string extract(string scrambeledword){
unsigned int index;
string output;
string input= " ";
for (index=0; index <= scrambeledword.length() ; index++);
{
if (index%2==0)
{
output+=input ;
cout << output;
}
}
return output;}
I want to extract the even numbered indexed letters from the 40 letter long word inputted by users. does this make sense? i have not taken arrays yet and do not want to include them.
Problems:
1. You have a ; after your for loop, the loop body is never run.
2. <= is wrong here since scrambeledword.length() is out of range. Use != or < instead.
3. You need to either assign something to input before adding it to output or get rid of it altogether.
4. As #Aconcagua pointed out, it is worth noting that I removed your declaration of index from the function scope and added it only to the for loop scope. If you also considered doing so, compiler would throw an error (since it'd be undeclared outside of the scope of for) and you'd be noted about the ; problem.
Fixed version:
string extract(const string &scrambeledword){ // copying strings is expensive
// unsigned int index; // obsolete
string output;
// string input= " "; // obsolete
for (size_t index = 0; index != scrambeledword.length(); ++index) // `<=` would be wrong since scrambeledword.length() is out of range
{
if (index % 2 == 0)
{
output += scrambeledword[index];
// cout << output; // obsolete. If you just want the characters, print scrambeledword[index]
cout << scrambeledword[index];
}
}
cout << endl; // break the line for better readability
return output;
}
Your code won't run the block under the for because there is a ; at the end of the line. That means the for runs without block. Basically it will count to the length of the given word.
In the for index <= scrambeledword.length() can cause an out of bound exception because you can index out of the string-array. Use index < scrambeledword.length() instead.
This can be a good solution for the problem:
string extract(const string& scrambeledword)
{
string output;
for (unsigned int index = 0; index < scrambeledword.length(); index++)
{
if (index % 2 == 0)
{
output += scrambeledword[index];
}
}
return output;
}
auto str = "HelloWorld"s;
int i = 0;
for_each(str.cbegin(), str.cend(), [&i](char const & c) { if (i++ % 2 == 0) cout << c; });
output: Hlool
You could go with something like this:
for(int i = 0; i < scrambleword.length(); i+=2){
output += scrambleword.at(i);
}
Related
i already studied c++ in school and during the last days i have been doing the beginner c++ course of codecademy. On codecademy there is an exercise in which i have to identify palindrome words and return true or false. I haven't been able to resolve it so i saw the solution and it was:
#include <iostream>
// Define is_palindrome() here:
bool is_palindrome(std::string text) {
std::string reversed_text = "";
for (int i = text.size() - 1; i >= 0; i--) {
reversed_text += text[i];
}
if (reversed_text == text) {
return true;
}
return false;
}
int main() {
std::cout << is_palindrome("madam") << "\n";
std::cout << is_palindrome("ada") << "\n";
std::cout << is_palindrome("lovelace") << "\n";
}
My only doubt is with this line:
for (int i = text.size() - 1; i >= 0; i--) {
reversed_text += text[i];
i know it has to do with index values but i can't understand why it has a -1.
Could somebody explain this to me?
i thanks in advance whoever read this post. i'm sorry for my english or my poor using of stacksoverflow, i'm italian and that's my first time using this site.
for (int i = text.size() - 1; i >= 0; i--) {
reversed_text += text[i];
text is basically the string that you receive as input via function. size() is function that returns the size of the string i.e text.size() so in our test cases it will return
5 for madam
3 for ada
8 for lovelace
If you think about the strings as an array with exact above size then the index range will become
0-4 for madam
0-2 for ada
0-7 for lovelace
So that's why the text.size()-1 is using as the starting index of loop. text.size() will return the actual size of string and then minus 1 to get the index of last character in string.
so behind the scene, your loop iteration will look something like below
for (int i = 4; i >= 0; i--) { //for madam
}
//aca
for (int i = 2; i >= 0; i--) {
}
//lovelace
for (int i = 7; i >= 0; i--) {
}
I hope it clear out your confusion.
