Arraychallenge (strArr) read the array of
strings stored in strArr which will contain
only two elements, both of which will represent
an array of positive integers. For example: if
strArr is ["[1, 2, 5, 6)", "[5, 2, 8, 11]'), then both
elements in the input represent two integer
arrays, and your goal for this challenge is to
add the elements in corresponding locations
from both arrays. For the example input your
program should do the following additions: [(1
+ 5), (2 + 2), (5+ 8). (6 + 11)] which then eguals
[6, 4, 13, 17). Your program should finally return
this resulting array in' a string format with each
element separated by a hyphen. 6-4-13-17.
If the two 'arrays, do not have the same amouni
of elements,
then simply append the remaining
elements onto the new
arraykexample shown
"below). Both arrays will be in the format (e 1
e2, e3,..] where at least one element will exist
in each array
Cevamles
Answers
Answer:
#include <iostream>
#include <string>
#include <vector>
#include <sstream>
using namespace std;
string ArrayMatching(string strArr[], int size)
{
int length = 0, length2 = 100 * 100, total, high;
vector <vector<int> > list(size);
vector <int> newArray;
string result;
for (int row = 0; row < size; row++)
{
for (int col = 0; col < strArr[row].length(); col++)
{
if (strArr[row][col] == '[' || strArr[row][col] == ' ') // Removing unnecessary characters
{
strArr[row].erase(strArr[row].begin() + col);
}
}
// Passing our string values to a new int list
// This will make it easier for calculation and other requirements
string temp;
for (int x = 0; x < strArr[row].length(); x++)
{
if (strArr[row][x] == ',' || x == strArr[row].length() - 1)
{
int num;
istringstream(temp) >> num;
list[row].push_back(num);
temp.clear();
}
else
{
temp.push_back(strArr[row][x]);
}
}
// Condition to keep track of the largest and shortest sub array
if (list[row].size() > length)
{
length = list[row].size();
high = row; // will determine from which sub-array we need to append if need to
}
if (list[row].size() < length2)
{
length2 = list[row].size();
}
}
// Loop to perform the calculation
for (int x = 0; x < length2; x++)
{
total = 0;
total += list[0][x];
total += list[1][x];
newArray.push_back(total);
}
// Condition to check if we need to append remaining elements
if (newArray.size() < length)
{
for (int x = length2; x < length; x++)
{
newArray.push_back(list[high][x]);
}
}
// Loop for formatting
for (int x = 0; x < newArray.size(); x++)
{
stringstream convert;
convert << newArray[x];
result += convert.str();
if (x != newArray.size() - 1)
{
result += "-";
}
}
return result;
}
int main()
{
string A[] = { "[1, 2, 5, 6]", "[5, 2, 8, 11]" };
string B[] = { "[5, 2, 3]", "[2, 2, 3, 10, 6]" };
cout << ArrayMatching(A, sizeof(A) / sizeof(A[0])) << endl; // 6-4-13-17
cout << ArrayMatching(B, sizeof(B) / sizeof(B[0])) << endl; // 7-4-6-10-6
return 0;
}
Explanation: