Amazon Interview Question Software Engineer / Developers


Country: India
Interview Type: In-Person


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2
of 2 vote

static int test1(string str1, string str2)
        {
            int i = 0;
            int j = 0;
            int count = 0;
            while (i < str1.Length)
            {
                if (str1[i] == str2[j])
                {
                    j++;
                    if (j == str2.Length)
                    {
                        count++;
                        j = 0;
                    }
                }
                else
                {
                    j = 0;
                }
                i++;
            }

            return count;
        }

- Sami Alam on April 14, 2012 | Flag Reply
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0
of 0 votes

can only find 1 when compare 'adffuckddfuckdffuck' with 'fuck'

- Anonymous on December 13, 2013 | Flag
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1
of 1 vote

One solution which would be useful in case the string is quite large is to use a suffix array
1. Create the suffix array in O(n)
2. Do a binary search on the suffix array for the string in O(lgn)

Due to the suffix array structure, all the suffixes which start with the given substring will be bunched together

So finally the time complexity is O(n)

- Renjith on April 14, 2012 | Flag Reply
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0
of 0 votes

Its a suggestion to u that never take the word "suffix array" or "suffix tree" in an interview until u r confident enough to code it.

- Anonymous on April 14, 2012 | Flag
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0
of 0 votes

Can you elaborate on how the get the occurrences of the substring ?

- GingerBreadMan on April 18, 2012 | Flag
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0
of 0 vote

can you give a example?
Not quit get the quesiton

- caxieyou on April 13, 2012 | Flag Reply
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0
of 0 votes

suppose in amazonam ,how much time am is occur..........not a big deal.

- calgary on April 13, 2012 | Flag
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0
of 0 votes

you are right!!

- vivekanand3435 on April 13, 2012 | Flag
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0
of 0 votes

can be done with KMP

- gimli on April 13, 2012 | Flag
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0
of 0 votes

One doubt i have is, for ex if we have the String "aaaa" and the subString is "aa", should the final output be 2 or 3

i.e do we need to count chatAt 1,2 as a valid sequence?

- balakumar7 on April 17, 2012 | Flag
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0
of 0 vote

#include<stdio.h>
#include<string.h>
#include<conio.h>
int main()
{char a[50];
char b[50];
int len1,len2,i,j,k;
gets(a);     

gets(b);
len1=strlen(a);
len2=strlen(b);
i=0;
while(i<len1)
{          j=0;
           if(a[i]==b[j])
           { k=i;
             while(a[k]==b[j]&&j<len2&&k<len1)
             {k++;j++;continue;}
             if(j==len2)
              printf("found at %d\n",i+1); 
             }
             i++;
             
             }
             getch();
             return 0;
             }

- mac on April 13, 2012 | Flag Reply
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0
of 0 votes

The complexity for the above code seems to be o(mn).
In this brute force itself we can put a check to run the loop only till (m-n+1). So complexity can be O((m-n+1).n).

- Raj on April 13, 2012 | Flag
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0
of 0 vote

#include<stdio.h>
#include<string.h>
#include<conio.h>
int main()
{char a[50];
char b[50];
int len1,len2,i,j,k;
gets(a);     

gets(b);
len1=strlen(a);
len2=strlen(b);
i=0;
while(i<len1)
{          j=0;
           if(a[i]==b[j])
           { k=i;
             while(a[k]==b[j]&&j<len2&&k<len1)
             {k++;j++;continue;}
             if(j==len2)
              printf("found at %d\n",i+1); 
             }
             i++;
             
             }
             getch();
             return 0;
             }

- mac on April 13, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Any better solution ??

- Tapesh Maheshwari on April 13, 2012 | Flag Reply
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0
of 0 votes

public class SubStringChecker {

	int count = 0;
	
	public static void main(String[] args) {
		
		SubStringChecker ssc = new SubStringChecker();
		ssc.go();
	}
	
	public void go(){
		
		String s1s = "asdbqasdhvasdfasd";
		String s2s = "asd";
		
		
		char[] s1 = s1s.toCharArray();
		char[] s2 = s2s.toCharArray();
		int length = s2.length;
		
		for(int i=0; i<s1.length; i++){
			if(s1[i] == s2[0] && i+length<=s1.length){
				if(s1s.substring(i, i + length).equals(s2s)){
					count++;
				}
			}
		}
		System.out.print(count);
	}

}

- yeswanth.devisetty on April 17, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

Trivial program to write in java...

public static void getOccurances (String a, String b) {
	int index = a.indexOf(b);
	while(index > -1) {
		System.out.println("String found at index: "+index);
		index = a.indexOf(b, index+b.length());
	}
}

- Anonymous on April 13, 2012 | Flag Reply
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0
of 0 votes

Fails for repeating substrings. getOccurances("aaa","aa") will only print 0. It should also print 1.