Thanks,
i know it has to do with index values but i can't understand why it has a -1.
If a string is n characters long, the characters in it are indexed from 0 to n−1.
Since the loop works with characters from the end of the string to the beginning, it starts with index text.size() - 1.
However, the solution you have shown is nominally inefficient. There is no reason to make a reversed copy of the string. It suffices merely to test whether each character in the first half of the string equals the character in the reflected position:
bool is_palindrome(std::string text)
{
size_t e = text.size();
for (int i = 0; i < e/2; ++i)
if (text[i] != text[e-1-i])
return false;
return true;
}
If using a for loop to reverse the string is confusing, you could also use the reverse function
std::string reversed_text = text;
reverse(reversed_text.begin(),reversed_text.end());
which just helps flip the entire string reversed_text and can achieve the same result in a simpler way.
The goal of the program is to take a string like "kcck" and delete the consecutive duplicates. It should first iterate through the string and delete cc leaving kk; then go through again and delete kk; then return "empty" since there are no characters left in the string.
Another example, "aabggtcc" should return "bt".
int i;
int j = i+1;
string deduplicate(string input) {
for(i=0; i<input.length(); ++i) {
while(j <input.length()) {
if(input[i] == input[j]) {
input.erase(i);
input.erase(j);
}
else if (input[i] != input[j]) {
++i; ++j;
}
if(input[i] == '\0') {
cout<<"empty";
}
}
}
return 0;
}
int main () {
cout<<deduplicate("aabg")<<endl;
cout<<deduplicate("ag")<<endl;
cout<<deduplicate("btaabb")<<endl;
return 0;
}
When I run the code it gives me:
libc++abi.dylib: terminating with uncaught exception of type std::out_of_range: basic_string
There are couple of issues with your snippet,
deduplicate function is returning zero(0) all the time
j is initialized in global scope and is never reset for new string
As you erase std::string::length() is calculated on new string hence your index i and j are won't point to same laocation.
Here is the snippet with rectified error,
string deduplicate(string input) {
int i = 0;
int j = 0;
while (i < input.length()) {
j = i + 1;
bool isRepeated = false;
while (j < input.length()) {
if (input[i] == input[j]) {
input.erase(j,1);
--j; //as string length is reduced by 1
isRepeated = true;
}
++j;
}
if (isRepeated) {
input.erase(i,1); //remove first letter as well
--i;//sting length is reduced by one
}
++i;
}
return input;
}
int main() {
std::cout << deduplicate("aabg") << endl;
std::cout << deduplicate("ag") << endl;
std::cout << deduplicate("btaabb") << endl;
return 0;
}
output:
bg
ag
t
which can be even simplified as,
std::string deduplicate(std::string input) {
std::string s ="";
for (auto c : input) //loop through all char
{
int f = 0;
for (auto c1 : input)
{
if (c1 == c)
{
f++; //increment if char is found
}
}
if (f == 1)//append char only if it present ones
s += c;
}
return s;
}
You are decreasing the size of the string whenever you call string.erase() that's why the variable i eventually exceeds the "current" string size input.length() in the while loop, and you get an error std::out_of_range: basic_string when you try to access input[i] in the if and else if conditions of the loop.
Try to manually go through the loop on the string on which you got the error and you will see that i has gone out of bound (i.e. i >= input.length()) in the while loop
With C++11 and on, instead of iterating over each character and making the comparison manually, you can use std::basic_string::find_first_not_of to look forward from a position in the string and find the first character not of the current character. If the position returned by .find_first_not_of is more than 1 from the current position, you can use .erase to erase that number characters. If the return is 1, then just increment your current position and repeat.
To operate on all duplicates characters in the modified string, you simply wrap it all in an outer-loop, keep a copy of the string before before entering the inner-loop to remove duplicate characters, and compare if the modified string is equal to your copy or the .length() is zero for your exit condition.
