Related
This function is meant to remove all special characters, numbers, and whitespace from the char array.
// Michael E. Torres II
// Vigenere Cipher
// February 4, 2018
// C++ code to implement Vigenere Cipher
#include <iostream>
#include <string>
#include <iomanip>
#include <fstream>
#include <algorithm>
#include <cctype>
#include <iterator>
#include <sstream>
#include <functional>
using namespace std;
// This function generates the key in
// a cyclic manner until it's length isi'nt
// equal to the length of original text
string generateKey(string str, string key)
{
int x = str.size();
for (int i = 0; ; i++)
{
if (x == i)
i = 0;
if (key.size() == str.size())
break;
key.push_back(key[i]);
}
return key;
}
// This function returns the encrypted text
// generated with the help of the key
string cipherText(string str, string key)
{
string cipher_text;
for (int i = 0; i < str.size(); i++)
{
// converting in range 0-25
int x = (str[i] + key[i]) % 26;
// convert into alphabets(ASCII)
x += 'A';
cipher_text.push_back(x);
}
return cipher_text;
}
// This function decrypts the encrypted text
// and returns the original text
string originalText(string cipher_text, string key)
{
string orig_text;
for (int i = 0; i < cipher_text.size(); i++)
{
// converting in range 0-25
int x = (cipher_text[i] - key[i] + 26) % 26;
// convert into alphabets(ASCII)
x += 'A';
orig_text.push_back(x);
transform(orig_text.begin(), orig_text.end(), orig_text.begin(), ::tolower);
}
return orig_text;
}
string removeNonAlpha(char *str)
{
unsigned long i = 0;
unsigned long j = 0;
char c;
while ((c = str[i++]) != '\0')
{
if (isalpha(c)) // this is where the breakpoint is automatically placed
{
str[j++] = c;
}
}
str[j] = '\0';
return str;
}
// Driver program to test the above function
int main(int argc, char *argv[])
{
string keyword = "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb";
stringstream ss;
char a[] = "“I think and think for months and years. Ninety-nine times, the conclusion is false. The hundredth time I am right.” – Albert Einstein “Imagination is more important than knowledge. For knowledge is limited, whereas imagination embraces the entire world, stimulating progress, giving birth to evolution.” – Albert Einstein";
int i = 0;
string str = removeNonAlpha(a);
str.append(512 - str.length(), 'X');
transform(str.begin(), str.end(), str.begin(), ::toupper);
transform(keyword.begin(), keyword.end(), keyword.begin(), ::toupper);
string key = generateKey(str, keyword);
string cipher_text = cipherText(str, key);
transform(cipher_text.begin(), cipher_text.end(), cipher_text.begin(), ::tolower);
transform(key.begin(), key.end(), key.begin(), ::tolower);
string orig = originalText(cipher_text, key);
cout << "Original/Decrypted Text : " << "\n";
for (int i = 0; i < orig.size(); i += 81)
orig.insert(i, "\n");
cout << orig;
cout << "\n\n" << "Ciphertext : " << "\n";
for (int i = 0; i < cipher_text.size(); i += 81)
cipher_text.insert(i, "\n");
cout << cipher_text;
cout << "\n\nPress ENTER key to Continue\n";
getchar();
return 0;
}
The char array works fine with this while loop, so long as there are no special characters [.,%$#!^]. As soon as there are any special characters in the char array, it gives me the debug assertion:
"Program: ...\Projects\ConsoleApplication17\Debug\ConsoleApplication17.exe
File: minkernel\crts\ucrt\src\appcrt\convert\isctype.cpp
Line: 42
Expression: c >= -1 && c <= 255
...
The program '[11048] ConsoleApplication17.exe' has exited with code 3 (0x3)."
If I run this on repl.it or cpp.sh, I get no issues though. I appreciate any help. Thank you.
It isn't done at all. It needs to be cleaned up a lot, but I'm just trying to test it as is.
see https://msdn.microsoft.com/en-us/library/xt82b8z8.aspx
isalpha expects a number between 0 and 0xFF:
The behavior of isalpha and _isalpha_l is undefined if c is not EOF or
in the range 0 through 0xFF, inclusive. When a debug CRT library is
used and c is not one of these values, the functions raise an
assertion.
You need to cast you char to an unsigned char before passing to isalpha.
As I'm familiarizing myself with the I/O aspect of C++, I'm trying to write a program to read some lines of integers from std::cin. Say the input looks like this:
1 2 3
4 5 6
7 8 9
10 11 12
How can I read the above lines into a 2D vector?
vector<vector<int>> nums;
/*
... some code here and nums will look like the following:
nums = {
{1,2,3},
{4,5,6},
{7,8,9},
{10,11,12}
}
*/
I've also tried to read the above lines of integers to a 1D vector, but I'm having some issues dealing with the '\n' character. My code is:
string rawInput;
vector<int> temp;
while(getline(cin, rawInput, ' ') ){
int num = atoi( rawInput.c_str() );
temp.push_back(num);
}
And the final result I got by printing out all the elements in the "temp" vector is:
1 2 3 5 6 8 9 11 12 // 4, 7, 10 went missing
Any help is appreciated. Thank you.
