I am trying to create a program that will take in an expression (example: "10*2+1") and solve it. My code is:
#include <iostream>
#include <string>
#include <vector>
void calculateString(std::string str);
int main() {
calculateString("10*2+2");
system("pause");
}
void calculateString(std::string str) {
int total = 0;
std::string temp1 = "";
std::string temp2 = "";
std::string add = "";
std::string *ray = new std::string[str.length()];
std::vector<int> newRay;
for (int i = 0; i < str.length(); i++) {
if (str.at(i) != '*' && str.at(i) != '/' && str.at(i) != '+' && str.at(i) != '-') {
add += str.at(i);
}
else {
ray[i] = add;
std::cout << ray[i] << "\n";
add = "";
}
}
for (int i = 0; i < str.length(); i++) {
if (ray[i].compare("*")) {
total = atoi(ray[i - 1].c_str()) * atoi(ray[i + 1].c_str());
newRay.push_back(total);
}
}
for (int i = 0; i < str.length(); i++) {
if (ray[i] == "+") {
newRay.push_back(atoi(ray[i - 1].c_str()) + atoi(ray[i + 1].c_str()));
}
}
for (int i = 0; i < newRay.size(); i++) {
std::cout << newRay[i] << "\n";
total += newRay[i];
}
std::cout << str << "=" << total << "\n";
}
However, whenever I run this I keep running into an access violation error that reads:
Exception thrown at 0x0F1CD4A0 (ucrtbased.dll) in CalcString.exe:
0xC0000005: Access violation reading location 0x01BE0FEE.
And it points to line 34 which is this:
total = atoi(ray[i - 1].c_str()) * atoi(ray[i + 1].c_str());
which is basically calculating the multiplication part of the expression then storing the asnwer in one variable. I have tried everything from changing my array to a vector to trying to rewrite all the methods and nothing seems to work. Please help
In
if (ray[i].compare("*"))
compare is misused. According to cppreference compare returns < 0 for smaller than, 0 for equal or > 0 for larger. As an if condition, 0 is false and everything else is true, so this resolves to
if (ray[i] != "*")
Probably the opposite of what is desired.
This allows entry into the if's body when ray[0] is "10" and i is 0, resulting in
total = atoi(ray[0 - 1].c_str()) * atoi(ray[0 + 1].c_str());
or
total = atoi(ray[-1].c_str()) * atoi(ray[1].c_str());
and accessing negative array indexes is undefined behaviour. Looks like it manifested as a crash in this case.
Solution:
All we care about is equality in this case so we can get away with
if (ray[i] == "*")
just as was done for
if (ray[i] == "+")
I also recommend a check to ensure an operator is never the first element of ray.
Related
My string is "AAABBAABABB",
and I want to get the result as
A = 3
B = 2
A = 2
B = 1
A = 1
B = 2
I have tried to use
for (int i = 0; i < n - 1; i++) {
if (msg[i] == msg[i + 1]) {
if(msg[i]==A)
a++;
else
b++;
}
}
I tried this by it didn't work for me. And I don't understand if there any other ways to find it out. Please help me out.
Iterate through the array by followings:
If i = 0, we can set a variable as 0th character and counter by 1.
If ith character is equal to the previous character, we can increase the counter.
If ith character is not equal to the (i-1)th character we can print the character, counter and start counting the new character.
Try the following snippet:
char ch = msg[0];
int cnt = 1;
for (int i = 1; i < n; i ++){
if(msg[i] != msg[i-1]){
cout<<ch<<" "<<cnt<<endl;
cnt = 1;
ch = msg[i];
}
else {
cnt++;
}
}
cout<<ch<<" "<<cnt<<endl;
You can use std::vector<std::pair<char, std::size_t>> to store character occurrences.