You can do something similar to the following:
#include <iostream>
#include <string>
int main (void) {
std::string s;
while (getline (std::cin, s)) {
std::string current;
do {
size_t pos = 0;
current = s;
while (pos < s.length()) {
size_t duplicates = s.find_first_not_of (s.at(pos), pos);
if (duplicates != std::string::npos && duplicates > pos + 1)
s.erase(s.begin() + pos, s.begin() + duplicates);
else if (duplicates == std::string::npos &&
(s.end() - s.begin() - pos) > 1)
s.erase(s.begin() + pos, s.end());
else
pos++;
}
} while (current != s && s.length());
std::cout << "'" << s << "'\n";
}
}
Example Use/Output
$ echo "kcck" | ./bin/ddcpp
''
$ echo "aabggtcc" | ./bin/ddcpp
'bt'
$ echo "aabg" | ./bin/ddcpp
'bg'
$ echo "ag" | ./bin/ddcpp
'ag'
$ echo "btaabb" | ./bin/ddcpp
'bt'
There are a number of ways to approach the problem and as long as they are reasonably efficient there isn't any one right/wrong way. If you have a modern C++ compiler, letting some of the built-in container functions handle the work is generally a bit more robust than reinventing it on your own. Look things over and let me know if you have questions.
Write a function countMatches that searches the substring in the given string and returns how many times the substring appears in the string.
I've been stuck on this awhile now (6+ hours) and would really appreciate any help I can get. I would really like to understand this better.
int countMatches(string str, string comp)
{
int small = comp.length();
int large = str.length();
int count = 0;
// If string is empty
if (small == 0 || large == 0) {
return -1;
}
// Increment i over string length
for (int i = 0; i < small; i++) {
// Output substring stored in string
for (int j = 0; j < large; j++) {
if (comp.substr(i, small) == str.substr(j, large)) {
count++;
}
}
}
cout << count << endl;
return count;
}
When I call this function from main, with countMatches("Hello", "Hello"); I get the output of 5. Which is completely wrong as it should return 1. I just want to know what I'm doing wrong here so I don't repeat the mistake and actually understand what I am doing.
I figured it out. I did not need a nested for loop because I was only comparing the secondary string to that of the string. It also removed the need to take the substring of the first string. SOOO... For those interested, it should have looked like this:
int countMatches(string str, string comp)
{
int small = comp.length();
int large = str.length();
int count = 0;
// If string is empty
if (small == 0 || large == 0) {
return -1;
}
// Increment i over string length
for (int i = 0; i < large; i++) {
// Output substring stored in string
if (comp == str.substr(i, small)) {
count++;
}
}
cout << count << endl;
return count;
}
The usual approach is to search in place:
std::string::size_type pos = 0;
int count = 0;
for (;;) {
pos = large.find(small, pos);
if (pos == std::string::npos)
break;
++count;
++pos;
}
That can be tweaked if you're not concerned about overlapping matches (i.e., looking for all occurrences of "ll" in the string "llll", the answer could be 3, which the above algorithm will give, or it could be 2, if you don't allow the next match to overlap the first. To do that, just change ++pos to pos += small.size() to resume the search after the entire preceding match.
The problem with your function is that you are checking that:
Hello is substring of Hello
ello is substring of ello
llo is substring of llo
...
of course this matches 5 times in this case.
What you really need is:
For each position i of str
check if the substring of str starting at i and of length = comp.size() is exactly comp.
The following code should do exactly that:
size_t countMatches(const string& str, const string& comp)
{
size_t count = 0;
for (int j = 0; j < str.size()-comp.size()+1; j++)
if (comp == str.substr(j, comp.size()))
count++;
return count;
}
First of all, sorry for the mis-worded title. I couldn't imagine a better way to put it.