First use getline to grab an entire line, then you can use a istringstream to create a stream of ints just for that line.
At that point it's just a matter of creating each subvector of ints using the vector constructor that takes two iterators. An istream_iterator<int> on your istringstream gets this done:
std::vector<std::vector<int>> nums;
std::string line;
while (std::getline(std::cin, line)) {
std::istringstream ss(line);
nums.emplace_back(std::istream_iterator<int>{ss}, std::istream_iterator<int>{});
}
What is happening is since you are using only ' '(space) as deliminator, the input happens to be
1
2
3\n4 //<------ Newline also comes with the input
...
So, you are passing 3\n4, 6\n7 etc to atoi it returns 3,6 etc(atoi parses the input till first non-digit input) and the 4,7 is lost.
To achieve want you want you can use getline with istringstream (keeping the default deliminator as newline)
string rawInput;
vector<vector<int>> temp;
while(getline(cin, rawInput) ){
istringstream bufferInput(rawInput);
temp.push_back(vector<int>{std::istream_iterator<int>{bufferInput}, std::istream_iterator<int>{}});
}
you can use stringstream
string rawInput;
vector<int> temp;
stringstream ss;
while(getline(cin,rawInput)){
ss<<rawInput;
vector<int> temp;
int x;
while(ss>>x){
temp.push_back(x);
}
num.push_back(temp)
}
I recently wrote an answer to another question but with a few adaptations it achieves exactly what you are looking for (I hope):
#ifndef _IOSTREAM_H
#include <iostream>
#endif
#ifndef _STRING_H
#include <string>
#endif
#ifndef _VECTOR_H
#include <vector>
#endif
using namespace std;
enum XYZ { X = 0, Y = 1, Z = 2 };
struct Vector {
float x, y, z;
Vector(float _x=0, float _y=0, float _z=0) {
x = _x;
y = _y;
z = _z;
}
float& operator[](size_t index) {
if (index == XYZ::X) return x;
if (index == XYZ::Y) return y;
if (index == XYZ::Z) return z;
throw new exception;
}
};
#define min(a, b) (((a) < (b)) ? (a) : (b))
bool isCharNumeric(char c) {
const char* numbers = "0123456789";
for (size_t index = 0; index < strlen(numbers); index++)
if (c == numbers[index]) return true; return false;
}
vector<Vector> parseNumbers(string str_in) {
str_in += " "; //safe, no out of bounds
vector<Vector> results = {};
char currentChar;
char skipChar = ' ';
bool found_period = false;
size_t count_len = 0;
Vector vector_buffer(0,0,0);
XYZ current_axis = (XYZ)0;
for (size_t index = 0; index < str_in.length(); index++) {
currentChar = str_in[index];
if (currentChar == skipChar || currentChar == '\n' || currentChar == '\t')
continue;
else if (isCharNumeric(currentChar)) {
string word = ""; //word buffer
size_t word_len = min(min(str_in.find_first_of(' ', index + 1) - (index), str_in.find_first_of('\n', index + 1) - (index)), str_in.find_first_of('\t', index + 1) - (index)); //whatever char comes first; newline, tab or space
//append chars of following word checking if it is still valid number char
if (word_len > 0) {
size_t count_word_len = 0;
for (count_word_len = 0; count_word_len < word_len; count_word_len++)
if (isCharNumeric(str_in[index + count_word_len])) {
word += str_in[index + count_word_len];
}
else if (str_in[index + count_word_len] == '.' && isCharNumeric(str_in[index + count_word_len + 1])) {
//Floating-point numbers
word += '.';
found_period = true;
continue;
}
else {
word = "";
continue;
}
vector_buffer[current_axis] = stof(word);
if (current_axis == XYZ::Z) {
current_axis = XYZ::X;
results.push_back(vector_buffer);
}
else {
current_axis = (XYZ)(current_axis + 1);
}
index += count_word_len;
word = "";
continue;
}
}
}
return results;
}
Example implementation:
int main(int argc, char** argv) {
string user_input;
cin >> user_input;
vector<Vector> numbers = parseNumbers(user_input);
for each (Vector v in numbers) {
cout << "X=" << v.X << "\n";
cout << "Y=" << v.Y << "\n";
cout << "Z=" << v.Z << "\n\n";
}
}
Suprisingly none of the answers use the istream stream operator:
http://www.cplusplus.com/reference/istream/istream/operator%3E%3E/
When stream is empty eofbit is set, so run a while loop on that.
Works great for all types, and can be overloaded for custom types (such as 2D texture).
The program below I came up with for checking whether two strings are anagrams. Its working fine for small string but for larger strings ( i tried : listened , enlisted ) Its giving me a 'no !'
Help !
