Displaying first element user input in an Array - c++

Question

We can define the term 'value of a name' as the average position of
the letters in the name, calculating 'A' as 1, 'B' as 2, 'C' as 3, and
so on. The value of "BOB" would be (2 + 15 + 2)/ 3 = 6. According to
this value, the names will be arranged from the smallest towards the
biggest in the output. When two or more names have the same value,
the name which is in the first position in the original list (the
first one the user inputs) should show up first in the sorted list
(the output).
Input In the first line we have an integer N (1 <= N <= 100), which is
the number of names. In every of the N lines we have one name ([A-Z],
no empty spaces). Names contain 1 - 200 letters.
Output Print out the sorted list (one name in a line).
Test-case
Input: 3 BOB AAAAAAA TOM Output: AAAAAAA BOB TOM
I tried something, and the code seemed to work, I just had a problem with the output. I couldn't find a way to arrange the names with the same value, according to their position in the original list. Here's the other test-case I tried, but didn't figure out:
Input:
10
COSOPYILSPKNKZSTUZVMEERQDL
RRPPNG
PQUPOGTJETGXDQDEMGPNMJEBI
TQJZMOLQ
BKNGFEJZWMJNJLSTUBHCFHXWMYUPZM
YNWEPZKNBOOXNZVWKIUS
LV
CJDFYDMYZVOEW
TMHEJLIDEHT
KGTGFIFWYTKPWTYQQPGKRRYFXN
Output:
TMHEJLIDEHT
PQUPOGTJETGXDQDEMGPNMJEBI
BKNGFEJZWMJNJLSTUBHCFHXWMYUPZM
CJDFYDMYZVOEW
RRPPNG
COSOPYILSPKNKZSTUZVMEERQDL
KGTGFIFWYTKPWTYQQPGKRRYFXN
TQJZMOLQ
YNWEPZKNBOOXNZVWKIUS
LV
My output:
TMHEJLIDEHT
PQUPOGTJETGXDQDEMGPNMJEBI
CJDFYDMYZVOEW // these two
BKNGFEJZWMJNJLSTUBHCFHXWMYUPZM // should be arranged with their places switched
RRPPNG
COSOPYILSPKNKZSTUZVMEERQDL
KGTGFIFWYTKPWTYQQPGKRRYFXN
TQJZMOLQ
YNWEPZKNBOOXNZVWKIUS
LV
#include <iostream>
#include <string>
using namespace std;
int main() {
int N;
cin >> N;
string words[N];
int res[N];
for (int i = 0; i < N; i++) {
int sum = 0;
int value = 0;
int temp = 0;
string word;
cin >> words[i];
word = words[i];
for (int j = 0; j < word.length(); j++) {
sum += (int)word[j] - 64;
}
value = sum / word.length();
res[i] = value;
}
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (res[i] < res[j]) {
swap(res[i], res[j]);
swap(words[i], words[j]);
}
}
}
for (int i = 0; i < N; i++) {
cout << words[i] << endl;
}
return 0;
}

Answer 1

string words[N];
int res[N];
This here is not valid C++, you can not size a stack array using a runtime variable, although some compilers might support such a feature. You might use say std::vector instead, which behaves much like an array.
vector<string> words;
vector<int> res;
for (int i = 0; i < N; i++) {
int sum = 0;
int value = 0;
int temp = 0;
string word;
cin >> word;
words.push_back(word);
for (int j = 0; j < word.length(); j++) {
sum += (int)word[j] - 64;
}
value = sum / word.length();
res.push_back(value);
}
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (res[i] < res[j]) {
swap(res[i], res[j]);
swap(words[i], words[j]);
}
}
}
The ordering is because your sorting algorithm is not stable. Stable means that items with equal values will maintain the same order relative to each other.
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (res[i] < res[j]) {
swap(res[i], res[j]);
swap(words[i], words[j]);
}
}
}
What you have is very close to bubble sort, which is stable.
for (int i = 0; i < N; i++) {
for (int j = 0; j < N - i - 1; j++) { // i elements sorted so far
if (res[j] > res[j + 1]) {
swap(res[j], res[j + 1]);
swap(words[j], words[j + 1]);
}
}
}
C++ also provides a stable sort in <algorithm>, but it can't function directly on two arrays like this unfortunately, one option is to compute the value on the fly, another could be to make a class holding both items and sort that, or another to sort the indices.
std::stable_sort(words.begin(), words.end(), [&](auto &a, auto &b)
{
int suma = 0, sumb = 0; // better yet, make a "int value(const string &str)" function.
for (int j = 0; j < a.length(); j++) {
suma += (int)a[j] - 64;
}
for (int j = 0; j < b.length(); j++) {
sumb += (int)b[j] - 64;
}
int valuea = suma / a.length();
int valueb = sumb / b.length();
return valuea < valueb;
});
A class containing both items is pretty straight forward, for indices, make a 3rd array and sort that.
vector<size_t> indices;
...
string word;
cin >> word;
indices.push_back(words.size());
words.push_back(word);
...
std::stable_sort(indices.begin(), indices.end(), [&](auto a, auto b){ return res[a] < res[n]; });
for (int i = 0; i < N; i++) {
cout << words[indices[i]] << endl;
}