Eventually, you would have something like:
#include <iostream>
#include <utility>
#include <vector>
#include <string>
int main() {
std::vector<std::pair<char, std::size_t>> occurrences;
std::string str{ "AAABBAABABB" };
for (auto const c : str) {
if (!occurrences.empty() && occurrences.back().first == c) {
occurrences.back().second++;
} else {
occurrences.emplace_back(c, 1);
}
}
for (auto const& it : occurrences) {
std::cout << it.first << " " << it.second << std::endl;
}
return 0;
}
It will output:
A 3
B 2
A 2
B 1
A 1
B 2
Demo
This is very similar to run length encoding. I think the simplest way (less line of codes) I can think of is like this:
void runLength(const char* msg) {
const char *p = msg;
while (p && *p) {
const char *start = p++; // start of a run
while (*p == *start) p++; // move p to next run (different run)
std::cout << *start << " = " << (p - start) << std::endl;
}
}
Please note that:
This function does not need to know the length of input string before hand, it will stop at the end of string, the '\0'.
It also works for empty string and NULL. Both these work: runLength(""); runLength(nullptr);
I can not comment yet, if you look carefully, mahbubcseju's code does not work for empty msg.
With std, you might do:
void print_sequence(const std::string& s)
{
auto it = s.begin();
while (it != s.end()) {
auto next = std::adjacent_find(it, s.end(), std::not_equal_to<>{});
next = next == s.end() ? s.end() : next + 1;
std::cout << *it << " = " << std::distance(it, next) << std::endl;
it = next;
}
}
Demo
Welcome to stackoverflow. Oooh, an algorithm problem? I'll add a recursive example:
#include <iostream>
void countingThing( const std::string &input, size_t index = 1, size_t count = 1 ) {
if( input.size() == 0 ) return;
if( input[index] != input[index - 1] ) {
std::cout << input[index - 1] << " = " << count << std::endl;
count = 0;
}
if( index < input.size() ) return countingThing( input, index + 1, count + 1 );
}
int main() {
countingThing( "AAABBAABABB" );
return 0;
}
To help work out algorithms and figuring out what to write in your code, I suggest a few steps:
First, write out your problem in multiple ways, what sort of input it expects and how you would like the output to be.
Secondly, try and solve it on paper, how the logic would work - a good tip to this is try to understand how YOU would solve it. Your brain is a good problem solver, and if you can listen to what it does, you can turn it into code (it isn't always the most efficient, though).
Thirdly, work it out on paper, see if your solution does what you expect it to do by following your steps by hand. Then you can translate the solution to code, knowing exactly what you need to write.
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.
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
*Sorry about my poor English. If there is anything that you don't understand, please tell me so that I can give you more information that 'make sence'.
**This is first time asking question in Stackoverflow. I've searched some rules for asking questions correctly here, but there should be something I missed. I welcome all feedback.
I'm currently solving algorithm problems to improve my skill, and I'm struggling with one question for three days. This question is from https://algospot.com/judge/problem/read/RESTORE , but since this page is in KOREAN, I tried to translate it in English.
Question
If there are 'k' pieces of partial strings given, calculate shortest string that includes all partial strings.
All strings consist only lowercase alphabets.
If there are more than 1 result strings that satisfy all conditions with same length, choose any string.
Input
In the first line of input, number of test case 'C'(C<=50) is given.
For each test case, number of partial string 'k'(1<=k<=15) is given in the first line, and in next k lines partial strings are given.
Length of partial string is between 1 to 40.
Output
For each testcase, print shortest string that includes all partial strings.
Sample Input
3
3
geo
oji
jing
2
world
hello
3
abrac
cadabra
dabr
Sample Output
geojing
helloworld
cadabrac
And here is my code. My code seems to work perfect with Sample Inputs, and when I made test inputs for my own and tested, everything worked fine. But when I submit this code, they say my code is 'wrong'.
Please tell me what is wrong with my code. You don't need to tell me whole fixed code, I just need sample inputs that causes error with my code. Added code description to make my code easier to understand.
Code Description
Saved all input partial strings in vector 'stringParts'.
Saved current shortest string result in global variable 'answer'.
Used 'cache' array for memoization - to skip repeated function call.