#include<iostream.h>
#include<string.h>
#include<stdio.h>
int main()
{
char str1[100], str2[100];
gets(str1);
gets(str2);
int i,j;
int n1=strlen(str1);
int n2=strlen(str2);
int c=0;
if(n1!=n2)
{
cout<<"\nThey are not anagrams ! ";
return 0;
}
else
{
for(i=0;i<n1;i++)
for(j=0;j<n2;j++)
if(str1[i]==str2[j])
++c;
}
if(c==n1)
cout<<"yes ! anagram !! ";
else
cout<<"no ! ";
system("pause");
return 0;
}
I am lazy, so I would use standard library functionality to sort both strings and then compare them:
#include <string>
#include <algorithm>
bool is_anagram(std::string s1, std::string s2)
{
std::sort(s1.begin(), s1.end());
std::sort(s2.begin(), s2.end());
return s1 == s2;
}
A small optimization could be to check that the sizes of the strings are the same before sorting.
But if this algorithm proved to be a bottle-neck, I would temporarily shed some of my laziness and compare it against a simple counting solution:
Compare string lengths
Instantiate a count map, std::unordered_map<char, unsigned int> m
Loop over s1, incrementing the count for each char.
Loop over s2, decrementing the count for each char, then check that the count is 0
The algorithm also fails when asked to find if aa and aa are anagrams. Try tracing the steps of the algorithm mentally or in a debugger to find why; you'll learn more that way.
By the way.. The usual method for finding anagrams is counting how many times each letter appears in the strings. The counts should be equal for each letter. This approach has O(n) time complexity as opposed to O(n²).
bool areAnagram(char *str1, char *str2)
{
// Create two count arrays and initialize all values as 0
int count1[NO_OF_CHARS] = {0};
int count2[NO_OF_CHARS] = {0};
int i;
// For each character in input strings, increment count in
// the corresponding count array
for (i = 0; str1[i] && str2[i]; i++)
{
count1[str1[i]]++;
count2[str2[i]]++;
}
// If both strings are of different length. Removing this condition
// will make the program fail for strings like "aaca" and "aca"
if (str1[i] || str2[i])
return false;
// Compare count arrays
for (i = 0; i < NO_OF_CHARS; i++)
if (count1[i] != count2[i])
return false;
return true;
}
I see 2 main approaches below:
Sort then compare
Count the occurrences of each letter
It's interesting to see that Suraj's nice solution got one point (by me, at the time of writing) but a sort one got 22. The explanation is that performance wasn't in people's mind - and that's fine for short strings.
The sort implementation is only 3 lines long, but the counting one beats it square for long strings. It is much faster (O(N) versus O(NlogN)).
Got the following results with 500 MBytes long strings.
Sort - 162.8 secs
Count - 2.864 secs
Multi threaded Count - 3.321 secs
The multi threaded attempt was a naive one that tried to double the speed by counting in separate threads, one for each string. Memory access is the bottleneck and this is an example where multi threading makes things a bit worse.
I would be happy to see some idea that would speed up the count solution (think by someone good with memory latency issues, caches).
#include<stdio.h>
#include<string.h>
int is_anagram(char* str1, char* str2){
if(strlen(str1)==strspn(str1,str2) && strlen(str1)==strspn(str2,str1) &&
strlen(str1)==strlen(str2))
return 1;
return 0;
}
int main(){
char* str1 = "stream";
char* str2 = "master";
if(is_anagram(str1,str2))
printf("%s and %s are anagram to each other",str1,str2);
else
printf("%s and %s are not anagram to each other",str1,str2);
return 0;
}
#include<iostream>
#include<unordered_map>
using namespace std;
int checkAnagram (string &str1, string &str2)
{
unordered_map<char,int> count1, count2;
unordered_map<char,int>::iterator it1, it2;
int isAnagram = 0;
if (str1.size() != str2.size()) {
return -1;
}
for (unsigned int i = 0; i < str1.size(); i++) {
if (count1.find(str1[i]) != count1.end()){
count1[str1[i]]++;
} else {
count1.insert(pair<char,int>(str1[i], 1));
}
}
for (unsigned int i = 0; i < str2.size(); i++) {
if (count2.find(str2[i]) != count2.end()) {
count2[str2[i]]++;
} else {
count2.insert(pair<char,int>(str2[i], 1));
}
}
for (unordered_map<char, int>::iterator itUm1 = count1.begin(); itUm1 != count1.end(); itUm1++) {
unordered_map<char, int>::iterator itUm2 = count2.find(itUm1->first);
if (itUm2 != count2.end()) {
if (itUm1->second != itUm2->second){
isAnagram = -1;
break;
}
}
}
return isAnagram;
}
int main(void)
{
string str1("WillIamShakespeare");
string str2("IamaWeakishSpeller");
cout << "checkAnagram() for " << str1 << "," << str2 << " : " << checkAnagram(str1, str2) << endl;
return 0;
}
It's funny how sometimes the best questions are the simplest.
The problem here is how to deduce whether two words are anagrams - a word being essentially an unsorted multiset of chars.
We know we have to sort, but ideally we'd want to avoid the time-complexity of sort.
