Lonely Integer Program in C++

Lonely Integer.cpp

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+/* Hackerrank Problem: You will be given an array of integers. All of the integers except one occur twice. That one is unique in the array.
+
+Given an array of integers, find and print the unique element.
+
+For example, , the unique element is .
+
+Function Description
+
+Complete the lonelyinteger function in the editor below. It should return the integer which occurs only once in the input array.
+
+lonelyinteger has the following parameter(s):
+
+a: an array of integers
+Input Format
+
+The first line contains a single integer, , denoting the number of integers in the array.
+The second line contains  space-separated integers describing the values in .
+
+Constraints
+
+It is guaranteed that  is an odd number and that there is one unique element.
+, where .
+Output Format
+
+Print the unique integer in the array.*/
+
+#include <bits/stdc++.h>
+
+using namespace std;
+
+vector<string> split_string(string);
+int lonelyinteger(vector<int> a) {
+int c=0;
+int uni;
+for(int i=0;i<a.size();i++){
+c=count(a.begin(),a.end(),a[i]);
+if(c==1){uni=a[i];}
+}
+return uni;
+}
+
+int main()
+{
+    ofstream fout(getenv("OUTPUT_PATH"));
+
+    int n;
+    cin >> n;
+    cin.ignore(numeric_limits<streamsize>::max(), '\n');
+
+    string a_temp_temp;
+    getline(cin, a_temp_temp);
+
+    vector<string> a_temp = split_string(a_temp_temp);
+
+    vector<int> a(n);
+
+    for (int i = 0; i < n; i++) {
+        int a_item = stoi(a_temp[i]);
+
+        a[i] = a_item;
+    }
+
+    int result = lonelyinteger(a);
+
+    fout << result << "\n";
+
+    fout.close();
+
+    return 0;
+}
+
+vector<string> split_string(string input_string) {
+    string::iterator new_end = unique(input_string.begin(), input_string.end(), [] (const char &x, const char &y) {
+        return x == y and x == ' ';
+    });
+
+    input_string.erase(new_end, input_string.end());
+
+    while (input_string[input_string.length() - 1] == ' ') {
+        input_string.pop_back();
+    }
+
+    vector<string> splits;
+    char delimiter = ' ';
+
+    size_t i = 0;
+    size_t pos = input_string.find(delimiter);
+
+    while (pos != string::npos) {
+        splits.push_back(input_string.substr(i, pos - i));
+
+        i = pos + 1;
+        pos = input_string.find(delimiter, i);
+    }
+
+    splits.push_back(input_string.substr(i, min(pos, input_string.length()) - i + 1));
+
+    return splits;
+}
+