Algorithm I designed to solve this problem is divided into two function -
restore() & eraseOverlapped().
restore() function calculates shortest string that includes all partial strings in 'stringParts'.
Result of resotre() is saved in 'answer'.
For restore(), there are three parameters - 'curString', 'selected' and 'last'.
'curString' stands for currently selected and overlapped string result.
'selected' stands for currently selected elements of 'stringParts'. Used bitmask to make my algorithm concise.
'last' stands for last selected element of 'stringParts' for making 'curString'.
eraseOverlapped() function does preprocessing - it deletes elements of 'stringParts' that can be completly included to other elements before executing restore().
#include <algorithm>
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#define MAX 15
using namespace std;
int k;
string answer; // save shortest result string
vector<string> stringParts;
bool cache[MAX + 1][(1 << MAX) + 1]; //[last selected string][set of selected strings in Bitmask]
void restore(string curString, int selected=0, int last=0) {
//base case 1
if (selected == (1 << k) - 1) {
if (answer.empty() || curString.length() < answer.length())
answer = curString;
return;
}
//base case 2 - memoization
bool& ret = cache[last][selected];
if (ret != false) return;
for (int next = 0; next < k; next++) {
string checkStr = stringParts[next];
if (selected & (1 << next)) continue;
if (curString.empty())
restore(checkStr, selected + (1 << next), next + 1);
else {
int check = false;
//count max overlapping area of two strings and overlap two strings.
for (int i = (checkStr.length() > curString.length() ? curString.length() : checkStr.length())
; i > 0; i--) {
if (curString.substr(curString.size()-i, i) == checkStr.substr(0, i)) {
restore(curString + checkStr.substr(i, checkStr.length()-i), selected + (1 << next), next + 1);
check = true;
break;
}
}
if (!check) { // if there aren't any overlapping area
restore(curString + checkStr, selected + (1 << next), next + 1);
}
}
}
ret = true;
}
//check if there are strings that can be completely included by other strings, and delete that string.
void eraseOverlapped() {
//arranging string vector in ascending order of string length
int vectorLen = stringParts.size();
for (int i = 0; i < vectorLen - 1; i++) {
for (int j = i + 1; j < vectorLen; j++) {
if (stringParts[i].length() < stringParts[j].length()) {
string temp = stringParts[i];
stringParts[i] = stringParts[j];
stringParts[j] = temp;
}
}
}
//deleting included strings
vector<string>::iterator iter;
for (int i = 0; i < vectorLen-1; i++) {
for (int j = i + 1; j < vectorLen; j++) {
if (stringParts[i].find(stringParts[j]) != string::npos) {
iter = stringParts.begin() + j;
stringParts.erase(iter);
j--;
vectorLen--;
}
}
}
}
int main(void) {
int C;
cin >> C; // testcase
for (int testCase = 0; testCase < C; testCase++) {
cin >> k; // number of partial strings
memset(cache, false, sizeof(cache)); // initializing cache to false
string inputStr;
for (int i = 0; i < k; i++) {
cin >> inputStr;
stringParts.push_back(inputStr);
}
eraseOverlapped();
k = stringParts.size();
restore("");
cout << answer << endl;
answer.clear();
stringParts.clear();
}
}
After determining which string-parts can be removed from the list since they are contained in other string-parts, one way to model this problem might be as the "taxicab ripoff problem" problem (or Max TSP), where each potential length reduction by overlap is given a positive weight. Considering that the input size in the question is very small, it seems likely that they expect a near brute-force solution, with possibly some heuristic and backtracking or other form of memoization.
Thanks Everyone who tried to help me solve this problem. I actually solved this problem with few changes on my previous algorithm. These are main changes.
In my previous algorithm I saved result of restore() in global variable 'answer' since restore() didn't return anything, but in new algorithm since restore() returns mid-process answer string I no longer need to use 'answer'.
Used string type cache instead of bool type cache. I found out using bool cache for memoization in this algorithm was useless.