It turns out that in many cases we can eliminate many words that are dissimilar in linear time by running through them both and XOR-ing the character values into an accumulator. The total XOR of all characters in both strings must be zero if both strings are anagrams, regardless of ordering. This is because anything xored with itself becomes zero.
Of course the inverse is not true. Just because the accumulator is zero does not mean we have an anagram match.
Using this information, we can eliminate many non-anagrams without a sort, short-circuiting at least the non-anagram case.
#include <iostream>
#include <string>
#include <algorithm>
//
// return a sorted copy of a string
//
std::string sorted(std::string in)
{
std::sort(in.begin(), in.end());
return in;
}
//
// check whether xor-ing the values in two ranges results in zero.
// #pre first2 addresses a range that is at least as big as (last1-first1)
//
bool xor_is_zero(std::string::const_iterator first1,
std::string::const_iterator last1,
std::string::const_iterator first2)
{
char x = 0;
while (first1 != last1) {
x ^= *first1++;
x ^= *first2++;
}
return x == 0;
}
//
// deduce whether two strings are the same length
//
bool same_size(const std::string& l, const std::string& r)
{
return l.size() == r.size();
}
//
// deduce whether two words are anagrams of each other
// I have passed by const ref because we may not need a copy
//
bool is_anagram(const std::string& l, const std::string& r)
{
return same_size(l, r)
&& xor_is_zero(l.begin(), l.end(), r.begin())
&& sorted(l) == sorted(r);
}
// test
int main() {
using namespace std;
auto s1 = "apple"s;
auto s2 = "eppla"s;
cout << is_anagram(s1, s2) << '\n';
s2 = "pppla"s;
cout << is_anagram(s1, s2) << '\n';
return 0;
}
expected:
1
0
Try this:
// Anagram. Two words are said to be anagrams of each other if the letters from one word can be rearranged to form the other word.
// From the above definition it is clear that two strings are anagrams if all characters in both strings occur same number of times.
// For example "xyz" and "zxy" are anagram strings, here every character 'x', 'y' and 'z' occur only one time in both strings.
#include <map>
#include <string>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
bool IsAnagram_1( string w1, string w2 )
{
// Compare string lengths
if ( w1.length() != w2.length() )
return false;
sort( w1.begin(), w1.end() );
sort( w2.begin(), w2.end() );
return w1 == w2;
}
map<char, size_t> key_word( const string & w )
{
// Declare a map which is an associative container that will store a key value and a mapped value pairs
// The key value is a letter in a word and the maped value is the number of times this letter appears in the word
map<char, size_t> m;
// Step over the characters of string w and use each character as a key value in the map
for ( auto & c : w )
{
// Access the mapped value directly by its corresponding key using the bracket operator
++m[toupper( c )];
}
return ( m );
}
bool IsAnagram_2( const string & w1, const string & w2 )
{
// Compare string lengths
if ( w1.length() != w2.length() )
return false;
return ( key_word( w1 ) == key_word( w2 ) );
}
bool IsAnagram_3( const string & w1, const string & w2 )
{
// Compare string lengths
if ( w1.length() != w2.length() )
return false;
// Instantiate a count map, std::unordered_map<char, unsigned int> m
unordered_map<char, size_t> m;
// Loop over the characters of string w1 incrementing the count for each character
for ( auto & c : w1 )
{
// Access the mapped value directly by its corresponding key using the bracket operator
++m[toupper(c)];
}
// Loop over the characters of string w2 decrementing the count for each character
for ( auto & c : w2 )
{
// Access the mapped value directly by its corresponding key using the bracket operator
--m[toupper(c)];
}
// Check to see if the mapped values are all zeros
for ( auto & c : w2 )
{
if ( m[toupper(c)] != 0 )
return false;
}
return true;
}
int main( )
{
string word1, word2;
cout << "Enter first word: ";
cin >> word1;
cout << "Enter second word: ";
cin >> word2;
if ( IsAnagram_1( word1, word2 ) )
cout << "\nAnagram" << endl;
else
cout << "\nNot Anagram" << endl;
if ( IsAnagram_2( word1, word2 ) )
cout << "\nAnagram" << endl;
else
cout << "\nNot Anagram" << endl;
if ( IsAnagram_3( word1, word2 ) )
cout << "\nAnagram" << endl;
else
cout << "\nNot Anagram" << endl;
system("pause");
return 0;
}
In this approach I took care of empty strings and repeated characters as well. Enjoy it and comment any limitation.
