Amazon Interview Question for Quality Assurance Engineers


Country: India
Interview Type: Written Test




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Always remember that whenever a string question is thrown at you, just remember this "A string is noting but an array of character that ranges from 0-256 chars(assuming all the char are alphabets)"

Now to the question,
you can sort the first and sort the second using library function,
and compare each char by char
Time Complexity -> O(n*logn)
## If len(str2) != len(s1) -> return false, strings of uneven lengths can never be anagrams

- Gaurav May 11, 2019 | Flag Reply
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A byte, 8 bits, can assume values from 0 to 255. Therefore having 256 unique possible values. Not "0-256" as mentioned

- Felipe May 28, 2019 | Flag
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you can just compute the hash function if they have the same characters they will result in the same hash value. which can be done in O(n)

- brahma August 13, 2019 | Flag
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import java.util.*;
class Anagrams {

  static boolean isAnagram(String a, String b) {
        char c[] = a.toLowerCase().toCharArray();
        char d[] = b.toLowerCase().toCharArray();
        Arrays.sort(c);
        Arrays.sort(d);
        if(Arrays.equals(c,d)){
            return true;
        }
        return false;
    }



  public static void main(String[] args) {
     Scanner scan = new Scanner(System.in);
    System.out.println("Enter first String: ");
    String a = scan.next();
    System.out.println("Enter seconf jumbled String");
    String b = scan.next();
    scan.close();
    boolean ret = isAnagram(a, b);
    System.out.println( (ret) ? "Anagrams" : "Not Anagrams" );
    }
}

- Sameer May 11, 2019 | Flag Reply
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this would increase the time complexity to O(nlogn).

- Naren May 30, 2019 | Flag
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For checking that two strings are an anagram of each other, we can use an unordered set in c++ and insert all the characters of the first string and then iterate over the second string and keep checking that the characters in second string must be present in the set if not then both strings are not anagrams of each other else they are.

Implementation:

#include<bits/stdc++.h>
using namespace std;
bool findanagram(string s1, string s2){
unordered_set<char> s;
int n1 = s1.length();
int n2 = s2.length();
transform(s1.begin(), s1.end(), s1.begin(), ::tolower);
transform(s2.begin(), s2.end(), s2.begin(), ::tolower);
for(int i = 0; i < n1; i++)
s.insert(s1[i]);
for(int j = 0; j < n2; j++){
if(s.find(s2[j]) == s.end())
return false;
}
return true;
}
int main()
{
string str1 = "LISTEN";
string str2 = "silent";
cout<<findanagram(str1, str2)<<endl;

return 0;
}

- swapnilkant11 May 29, 2019 | Flag Reply
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public boolean isAnagram(String s, String t) {
        if (s == null || t == null || s.length() != t.length()) {
            return false;
        }
        char[] map = new char[256];
        for (char c : s.toCharArray()) {
            map[c]++;
        }

        for (char c : t.toCharArray()) {
            map[c]--;
            if (map[c] < 0) {
                return false;
            }
        }
        return true;
    }

- naren May 30, 2019 | Flag Reply
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// Solution in Javascript

function anagramCheck(str,strg) {
   var arr =[];
   var arr2 = [];

   a = str.toLowerCase();
   b = strg.toLowerCase();

   arr = a.split('').sort();
   arr2 = b.split('').sort();

 if (arr.length === arr2.length) {
    for (i=0;i<arr.length; i++) {
      if (arr[i] !==arr2[i]) {
          return false;
      }
    }
    return true;
 } else {
     return false;
 }
}

- harshsheetal18 June 16, 2019 | Flag Reply
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#include<vector>
#include<algorithm>
#include<string>
#include<set>

using namespace std;
typedef vector <int> VI;
int main()
{
//code
int t;
cin >> t;
while (t--)
{
string s,st;
int f = 0;
cin >> s;
cin >> st;
VI v1(26, 0);
VI v2(26, 0);
for (int i = 0; i < s.length(); i++)
{
v1[s[i] - 'a']++;
v2[st[i] - 'a']++;
}
for (int i = 0; i < 26; i++)
{
if (v1[i] != v2[i])
{
f = 1;
break;
}
}
if (f == 1)
cout << "NO\n";
else
cout << "YES\n";
}
return 0;
}

- ravircit August 20, 2019 | Flag Reply


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