Deleted 'curString' parameter from restore(). Since what we only need during recursive call is one previously selected partial string, 'last' can replace role of 'curString'.
CODE
#include <algorithm>
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#define MAX 15
using namespace std;
int k;
vector<string> stringParts;
string cache[MAX + 1][(1 << MAX) + 1];
string restore(int selected = 0, int last = -1) {
if (selected == (1 << k) - 1) {
return stringParts[last];
}
if (last == -1) {
string ret = "";
for (int next = 0; next < k; next++) {
string resultStr = restore(selected + (1 << next), next);
if (ret.empty() || ret.length() > resultStr.length())
ret = resultStr;
}
return ret;
}
string& ret = cache[last][selected];
if (!ret.empty()) {
cout << "cache used in [" << last << "][" << selected << "]" << endl;
return ret;
}
string curString = stringParts[last];
for (int next = 0; next < k; next++) {
if (selected & (1 << next)) continue;
string checkStr = restore(selected + (1 << next), next);
int check = false;
string resultStr;
for (int i = (checkStr.length() > curString.length() ? curString.length() : checkStr.length())
; i > 0; i--) {
if (curString.substr(curString.size() - i, i) == checkStr.substr(0, i)) {
resultStr = curString + checkStr.substr(i, checkStr.length() - i);
check = true;
break;
}
}
if (!check)
resultStr = curString + checkStr;
if (ret.empty() || ret.length() > resultStr.length())
ret = resultStr;
}
return ret;
}
void EraseOverlapped() {
int vectorLen = stringParts.size();
for (int i = 0; i < vectorLen - 1; i++) {
for (int j = i + 1; j < vectorLen; j++) {
if (stringParts[i].length() < stringParts[j].length()) {
string temp = stringParts[i];
stringParts[i] = stringParts[j];
stringParts[j] = temp;
}
}
}
vector<string>::iterator iter;
for (int i = 0; i < vectorLen - 1; i++) {
for (int j = i + 1; j < vectorLen; j++) {
if (stringParts[i].find(stringParts[j]) != string::npos) {
iter = stringParts.begin() + j;
stringParts.erase(iter);
j--;
vectorLen--;
}
}
}
}
int main(void) {
int C;
cin >> C;
for (int testCase = 0; testCase < C; testCase++) {
cin >> k;
for (int i = 0; i < MAX + 1; i++) {
for (int j = 0; j < (1 << MAX) + 1; j++)
cache[i][j] = "";
}
string inputStr;
for (int i = 0; i < k; i++) {
cin >> inputStr;
stringParts.push_back(inputStr);
}
EraseOverlapped();
k = stringParts.size();
string resultStr = restore();
cout << resultStr << endl;
stringParts.clear();
}
}
This algorithm is much slower than the 'ideal' algorithm that the book I'm studying suggests, but it was fast enough to pass this question's time limit.
I'm creating a program that creates an array of objects in random positions in an array size 8. Once created, I need them to sort so that all the objects in the array are shifted up to the top, so no gaps exist between them. I'm almost there, but I cannot seem to get them to swap to index 0 in the array, and they instead swap to index 1. Any suggestions? (Must be done the way I'm doing it, not with other sorting algorithms or whatnot)
#include <iostream>
#include <string>
#include <ctime>
using namespace std;
struct WordCount {
string name = "";
int count = 0;
};
int main() {
cout << "Original random array: " << endl;
srand(static_cast<int>(time(0)));
int i = 0;
WordCount wordArr[8];
while (i < 4) {
int randomNum = 0 + (rand() % static_cast<int>(7 + 1));
if(wordArr[randomNum].name == "") {
wordArr[randomNum].name = "word" + static_cast<char>(i);
wordArr[randomNum].count = i;
i++;
}
}
int j = 0;
while (j < 8) {
cout << wordArr[j].name << " " << wordArr[j].count << endl;
j++;
}
cout << "\n\nSorted array: " << endl;
for (int i = 7; i >= 0; i--) {
for (int j = 0; j <= 7; j++) {
if (wordArr[i].name != "") {
if (wordArr[j].name == "") {
WordCount temp = wordArr[i];
wordArr[i] = wordArr[j];
wordArr[j] = temp;
}
}
}
}
int k = 0;
while (k < 8) {
cout << wordArr[k].name << " " << wordArr[k].count << endl;
k++;
}
return 0;
}
If I understand your requirement correctly, you want to move all the non-blank entries to the start of the array. To do this, you need an algorithm like this for example:
for i = 0 to 7
if wordArr[i].name is blank
for j = i + 1 to 7
if wordArr[j].name is not blank
swap [i] and [j]
break
So, starting from the beginning, if we encounter a blank entry, we look forward for the next non-blank entry. If we find such an entry, we swap the blank and non-blank entry, then break to loop again looking for the next blank entry.