#include <iostream>
#include <map>
#include <string>
using namespace std;
bool is_anagram( const string a, const string b ){
std::map<char, int> m;
int count = 0;
for (int i = 0; i < a.length(); i++) {
map<char, int>::iterator it = m.find(a[i]);
if (it == m.end()) {
m.insert(m.begin(), pair<char, int>(a[i], 1));
} else {
m[a[i]]++;
}
}
for (int i = 0; i < b.length(); i++) {
map<char, int>::iterator it = m.find(b[i]);
if (it == m.end()) {
m.insert(m.begin(), pair<char, int>(b[i], 1));
} else {
m[b[i]]--;
}
}
if (a.length() <= b.length()) {
for (int i = 0; i < a.length(); i++) {
if (m[a[i]] >= 0) {
count++;
} else
return false;
}
if (count == a.length() && a.length() > 0)
return true;
else
return false;
} else {
for (int i = 0; i < b.length(); i++) {
if (m[b[i]] >= 0) {
count++;
} else {
return false;
}
}
if (count == b.length() && b.length() > 0)
return true;
else
return false;
}
return true;
}
Check if the two strings have identical counts for each unique char.
bool is_Anagram_String(char* str1,char* str2){
int first_len=(int)strlen(str1);
int sec_len=(int)strlen(str2);
if (first_len!=sec_len)
return false;
int letters[256] = {0};
int num_unique_chars = 0;
int num_completed_t = 0;
for(int i=0;i<first_len;++i){
int char_letter=(int)str1[i];
if(letters[char_letter]==0)
++num_unique_chars;
++letters[char_letter];
}
for (int i = 0; i < sec_len; ++i) {
int c = (int) str2[i];
if (letters[c] == 0) { // Found more of char c in t than in s.
return false;
}
--letters[c];
if (letters[c] == 0) {
++num_completed_t;
if (num_completed_t == num_unique_chars) {
// it’s a match if t has been processed completely
return i == sec_len - 1;
}
}
}
return false;}
#include <iostream>
#include <string.h>
using namespace std;
const int MAX = 100;
char cadA[MAX];
char cadB[MAX];
bool chrLocate;
int i,m,n,j, contaChr;
void buscaChr(char [], char []);
int main() {
cout << "Ingresa CadA: ";
cin.getline(cadA, sizeof(cadA));
cout << "Ingresa CadB: ";
cin.getline(cadB, sizeof(cadA));
if ( strlen(cadA) == strlen(cadB) ) {
buscaChr(cadA,cadB);
} else {
cout << "No son Anagramas..." << endl;
}
return 0;
}
void buscaChr(char a[], char b[]) {
j = 0;
contaChr = 0;
for ( i = 0; ( (i < strlen(a)) && contaChr < 2 ); i++ ) {
for ( m = 0; m < strlen(b); m++ ) {
if ( a[i] == b[m]) {
j++;
contaChr++;
a[i] = '-';
b[m] = '+';
} else { contaChr = 0; }
}
}
if ( j == strlen(a)) {
cout << "SI son Anagramas..." << endl;
} else {
cout << "No son Anagramas..." << endl;
}
}
Your algorithm is incorrect. You're checking each character in the first word to see how many times that character appears in the second word. If the two words were 'aaaa', and 'aaaa', then that would give you a count of 16. A small alteration to your code would allow it to work, but give a complexity of N^2 as you have a double loop.
for(i=0;i<n1;i++)
for(j=0;j<n2;j++)
if(str1[i]==str2[j])
++c, str2[j] = 0; // 'cross off' letters as they are found.
I done some tests with anagram comparisons. Comparing two strings of 72 characters each (the strings are always true anagrams to get maximum number of comparisons), performing 256 same-tests with a few different STL containers...
template<typename STORAGE>
bool isAnagram(const string& s1, const string& s2, STORAGE& asciiCount)
{
for(auto& v : s1)
{
asciiCount[v]++;
}
for(auto& v : s2)
{
if(--asciiCount[static_cast<unsigned char>(v)] == -1)
{
return false;
}
}
return true;
}
Where STORAGE asciiCount =
map<char, int> storage; // 738us
unordered_map<char, int> storage; // 260us
vector<int> storage(256); // 43us
// g++ -std=c++17 -O3 -Wall -pedantic
This is the fastest I can get.
These are crude tests using coliru online compiler + and std::chrono::steady_clock::time_point for measurements, however they give a general idea of performance gains.
vector has the same performance, uses only 256 bytes, although strings are limited to 255 characters in length (also change to: --asciiCount[static_cast(v)] == 255 for unsigned char counting).
Assuming vector is the fastest. An improvement would be to just allocate a C style array unsigned char asciiCount[256]; on the stack (since STL containers allocate their memory dynamically on the heap)
You could probably reduce this storage to 128 bytes, 64 or even 32 bytes (ascii chars are typically in range 0..127, while A-Z+a-z 64.127, and just upper or lower case 64..95 or 96...127) although not sure what gains would be found from fitting this inside a cache line or half.
Any better ways to do this? For Speed, Memory, Code Elegance?
1. Simple and fast way with deleting matched characters
bool checkAnagram(string s1, string s2) {
for (char i : s1) {
unsigned int pos = s2.find(i,0);
if (pos != string::npos) {
s2.erase(pos,1);
} else {
return false;
}
}
return s2.empty();
}
2. Conversion to prime numbers. Beautiful but very expensive, requires special Big Integer type for long strings.
// https://en.wikipedia.org/wiki/List_of_prime_numbers
int primes[255] = {2, 3, 5, 7, 11, 13, 17, 19, ... , 1613};
bool checkAnagramPrimes(string s1, string s2) {
long c1 = 1;
for (char i : s1) {
c1 = c1 * primes[i];
}
long c2 = 1;
for (char i : s2) {
c2 = c2 * primes[i];
if (c2 > c1) {
return false;
}
}
return c1 == c2;
}
string key="listen";
string key1="silent";
string temp=key1;
int len=0;
//assuming both strings are of equal length
for (int i=0;i<key.length();i++){
for (int j=0;j<key.length();j++){
if(key[i]==temp[j]){
len++;
temp[j] = ' ';//to deal with the duplicates
break;
}
}
}
cout << (len==key.length()); //if true: means the words are anagrams
Instead of using dot h header which is deprecated in modern c++.