- yossee on April 14, 2012 | Flag
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0
of 0 vote

static int test1(string str1, string str2)
{
int i = 0;
int j = 0;
int count = 0;
while (i < str1.Length)
{
if (str1[i] == str2[j])
{
j++;
if (j == str2.Length)
{
count++;
j = 0;
}
}
else
{
j = 0;
}
i++;
}

return count;
}

- Anonymous on April 14, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

static int test1(string str1, string str2)
{
int i = 0;
int j = 0;
int count = 0;
while (i < str1.Length)
{
if (str1[i] == str2[j])
{
j++;
if (j == str2.Length)
{
count++;
j = 0;
}
}
else
{
j = 0;
}
i++;
}

return count;
}

- Sami Alam on April 14, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

//complexity n*m
public static void main(String args[]){
String s="ABCAABCDBCEBC";
String sub ="BC";

int m = s.length();
int n = sub.length();
int count=0;
for(int i=0;i<m-n;i++){
int j=0;
while(j<n && s.charAt(i+j)==sub.charAt(j)){
j++;
}
if(j==n){
count++;
}
}
System.out.println("no of occurrence"+count);
}

- Nitin Agarwal on April 14, 2012 | Flag Reply
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0
of 0 vote

public static void main(String[] args) 
    {
        try
        {
            String substring, text;
            int count = 0, length, i = 0, index, prevIndex = -1;
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            text = br.readLine();
            substring = br.readLine();
            length = text.length();            
            for(i = 0; i<length; i++)
            {
                if((index = text.indexOf(substring, i)) != -1)
                {                    
                    if(index != prevIndex)
                        count += 1;                   
                    prevIndex = index;
                }                
            }
            System.out.println(count);                  
        }
        catch(Exception e)
        {
            e.printStackTrace();
        }

}

- Jagdish on April 14, 2012 | Flag Reply
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0
of 0 vote

public static int count(String needle, String haystack) {
        int length = haystack.length();
        int needleLength = needle.length();
        int i = 0;
        int j = 0;
        int count = 0;
        while (i < length) {
            if (j >= needleLength - 1) {
                j = 0;
                count++;
            }
            if (needle.charAt(j) == haystack.charAt(i)) {
                j++;
            }
            i++;
        }
        return count;
    }

- Anonymous on April 15, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class SubStringChecker {

	int count = 0;
	
	public static void main(String[] args) {
		
		SubStringChecker ssc = new SubStringChecker();
		ssc.go();
	}
	
	public void go(){
		
		String s1s = "asdbqasdhvasdfasd";
		String s2s = "asd";
		
		
		char[] s1 = s1s.toCharArray();
		char[] s2 = s2s.toCharArray();
		int length = s2.length;
		
		for(int i=0; i<s1.length; i++){
			if(s1[i] == s2[0] && i+length<=s1.length){
				if(s1s.substring(i, i + length).equals(s2s)){
					count++;
				}
			}
		}
		System.out.print(count);
	}

}

- yeswanth.devisetty on April 17, 2012 | Flag Reply
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0
of 0 votes

public class SubStringOccur1 {
int count = 0;
public static void main(String[] args) {
SubStringOccur1 ssc = new SubStringOccur1();
ssc.go();
}
public void go(){
String mainStr="abcghjabcjkdabklabcklabcdab";
String subStr="abc";
boolean s1=mainStr.startsWith(subStr);
boolean s2=mainStr.endsWith(subStr);
int count=0;
String[] abc=mainStr.split(subStr);
if(s1 &&!s2){
count=abc.length-1;
}else{
count=abc.length;
}
System.out.println(Arrays.toString(abc));
System.out.println("occurances=="+(count));
}

}

- Anonymous on May 30, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

KMP
bayer moore
algoithm's direct implementation

- shivirocks2909 on October 02, 2012 | Flag Reply


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