Write C++ a function that determines if two strings are anagrams. The function s

Question

Write C++ a function that determines if two strings are anagrams. The function should not be case sensitive and should disregard any punctuation or spaces. Two strings are anagrams if the letters can be rearranged to form each other. For example, “Eleven plus two” is an anagram of “Twelve plus one”. Each string contains one “v”, three “e’s”, two “l’s”, etc… Test your function with several strings that are anagrams and non-anagrams. You may use either the string class or a C-style string.

Hint: Count the number of each letter for the two strings and compare the two “histograms”.

Hint: Use an array of size 26 for the histogram.

Sample output (bolded text denote input from user)

<program start>

Enter two strings.

eleven plus two

twelve plus one

They are anagrams!

<program end>

<program start>

Enter two strings.

Go Pokes

Go Sooners

They are not anagrams.

<program end>

Note: Your program must compile, Make sure that you are submitting your source code not only the excecutible file

Explanation / Answer

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

int   main()

{

char s1[20],s2[20];

cout<<"enter two strings"<<endl;

cin>>s1>>s2;

if(anagrams(s1,s2,strlen(s1),.strlen(s2)))

cout<<"given strings are anagrams"<<endl;

else

cout<<"given strings are not anagrams"<<endl;

}

int anagrams(char s1[],char s2[],int len1,int len2)

{
char ch;

int a[26]={0},b[26]={0},i,j,count;

for(i=0;i<len1;i++)

{

ch=s1[i];
ch=toupper(ch);

count=1;
for(j=1;j<len1;j++)

if(ch==s1[j])

count++;

a[toascii(ch)-65]=count;
}

for(i=0;i<len2;i++)

{

ch=s2[i];
ch=toupper(ch);

count=1;
for(j=1;j<len2;j++)

if(ch==s2[j])

count++;

a[toascii(ch)]=count;
}
for(i=0;i<26;i++)
if(a[i]!=b[i])
return 0;
return 1;
}

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