand.
0

This is assuming the string digits start from 0-9-A-Z.so its like this
0000....0009....000A.....000Z.....0010......001Z.....002Z............0ZZZ........1000...........ZZZZ........10000 etc....

Comment hidden because of low score. Click to expand.
0
of 0 vote

I think the poster should clarify whether he has hex or is it like the one I am taking about in the above code snippet

Comment hidden because of low score. Click to expand.
0
of 0 vote

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

int increment(int i)
{

if(i=='f')
return 0;
else
return (i+1);

}
int main()
{

string a;

int i=0,j;
cout<< "enter the string"<<endl;
cin>>a;

int k=a.length();

while(i<k)
{

j=increment((int)a[k-i-1]);

a[k-i-1]=(char)(j);
if(j==0)
{
a[k-i-1]=(char)48;
i++;
}
else break;
}
cout<<a<<endl;
return 0;
}

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````void Increment(string &s)
{
if(s.size()==0)
return;
int n=s.size()-1;

for(int i=n;i>=0;i--)
{
if(s[i]-'0'>=0 && s[i]-'0'<=8)
{
s[i]=((s[i]-'0')+1)+'0';
break;
}
else if(s[i]=='9')
{
s[i]='A';
break;
}
else if(s[i]>='A' && s[i]<='E')
{
s[i]++;
break;
}
else if(s[i]=='F')
{
s[i]='0';
if(i==0)
{
s='F'+s;
}
}
}
}

int main()
{
string s="0";//"AB0CD1010";

for(int i=0;i<300;i++)
{
Increment(s);
cout<<s<<endl;
}

return 1;``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

Number(parseInt("AB0CD1010",16)+1).toString(16)

Comment hidden because of low score. Click to expand.
0
of 0 vote

``Number(parseInt("AB0CD1010",16)+1).toString(16)``

Comment hidden because of low score. Click to expand.
0
of 0 vote

``Number(parseInt("AB0CD1010",16)+1).toString(16)``

Name:

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