The problem I'm facing is as follows: In a part of my program, the program counts occurences of different a-zA-Z letters and then tells how many of each letters can be found in an array. The problem, however, is this:
If I have an array that consists of A;A;F;A;D or anything similar, the output will be this:
A - 3
A - 3
F - 1
A - 3
D - 1
But I am required to make it like this:
A - 3
F - 1
D - 1
I could solve the problem easily, however I can't use an additional array to check what values have been already echoed. I know why it happens, but I don't know a way to solve it without using an additional array.
This is the code snippet (the array simply consists of characters, not worthy of adding it to the snippet):
n is the size of array the user is asked to choose at the start of the program (not included in the snippet).
initburts is the current array member ID that is being compared against all other values.
burts is the counter that is being reset after the loop is done checking a letter and moves onto the next one.
do {
for (i = 0; i < n; i++) {
if (array[initburts] == array[i]) {
burts++;
}
}
cout << "\n\n" << array[initburts] << " - " << burts;
initburts++;
burts = 0;
if (initburts == n) {
isDone = true;
}
}
while (isDone == false);
Do your counting first, then loop over your counts printing the results.
std::map<decltype(array[0]), std::size_t> counts;
std::for_each(std::begin(array), std::end(array), [&counts](auto& item){ ++counts[item]; });
std::for_each(std::begin(counts), std::end(counts), [](auto& pair) { std::cout << "\n\n" << pair.first << " - " pair.second; });
for (i = 0; i < n; i++)
{
// first check if we printed this character already;
// this is the case if the same character occurred
// before the current one:
bool isNew = true;
for (j = 0; j < i; j++)
{
// you find out yourself, do you?
// do not forget to break the loop
// in case of having detected an equal value!
}
if(isNew)
{
// well, now we can count...
unsigned int count = 1;
for(int j = i + 1; j < n; ++j)
count += array[j] == array[i];
// appropriate output...
}
}
That would do the trick and retains the array as is, however is an O(n²) algorithm. More efficient (O(n*log(n))) is sorting the array in advance, then you can just iterate over the array once. Of course, original array sequence gets lost then:
std::sort(array, array + arrayLength);
auto start = array;
for(auto current = array + 1; current != array + arrayLength; ++current)
{
if(*current != *start)
{
auto char = *start;
auto count = current - start;
// output char and count appropriately
}
}
// now we yet lack the final character:
auto char = *start;
auto count = array + arrayLength - start;
// output char and count appropriately
Pointer arithmetic... Quite likely that your teacher gets suspicious if you just copy this code, but it should give you the necessary hints to make up your own variant (use indices instead of pointers...).
I would do it this way.
#include <iostream>
#include <string>
#include <vector>
using namespace std;
int main()
{
string s;
vector<int> capCount(26, 0), smallCount(26, 0);
cout << "Enter the string\n";
cin >> s;
for(int i = 0; i < s.length(); ++i)
{
char c = s.at(i);
if(c >= 'A' && c <= 'Z')
++capCount[(int)c - 65];
if(c >= 'a' && c <= 'z')
++smallCount[(int)c - 97];
}
for(int i = 0; i < 26; ++i)
{
if(capCount[i] > 0)
cout << (char) (i + 65) << ": " << capCount[i] << endl;
if(smallCount[i] > 0)
cout << (char) (i + 97) << ": " << smallCount[i] << endl;
}
}
Note: I have differentiated lower and upper case characters.
Here's is the sample output:
output
I recently did this question
Specification:
Input Format The first line contains the number of test cases, T. Next,
T lines follow each containing a long string S.
Output Format For each long string S, display the number of times SUVO
and SUVOJIT appears in it.