Try this solution.
#include <iostream>
#include <string>
#include <map>
int main(){
std::string word_1 {};
std::cout << "Enter first word: ";
std::cin >> word_1;
std::string word_2 {};
std::cout << "Enter second word: ";
std::cin >> word_2;
if(word_1.length() == word_2.length()){
std::map<char, int> word_1_map{};
std::map<char, int> word_2_map{};
for(auto& c: word_1)
word_1_map[std::tolower(c)]++;
for(auto& c: word_2)
word_2_map[std::tolower(c)]++;
if(word_1_map == word_2_map){
std::cout << "Anagrams" << std::endl;
}
else{
std::cout << "Not Anagrams" << std::endl;
}
}else{
std::cout << "Length Mismatch" << std::endl;
}
}
#include <bits/stdc++.h>
using namespace std;
#define NO_OF_CHARS 256
int main()
{ bool ans = true;
string word1 = "rest";
string word2 = "tesr";
unordered_map<char,int>maps;
for(int i = 0 ; i <5 ; i++)
{
maps[word1[i]] +=1;
}
for(int i = 0 ; i <5 ; i++)
{
maps[word2[i]]-=1 ;
}
for(auto i : maps)
{
if(i.second!=0)
{
ans = false;
}
}
cout<<ans;
}
Well if you don't want to sort than this code will give you perfect output.
#include <iostream>
using namespace std;
int main(){
string a="gf da";
string b="da gf";
int al,bl;
int counter =0;
al =a.length();
bl =b.length();
for(int i=0 ;i<al;i++){
for(int j=0;j<bl;j++){
if(a[i]==b[j]){
if(j!=bl){
b[j]=b[b.length()-counter-1];
bl--;
counter++;
break;
}else{
bl--;
counter++;
}
}
}
}
if(counter==al){
cout<<"true";
}
else{
cout<<"false";
}
return 0;
}
Here is the simplest and fastest way to check for anagrams
bool anagram(string a, string b) {
int a_sum = 0, b_sum = 0, i = 0;
while (a[i] != '\0') {
a_sum += (int)a[i]; // (int) cast not necessary
b_sum += (int)b[i];
i++;
}
return a_sum == b_sum;
}
Simply adds the ASCII values and checks if the sums are equal.
For example:
string a = "nap" and string b = "pan"
a_sum = 110 + 97 + 112 = 319
b_sum = 112 + 97 + 110 = 319
I have a simple word wrap function that takes a long string as an input and then breaks that string into smaller strings and adds them to an array to be outputted later. Right now the last word or two isn't outputting. That's the main problem. However, I would also like to improve the function. I know it's kind of messy. I was wondering if there are any better ways of solving this problem. I think the array is unnecessary but I don't know how else to do it. After the array is filled with all the smaller strings, I just output them to a text file. Any suggestions would be greatly appreciated.
Here's the Word Wrap function:
void WordWrap(string inputString, string formatedAr[], const int SIZE)
{
unsigned int index;
unsigned int word;
unsigned int max = 65;
string outWord;
string outLine;
outWord = "";
outLine = "";
word = 0;
for(int i = 0; i < SIZE; i++)
{
formatedAr[i] = "";
}
for(index = 0; index <= inputString.length(); index++)
{
if(inputString[index] != ' ')
{
outWord += inputString[index];
}
else
{
if(outLine.length() + outWord.length() > max)
{
formatedAr[word] = outLine;
word++;
outLine.clear();
}
outLine += outWord + " ";
outWord.clear();
}
}
formatedAr[word] = outLine;
}
And this is where I call the function and output the array:
WordWrap(dvdPtr -> synopsis, formatedAr, SIZE);
index = 0;
while(index < SIZE && formatedAr[index] != "")
{
outFile << formatedAr[index] << endl;
index++;
}
Here is an example code.
#include <iostream>
#include <sstream>
#include <string>
#include <vector>
using namespace std;
void WordWrap(const string& inputString, vector<string>& outputString, unsigned int lineLength)
{
istringstream iss(inputString);
string line;
do
{
string word;
iss >> word;
if (line.length() + word.length() > lineLength)
{
outputString.push_back(line);
line.clear();
}
line += word + " ";
}while (iss);
if (!line.empty())
{
outputString.push_back(line);
}
}
/*
A simple test:
Input string: "aaa bbb ccccccc dddd d111111111111111 33333 4444444444 222222222 ajdkjklad 341343"
Length per line: 20
Output lines of strings:
Line 1: aaa bbb ccccccc dddd
Line 2: d111111111111111
Line 3: 33333 4444444444
Line 4: 222222222 ajdkjklad
*/
I was asked this during an interview and apparently it's an easy question but it wasn't and still isn't obvious to me.
