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Westkost [7]
3 years ago
13

Two strings, and , are called anagrams if they contain all the same characters in the same frequencies. For example, the anagram

s of CAT are CAT, ACT, TAC, TCA, ATC, and CTA. Complete the function in the editor. If and are case-insensitive anagrams, print "Anagrams"; otherwise, print "Not Anagrams" instead.
Computers and Technology
1 answer:
GenaCL600 [577]3 years ago
8 0

Answer:

static boolean isAnagram(String a, String b) {

<em> </em><em> </em><em>// 1 - Strings inequal in length can never be Anagram</em>

        if (a.length() != b.length()) {

             return false;

        }

<em> </em><em>        // 2 - Convert both Strings to Lower Case</em>

       a = a.toLowerCase();

        b = b.toLowerCase();

<em> </em><em>        // 3 - Create an Array to store character count</em>

        int charCount[] = new int[26];

<em>  </em><em>// 4 - Count Each Character</em>

        for (int i = 0; i < a.length(); i++) {

             charCount[a.charAt(i) - 97]++;

             charCount[b.charAt(i) - 97]--;

        }

       

<em>        </em><em> </em><em>// 5 - Check  for mismatching characters</em>

        for (int i = 0; i < charCount.length; i++) {

             if (charCount[i] != 0) {

                 return false;

            }

        }

       return true;

}

Explanation:

Complete Questions:

Two strings, a and b, are called anagrams if they contain all the same characters in the same frequencies. For example, the anagrams of CAT are CAT, ACT, TAC, TCA, ATC, and CTA. Complete the function in the editor. If and are case-insensitive anagrams, print "Anagrams"; otherwise, print "Not Anagrams" instead.

Function:

static boolean isAnagram(String a, String b) {

       // Complete the function

}

This algorithm has five steps to calculate whether two given strings are anagram or not.

  1. Check Strings length - If two strings don't have equal length, then they can never be anagrams. Because, anagrams contain <em>same characters in the same frequencies.</em>
  2. Convert both strings to lower case - in programming, <em>'A' is not equal to 'a'</em>. So, we need to make sure we have letters in same case to avoid such errors.
  3. Create an Array for character count - We need some method to ensure that the strings are anagrams. So, here we will count the characters and decide if the strings are anagrams (as discussed in next point)
  4. Count each character - If <u><em>String a</em></u><em> </em>has any character, we will <u><em>increase</em></u> the count of that character by 1. If <u><em>String b</em></u> has any character, we will <u><em>decrease</em></u> the count of that character by 1. At the end, if both strings have same characters in the same frequencies, all our character counts will be equal to 0.
  5. Check for mismatching characters - We check if all our character counts are equal to 0. If not, the function will return false, otherwise it will return true.
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Write a loop that reads c-strings from standard input where the c-string is either "land", "air", or "water". The loop terminate
Anvisha [2.4K]

Answer:

I am writing a C++ code.        

#include <iostream> // for input output functions

using namespace std; // identifies objects like cin cout

int main() { //start of main() function body

string c_string; // stores the string entered by user from land water or air

int land=0; // contains the number of times land string is read in

int air=0; //contains the number of times air string is read in

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/*while loop continues to execute until user enters xxxxx or maximum length of a string exceeds 8 */

while(c_string!= "xxxxx" && c_string.length()<=8) {

if(c_string=="land"){ // if string entered by user is land

land = land + 1;} //counts and increments each occurrence of land string by 1

else if(c_string=="air"){// if string entered by user is air

air = air + 1;}// counts and increments each occurrence of string air by 1

else if(c_string=="water"){// if string entered by user is water

water = water + 1;}//counts and increments each occurrence of air by 1

cin>>c_string;}

/* keeps reading the string entered by user from land air or water, until the loop breaks after the user enters xxxxx or user enters a string whose length is greater than 8 */

//prints the number of times land, air and water are read in

cout << "land:"<<land;

cout << endl<<"air:"<<air;

cout << endl<< "water:"<<water; }                      

                                                                                     

Explanation:

Everything is well explained in the comments above.

The program prompts the user to input strings. These strings are either land air or water. The while loop continues to read the input strings until user enters xxxxx or the string entered by user exceeds the length 8. Both these terminating conditions are added in the while loop. After the loop terminates, the number of times land, air and water strings are read is displayed on the output screen. Any other string entered by user other than these 3 is ignored. The program along with the output is attached.

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</span>
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I hope the answer will help you. Thank you. 

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