I wrote the following code for this :
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int suvo = 0;
int suvojit = 0;
string s;
cin >> s;
for (int i = 0; i <= s.size() - 7; i++) {
if (s.substr(i, 7) == "SUVOJIT")
suvojit++;
}
for (int i = 0; i <= s.size() - 4; i++) {
if (s.substr(i, 4) == "SUVO")
suvo++;
}
cout << "SUVO = " << suvo - suvojit << ", SUVOJIT = " << suvojit << "\n";
}
return 0;
}
The code about gave out of bounds exception for substr() function for this test case:
15
RSUVOYDSUVOJITNSUVOUSUVOJITESUVOSUVOSGSUVOKSUVOJIT
SUVOJITWSUVOSUVOJITTSUVOCKSUVOJITNSUVOSUVOJITSUVOJITSUVOSUVOSUVOJITTSUVOJ
SUVOSUVOSUVOJITASUVOJITGCEBISUVOJITKJSUVORSUVOQCGVHRQLFSUVOOHPFNJTNSUVOJITKSSUVO
SUVOJITSUVOJITJGKSUVOJITISUVOJITKJLUSUVOJITUBSUVOX
MMHBSUVOFSUVOFMSUVOJITUMSUVOJITPSVYBYPMCSUVOJIT
OASUVOSUVOJITSUVOSTDYYJSUVOJITSUVOJITSUVO
RLSUVOCPSUVOJITYSUVOSUVOOGSUVOOESUVOJITMSUVO
WVLFFSUVOJITSUVOVSUVORLESUVOJITPSUVOJITSUVO
RSUVOSUVOJITQWSUVOUMASUVOSUVOJITXNNRRUNUSUVOJIT
HYLSSUVOSUVOSUVOJITPOSUVOJIT
DGMUCSSSUVOJITMJSUVOHSUVOCWTGSUVOJIT
OBNSSUVOYSUVOSUVOJITSUVOJITRHFDSUVODSUVOJITEGSUVOSUVOSUVOJITSUVOSUVOJITSSUVOSUVOSUVOSSUVOJIT
AG
NSUVOJITSUVOSUVOJIT
CGJGDSUVOEASUVOJITSGSUVO
However, when instead of using the s.size() function, I converted the string into a char constant and took the length of it using strlen, then the code caused no error and everything went smoothly.
So, my question is... Why did this happen?
This is my working code with the change:
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int suvo = 0;
int suvojit = 0;
string s;
cin >> s;
int le = strlen(&s[0]);
for (int i = 0; i <= le - 7; i++) {
if (s.substr(i, 7) == "SUVOJIT")
suvojit++;
}
for (int i = 0; i <= le - 4; i++) {
if (s.substr(i, 4) == "SUVO")
suvo++;
}
cout << "SUVO = " << suvo - suvojit << ", SUVOJIT = " << suvojit << "\n";
}
return 0;
}
In one case, you use size_t, in the other case you use int.
If the length is for example 6 characters, then s.size () - 7 is not -1, but one huge number and everything goes wrong. But if you write int len = strlen (...), then len - 7 is indeed -1 and everything is fine.
When I see a number subtracted from size_t, that's an immediate red flag. Write "i + 7 ≤ s.size()", not "i ≤ s.size() - 7".
First of all, in my testing your second leads to a problem as well:
Second, especially with older compilers (well, libraries, really) this can be horrendously inefficient, creating a huge number of temporary strings that you only use to compare with another string1.
So, let's consider how the job should be done instead. std::string has a member named find for situations like this. It returns the position of one string inside another, or std::string::npos if there is none. It allows you to specify a starting position at which to begin searching, when you don't want to start from the beginning.
We also, of course, have two instances of essentially identical code, once to search for SUVO, the other to search for SUVOJIT. The code would be much better off with the search code moved into a function, so we only have the search code in one place.
int count_pos(std::string const &haystack, std::string const &needle) {
size_t pos = 0;
int ret = 0;
while ((pos = haystack.find(needle, pos)) != std::string::npos) {
++ret;
++pos;
}
return ret;
}
Note that this also eliminates quite a bit more messy "stuff" like having to compute the maximum possible position at which at match could take place.
1. Why does compiler/library age matter? Older libraries often used a COW string that dynamically allocated storage for every string. More recent ones typically include what's called a "short string optimization", where storage for a short string is allocated inside the string object itself, avoiding the dynamic allocation.