Note, this isn't the most efficient of solutions, but it will get you started.
Note also I'd replace the 4 and 8 with definitions like:
#define MAX_ENTRIES (8)
#define TO_GENERATE_ENTRIES (4)
Finally:
wordArr[randomNum].name = "word" + static_cast<char>(i);
That will not do what you want it to do; try:
wordArr[randomNum].name = "word" + static_cast<char>('0' + i);
To append the digits, not the byte codes, to the end of the number. Or perhaps, if you have C++11:
wordArr[randomNum].name = "word" + std::to_string(i);
I see couple of problems.
The expression "word" + static_cast<char>(i); doesn't do what you are hoping to do.
It is equivalent to:
char const* w = "word";
char const* p = w + i;
When i is 2, p will be "rd". You need to use std::string("word") + std::to_string(i).
The logic for moving objects with the non-empty names to objects with empty names did not make sense to me. It obviously does not work for you. The following updated version works for me:
for (int i = 0; i <= 7; ++i) {
// If the name of the object at wordArr[i] is not empty, move on to the
// next item in the array. If it is empty, copy the next object that
// has a non-empty name.
if ( wordArr[i].name == "") {
// Start comparing from the object at wordArr[i+1]. There
// is no need to start at wordArr[i]. We know that it is empty.
for (int j = i+1; j <= 7; ++j) {
if (wordArr[j].name != "") {
WordCount temp = wordArr[i];
wordArr[i] = wordArr[j];
wordArr[j] = temp;
}
}
}
}
There was two problems as :
wordArr[randomNum].name = "word" + static_cast<char>(i); this is not what your are looking for, if you want that your names generate correctly you need something like this :
wordArr[randomNum].name = "word " + std::to_string(i);
Your sorting loop does not do what you want, it's just check for the "gaps" as you said, you need something like this :
for (int i = 0; i < 8; ++i) {
for (int j = i+1; j < 8; ++j) {
if (wordArr[i].name == "" || (wordArr[i].count < wordArr[j].count)) {
WordCount temp = wordArr[i];
wordArr[i] = wordArr[j];
wordArr[j] = temp;
}
}
}
Your algorithm sorts the array, but then looses the sorting again.
You want to swap elements only when i > j, in order to push elements to the top only. As a result, you need to change this:
if (wordArr[j].name == "")
to this:
if (wordArr[j].name == "" && i > j)
Consider this array example:
0
ord 1
0
0
rd 2
word 0
d 3
0
Your code will sort it to:
d 3
ord 1
word 0
rd 2
0
0
0
0
but when i = 3, it will try to populate the 5th cell, and it will swap it with rd 2, which is not what we want.
This will push rd 2 down, but we don't want that, we want gaps (zeroes) to go to the end of the array, thus we need to swap eleemnts only when they are going to go higher, not lower, which is equivalent to say when i > j.
PS: If you are a beginner skip that part.
You can optimize the inner loop by using one if statement and a break keyword, like this:
for (int j = 0; j <= 7; j++) {
if (wordArr[i].name != "" && wordArr[j].name == "" && i > j) {
WordCount temp = wordArr[i];
wordArr[i] = wordArr[j];
wordArr[j] = temp;
break;
}
}