Given a string, count all the words in it. Doesn't matter if they are repeated. Just the total count like in a text files word count. Words are anything separated by a space and punctuation doesn't matter, as long as it's part of a word.
For example:
A very, very, very, very, very big dog ate my homework!!!! ==> 11 words
My "algorithm" just goes through looking for spaces and incrementing a counter until I hit a null. Since i didn't get the job and was asked to leave after that I guess My solution wasn't good? Anyone have a more clever solution? Am I missing something?
Assuming words are white space separated:
unsigned int countWordsInString(std::string const& str)
{
std::stringstream stream(str);
return std::distance(std::istream_iterator<std::string>(stream), std::istream_iterator<std::string>());
}
Note: There may be more than one space between words. Also this does not catch other white space characters like tab new line or carriage return. So counting spaces is not enough.
The stream input operator >> when used to read a string from a stream. Reads one white space separated word. So they were probably looking for you to use this to identify words.
std::stringstream stream(str);
std::string oneWord;
stream >> oneWord; // Reads one space separated word.
When can use this to count words in a string.
std::stringstream stream(str);
std::string oneWord;
unsigned int count = 0;
while(stream >> oneWord) { ++count;}
// count now has the number of words in the string.
Getting complicated:
Streams can be treated just like any other container and there are iterators to loop through them std::istream_iterator. When you use the ++ operator on an istream_iterator it just read the next value from the stream using the operator >>. In this case we are reading std::string so it reads a space separated word.
std::stringstream stream(str);
std::string oneWord;
unsigned int count = 0;
std::istream_iterator loop = std::istream_iterator<std::string>(stream);
std::istream_iterator end = std::istream_iterator<std::string>();
for(;loop != end; ++count, ++loop) { *loop; }
Using std::distance just wraps all the above in a tidy package as it find the distance between two iterators by doing ++ on the first until we reach the second.
To avoid copying the string we can be sneaky:
unsigned int countWordsInString(std::string const& str)
{
std::stringstream stream;
// sneaky way to use the string as the buffer to avoid copy.
stream.rdbuf()->pubsetbuf (str.c_str(), str.length() );
return std::distance(std::istream_iterator<std::string>(stream), std::istream_iterator<std::string>());
}
Note: we still copy each word out of the original into a temporary. But the cost of that is minimal.
A less clever, more obvious-to-all-of-the-programmers-on-your-team method of doing it.
#include <cctype>
int CountWords(const char* str)
{
if (str == NULL)
return error_condition; // let the requirements define this...
bool inSpaces = true;
int numWords = 0;
while (*str != '\0')
{
if (std::isspace(*str))
{
inSpaces = true;
}
else if (inSpaces)
{
numWords++;
inSpaces = false;
}
++str;
}
return numWords;
}
You can use the std::count or std::count_if to do that. Below a simple example with std::count:
//Count the number of words on string
#include <iostream>
#include <string>
#include <algorithm> //count and count_if is declared here
int main () {
std::string sTEST("Text to verify how many words it has.");
std::cout << std::count(sTEST.cbegin(), sTEST.cend(), ' ')+1;
return 0;
}
UPDATE: Due the observation made by Aydin Özcan (Nov 16) I made a change to this solution. Now the words may have more than one space between them. :)
//Count the number of words on string
#include <string>
#include <iostream>
int main () {
std::string T("Text to verify : How many words does it have?");
size_t NWords = T.empty() || T.back() == ' ' ? 0 : 1;
for (size_t s = T.size(); s > 0; --s)
if (T[s] == ' ' && T[s-1] != ' ') ++NWords;
std::cout << NWords;
return 0;
}
Another boost based solution that may work (untested):
vector<string> result;
split(result, "aaaa bbbb cccc", is_any_of(" \t\n\v\f\r"), token_compress_on);
More information can be found in the Boost String Algorithms Library
This can be done without manually looking at every character or copying the string.
#include <boost/iterator/transform_iterator.hpp>
#include <cctype>
boost::transform_iterator
< int (*)(int), std::string::const_iterator, bool const& >
pen( str.begin(), std::isalnum ), end( str.end(), std::isalnum );
size_t word_cnt = 0;
while ( pen != end ) {
word_cnt += * pen;
pen = std::mismatch( pen+1, end, pen ).first;
}
return word_cnt;
I took the liberty of using isalnum instead of isspace.
This is not something I would do at a job interview. (It's not like it compiled the first time.)
Or, for all the Boost haters ;v)
if ( str.empty() ) return 0;
size_t word_cnt = std::isalnum( * str.begin() );
for ( std::string::const_iterator pen = str.begin(); ++ pen != str.end(); ) {
word_cnt += std::isalnum( pen[ 0 ] ) && ! std::isalnum( pen[ -1 ] );
}
return word_cnt;
An O(N) solution that is also very simple to understand and implement:
(I haven't checked for an empty string input. But I am sure you can do that easily.)
