Anagram Checks - Amazon Top Interview Questions


Problem Statement :


Given two strings s0 and s1, return whether they are anagrams of each other.

Constraints

n ≤ 100,000 where n is the length of s0
m ≤ 100,000 where m is the length of s1


Example 1


Input
s0 = "listen"
s1 = "silent"


Output
True



Example 2

Input
s0 = "bedroom"
s1 = "bathroom"


Output
False


Solution :



title-img



                        Solution in C++ :

int primes[26] = {2,  3,  5,  7,  11, 13, 16, 17, 23, 29, 31, 41, 43,
                  47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101};

bool solve(string s0, string s1) {
    long long firstCode = 0;
    long long secondCode = 0;

    for (auto &c : s0) {
        firstCode += (c - 'a' + 1) * primes[(c - 'a')];
    }

    for (auto &c : s1) {
        secondCode += (c - 'a' + 1) * primes[(c - 'a')];
    }

    return firstCode == secondCode;
}
                    

                        Solution in Java :

import java.util.*;

class Solution {
    public boolean solve(String s0, String s1) {
        if (s0.length() != s1.length())
            return false;

        int[] store = new int[256];

        for (int i = 0; i < s0.length(); i++) {
            store[s0.charAt(i)]++;
            store[s1.charAt(i)]--;
        }

        for (int n : store)
            if (n != 0)
                return false;

        return true;
    }
}
                    

                        Solution in Python : 
                            
class Solution:
    def solve(self, s0, s1):

        freq = {}
        freq2 = {}

        if len(s0) != len(s1):
            return False

        for ch in s0:
            if ch in freq:
                freq[ch] += 1
            else:
                freq[ch] = 1

        for ch in s1:
            if ch in freq2:
                freq2[ch] += 1
            else:
                freq2[ch] = 1

        for key in freq.keys():
            if key not in freq2 or freq[key] != freq2[key]:
                return False

        return True
                    

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