Added a HackerRank Problem Balanced Brackets

Yash2110-Yash/BalancedBrackets(HR)/BalancedBrackets.cpp

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+#include <bits/stdc++.h>
+
+using namespace std;
+
+string ltrim(const string &);
+string rtrim(const string &);
+
+/*
+ * Complete the 'isBalanced' function below.
+ *
+ * The function is expected to return a STRING.
+ * The function accepts STRING s as parameter.
+ */
+
+string isBalanced(string s) 
+{
+    stack<char> st;
+    int len = s.size();
+    int i;
+
+    for(i=0;i<len;i++)
+
+    {
+        if(s[i] == '(' || s[i] == '{' || s[i] == '[')
+        {
+            st.push(s[i]);
+        }        
+
+ else if( !st.empty() && 
+          (s[i] == ')' && st.top() == '(' || 
+           s[i] == '}' && st.top() == '{' || 
+           s[i] == ']' && st.top() == '[')) {
+            st.pop();
+        }
+
+        else 
+        return "NO";
+
+    }
+
+    if(st.empty())
+        return "YES";
+    else
+        return "NO";
+
+}
+
+int main()
+{
+    ofstream fout(getenv("OUTPUT_PATH"));
+
+    string t_temp;
+    getline(cin, t_temp);
+
+    int t = stoi(ltrim(rtrim(t_temp)));
+
+    for (int t_itr = 0; t_itr < t; t_itr++) {
+        string s;
+        getline(cin, s);
+
+        string result = isBalanced(s);
+
+        fout << result << "\n";
+    }
+
+    fout.close();
+
+    return 0;
+}
+
+string ltrim(const string &str) {
+    string s(str);
+
+    s.erase(
+        s.begin(),
+        find_if(s.begin(), s.end(), not1(ptr_fun<int, int>(isspace)))
+    );
+
+    return s;
+}
+
+string rtrim(const string &str) {
+    string s(str);
+
+    s.erase(
+        find_if(s.rbegin(), s.rend(), not1(ptr_fun<int, int>(isspace))).base(),
+        s.end()
+    );
+
+    return s;
+}

Yash2110-Yash/BalancedBrackets(HR)/README.md

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+Question:
+</br>
+        Link: https://www.hackerrank.com/challenges/balanced-brackets/problem?h_r=internal-search
+        </br>
+Answer:
+</br>
+        - Here we use stacks to first store the opening brackets and then compare the last added bracket with the closing brakcets being added.
+        </br>
+        - We keep poping out the brackets from the stack which are balanced i,e they find their opposite pair in sequence
+        </br>
+        - If in the end we are able to make the stack empty then the string has balanced brackets