#include <iostream>
#include <string>
using namespace std;
int countNumberOfWords(string sentence){
int numberOfWords = 0;
size_t i;
if (isalpha(sentence[0])) {
numberOfWords++;
}
for (i = 1; i < sentence.length(); i++) {
if ((isalpha(sentence[i])) && (!isalpha(sentence[i-1]))) {
numberOfWords++;
}
}
return numberOfWords;
}
int main()
{
string sentence;
cout<<"Enter the sentence : ";
getline(cin, sentence);
int numberOfWords = countNumberOfWords(sentence);
cout<<"The number of words in the sentence is : "<<numberOfWords<<endl;
return 0;
}
Here is a single pass, branchless (almost), locale-aware algorithm which handles cases with more than one space between words:
If the string is empty return 0
let transitions = number of adjacent char pairs (c1, c2) where c1 == ' ' and c2 != ' '
if the sentence starts with a space, return transitions else return transitions + 1
Here is an example with string = "A very, very, very, very, very big dog ate my homework!!!!"
i | 0123456789
c1 | A very, very, very, very, very big dog ate my homework!!!!
c2 | A very, very, very, very, very big dog ate my homework!!!!
| x x x x x x x x x x
Explanation
Let `i` be the loop counter.
When i=0: c1='A' and c2=' ', the condition `c1 == ' '` and `c2 != ' '` is not met
When i=1: c1=' ' and c2='A', the condition is met
... and so on for the remaining characters
Here are 2 solutions I came up with
Naive solution
size_t count_words_naive(const std::string_view& s)
{
if (s.size() == 0) return 0;
size_t count = 0;
bool isspace1, isspace2 = true;
for (auto c : s) {
isspace1 = std::exchange(isspace2, isspace(c));
count += (isspace1 && !isspace2);
}
return count;
}
If you think carefully, you will be able to reduce this set of operations into an inner product (just for fun, I don't recommend this as this is arguably much less readable).
Inner product solution
size_t count_words_using_inner_prod(const std::string_view& s)
{
if (s.size() == 0) return 0;
auto starts_with_space = isspace(s.front());
auto num_transitions = std::inner_product(
s.begin()+1, s.end(), s.begin(), 0, std::plus<>(),
[](char c2, char c1) { return isspace(c1) && !isspace(c2); });
return num_transitions + !starts_with_space;
}
I think that will help
the complexty O(n)
#include <iostream>
#include <string>
#include <ctype.h>
using namespace std;
int main()
{
int count = 0, size;
string sent;
getline(cin, sent);
size = sent.size();
check if the char is in alpha and the next char not in alpha
for (int i = 0; i < size - 1; ++i) {
if (isalpha(sent[i]) && !isalpha(sent[i+1])) {
++count;
}
}
if the word in the last of sentence didn't count above so it count here
if (isalpha(sent[size - 1]))++count;
cout << count << endl;
return 0;
}
A very concise O(N) approach:
bool is_letter(char c) { return c >= 'a' && c <= 'z' || c >= 'A' && c <= 'Z'; }
int count_words(const string& s) {
int i = 0, N = s.size(), count = 0;
while(i < N) {
while(i < N && !is_letter(s[i])) i++;
if(i == N) break;
while(i < N && is_letter(s[i])) i++;
count++;
}
return count;
}
A divide-and-conquer approach, complexity is also O(N):
int DC(const string& A, int low, int high) {
if(low > high) return 0;
int mid = low + (high - low) / 2;
int count_left = DC(A, low, mid-1);
int count_right = DC(A, mid+1, high);
if(!is_letter(A[mid]))
return count_left + count_right;
else {
if(mid == low && mid == high) return 1;
if(mid-1 < low) {
if(is_letter(A[mid+1])) return count_right;
else return count_right+1;
} else if(mid+1 > high) {
if(is_letter(A[mid-1])) return count_left;
else return count_left+1;
}
else {
if(!is_letter(A[mid-1]) && !is_letter(A[mid+1]))
return count_left + count_right + 1;
else if(is_letter(A[mid-1]) && is_letter(A[mid+1]))
return count_left + count_right - 1;
else
return count_left + count_right;
}
}
}
int count_words_divide_n_conquer(const string& s) {
return DC(s, 0, s.size()-1);
}
Efficient version based on map-reduce approach
#include <iostream>
#include <string_view>
#include <numeric>
std::size_t CountWords(std::string_view s) {
if (s.empty())
return 0;
std::size_t wc = (!std::isspace(s.front()) ? 1 : 0);
wc += std::transform_reduce(
s.begin(),
s.end() - 1,
s.begin() + 1,
std::size_t(0),
std::plus<std::size_t>(),
[](char left, char right) {
return std::isspace(left) && !std::isspace(right);
});
return wc;
}
int main() {
std::cout << CountWords(" pretty little octopus "sv) << std::endl;
return 0;
}