Question

Anagram Difference We define an anagram to be a word whose characters can be rearranged to create another word. Given two strings, we want to know the minimum number of characters in either string that we must modify to make the two strings anagrams. If it's not possible to make the two strings anagrams, we consider this number to be -t For example: tea and ate are anagrams, so we would need to modity 0 characters tea and toe are not anagrams, but we can modify 1 character in either string (o-a ora-o) to make them anagrams act and acts are not anagrams and cannot be converted to anagrams because they contain different numbers of characters Function Description Complete the function make the two strings anagrams If it's not possible, return -1 e in the editor below. The function must retun an array of integers which denote the minimum number of characters in either string that need to be modified to a: the first string b: the second string .1sns 100

Answer 1

A simple strategy is used for calculating the minimum difference to make anagram i.e. count the characters in the first string and then iterate through the second string and minus the similar characters and add the remaining characters . this strategy is used in the following program

Here is the code:

#include<bits/stdc++.h>
using namespace std;
vector<int> getMinimumDifference(std::vector<string> a,std::vector<string> b)
{
std::vector<string> ::iterator ita;
std::vector<string> ::iterator itb;
std::vector<int> difference;
for(ita=a.begin(),itb=b.begin();ita!=a.end() && itb!= b.end(); ita++ , itb++)
{
string s1=*ita;
string s2=*itb;
int count = 0;
  
// store the count of character
int char_count[26];
//intialize with the 0   
for (int i = 0; i < 26; i++){
char_count[i] = 0;
}
if(s1.length()!=s2.length()) // when lengths are not equal then we can not make anagram so push -1
{
difference.push_back(-1);
continue;
}
// iterate though the first String
// and update count
for (int i = 0; i < s1.length(); i++)
char_count[s1[i] - 'a']++; // this s1[i]-'a' will map a to 0
  
// iterate through the second string
// update char_count.
// if character is not found in
// char_count then increase count
for (int i = 0; i < s2.length(); i++){
char_count[s2[i] - 'a']--;
if (char_count[s2[i] - 'a'] < 0)
count++;
}
difference.push_back(count);

}
return difference;

}
int main()
{
std::vector<string> a;
std::vector<string> b;
// some dummy data for testing
a.push_back("tea");
b.push_back("toe");
a.push_back("ast");
b.push_back("astb");
a.push_back("anki");
b.push_back("hell");

std::vector<int> difference=getMinimumDifference(a,b);
std::vector<int>:: iterator it;
std::vector<string> ::iterator ita;
std::vector<string> ::iterator itb;
for(it=difference.begin(),ita=a.begin(),itb=b.begin();it!=difference.end();it++,ita++,itb++)
cout<<*ita<<"\t"<<*itb<<"\t"<<*it<<"\n";
}

here is the screenshot of the program

1 #include<bits/stdc++.h> 2 using namespace std; 3 vector<int getMinimumDifference(std::vector<string> a,std::vector<string> b) std::vector<string> ::iterator ita; std::vector<string> ::iterator itb; std: : vector<int> difference; 8 for (ita-a.begin),itb-b.begin();ita!-a.end() && itb!- b.end(); ita++, itb++) 6 string s1-*ita; string s2-*itb; int count = 0; //store the count of character int char_count [26]; //intialize with the 0 6 for (inti-e; i < 26; i++){ char_count[i] -; if (s1.length)!-s2.length()) // when lengths are not equal then we can not make anagram difference.push_back (-1); continue // iterate though the first string // and update count for (int í 0; i < s1. length(); i++) 6 char_count[s1[i] a ++; I/ this s1[i]-'a will map a to e 8 9 //iterate through the second string // update char_count. /if character is not found in // char count then increase count for (int i = 0; i < s2.length(); i++){ 4 char_count[s2[i] - a']--; if (char_count[s2[i] - a'] <0) 6 count++; 8

9 30 // iterate through the second string // update char count // if character is not found in / char count then increase count for (int İ-0; i < s2.length(); і++){ char_count[s2[i] a ]--; if (char_count[s2[i] ]< 0) count++; difference.push_back (count); 39 10 12 return difference; 14 15 int main() 16 std::vector<string> a; std::vector<string> b; //some dummy data for testing 0 a.push_back("tea") 1 b.push_back("toe"); 2 a.push_back("ast") b.push_back("astb"); 4 a.push_back("anki"); 5 b.push back("hell"); 56 7 std::vector<int> difference-getMinimumDifference (a,b); 58 59 50 1 for(it-difference.begin),ita-a.begin),itb-b.begin();it!-difference.end();it++,ita++,itb++) std::vector<int: iterator it; std; : vectorくstring» ;iterator ita; std::vector<string> ::iterator itb; cout<<*ita<<"\t"<<*itb<<"t"<<*it<<"\n"; Activate Windows Go to Settings to activate W

here is the output of the program:

PS D:\mingw-w64\1686-8.1.0-posix-dwarf-rt-v6_rev0\mingw32\bin> ./g++ digi .cpp PS D: \mingw-w64 1686-8.1.0-posix-dwarf-rt_v6-revo\ mingw32\bin> ./a.exe tea ast anki he11 PS D:\mingw-w64\1686-8.1.0-posix-dwarf-rt-v6_revo\mingw32\bin>- 1 toe astb-1 4

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