Answer 2

A possible solution could be order the result array during construction.
When you add the words in the result array, use the result obtained to add the word in the right place. In this way you can check if exist already the same value and add the new word after the previous with the same value.

Answer 3

After reading the next word use insertion sort (wiki) which is stable
read word
calculate value
insert in a right place in the array
go to 1 until i < N otherwise print out
Doesn't require additional sorting procedure.

Answer 4

in python:
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
times = int(input())
entries = []
ordered = []
for x in range(times):
entries.append(input())
for x in entries:
chars = []
for y in x:
chars.append(ord(y) - 96)
ordered.append(sum(chars))
print(sort_list(entries,ordered))

Answer 5

If you use a std::multimap<int, std::string>, there would be no need to sort, as the key would already serve as the sorting criteria.
Here is a solution using std::multimap:
#include <string>
#include <numeric>
#include <iostream>
#include <sstream>
#include <map>
// Test data
std::string test = "10\n"
"COSOPYILSPKNKZSTUZVMEERQDL\n"
"RRPPNG\n"
"PQUPOGTJETGXDQDEMGPNMJEBI\n"
"TQJZMOLQ\n"
"BKNGFEJZWMJNJLSTUBHCFHXWMYUPZM\n"
"YNWEPZKNBOOXNZVWKIUS\n"
"LV\n"
"CJDFYDMYZVOEW\n"
"TMHEJLIDEHT\n"
"KGTGFIFWYTKPWTYQQPGKRRYFXN\n";
int main()
{
std::istringstream strm(test);
// Read in the data
std::multimap<int, std::string> strmap;
int N;
strm >> N;
std::string word;
for (int i = 0; i < N; ++i)
{
strm >> word;
// get the average using std::accumulate and divide by the length of the word
int avg = std::accumulate(word.begin(), word.end(), 0,
[&](int total, char val) { return total + val - 'A' + 1; }) / word.length();
// insert this value in the map
strmap.insert({ avg, word });
}
// output results
for (auto& w : strmap)
std::cout << w.second << "\n";
}
Output:
TMHEJLIDEHT
PQUPOGTJETGXDQDEMGPNMJEBI
BKNGFEJZWMJNJLSTUBHCFHXWMYUPZM
CJDFYDMYZVOEW
RRPPNG
COSOPYILSPKNKZSTUZVMEERQDL
KGTGFIFWYTKPWTYQQPGKRRYFXN
TQJZMOLQ
YNWEPZKNBOOXNZVWKIUS
LV
The std::accumulate is used to add up the values to get the average.

Answer 6

Or just order them in the end (You won't need the 2nd array):
for (int i = 0; i < N; i++) {
for (int j = i + 1; j < N; j++) {
int sumA = 0, sumB = 0;
for (int k = 0; k < words[i].size(); k++)
sumA += words[i][k] - 'A' + 1;
for (int k = 0; k < words[j].size(); k++)
sumB += words[j][k] - 'A' + 1;
if (sumA / words[i].size() > sumB / words[j].size())
swap(words[i], words[j]);
}
}
As they shown above, it's way better to use a vector to store your data.