Libraray Fine Hackerrank solution in C || Balanced Brackets Hackerrank solution in C

 

Library Fine

Your local library needs your help! Given the expected and actual return dates for a library book, create a program that calculates the fine (if any). The fee structure is as follows:

1. If the book is returned on or before the expected return date, no fine will be charged (i.e.: .
2. If the book is returned after the expected return day but still within the same calendar month and year as the expected return date, .
3. If the book is returned after the expected return month but still within the same calendar year as the expected return date, the .
4. If the book is returned after the calendar year in which it was expected, there is a fixed fine of .

Charges are based only on the least precise measure of lateness. For example, whether a book is due January 1, 2017 or December 31, 2017, if it is returned January 1, 2018, that is a year late and the fine would be .

Function Description

Complete the libraryFine function in the editor below. It must return an integer representing the fine due.

libraryFine has the following parameter(s):

• d1, m1, y1: returned date day, month and year
• d2, m2, y2: due date day, month and year

Input Format

The first line contains  space-separated integers, , denoting the respective , , and  on which the book was returned.
The second line contains  space-separated integers, , denoting the respective , , and  on which the book was due to be returned.

Constraints

Output Format

Print a single integer denoting the library fine for the book received as input.

Sample Input

9 6 2015
6 6 2015

Sample Output

45

Hackerrank solution in C++:

#include<iostream>
using namespace std;
int main(){

    int d1,d2,m1,m2,y1,y2,fine = 0;
    cin >> d1 >> m1 >> y1 >> d2 >> m2 >> y2;

    if(d1 > d2 && m1 == m2 && y1 == y2)fine = 15 * (d1-d2);
    else if(m1 > m2 && y1 == y2)fine = 500 * (m1-m2);
    else if(y1 > y2)fine  = 10000;
    cout << fine << endl;
        
    return 0;
}

Balanced Brackets

A bracket is considered to be any one of the following characters: (, ), {, }, [, or ].
Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and ().
A matching pair of brackets is not balanced if the set of brackets it encloses are not matched. For example, {[(])} is not balanced because the contents in between { and } are not balanced. The pair of square brackets encloses a single, unbalanced opening bracket, (, and the pair of parentheses encloses a single, unbalanced closing square bracket, ].
By this logic, we say a sequence of brackets is balanced if the following conditions are met:
It contains no unmatched brackets.
The subset of brackets enclosed within the confines of a matched pair of brackets is also a matched pair of brackets.
Given  strings of brackets, determine whether each sequence of brackets is balanced. If a string is balanced, return YES. Otherwise, return NO.
Function Description
Complete the function isBalanced in the editor below. It must return a string: YES if the sequence is balanced or NO if it is not.
isBalanced has the following parameter(s):
s: a string of brackets
Input Format
The first line contains a single integer , the number of strings.
Each of the next  lines contains a single string , a sequence of brackets.
Constraints
, where  is the length of the sequence.
All chracters in the sequences ∈ { {, }, (, ), [, ] }.
Output Format
For each string, return YES or NO.
Sample Input
3
{[()]}
{[(])}
{{[[(())]]}}
Sample Output
YES
NO
YES
Solution in C++;

#include <iostream>
#include <algorithm>
#include <unordered_map>
#include <stack>
using namespace std;
string isBalanced(string s){
    stack <char> st;
    for(auto c:s){
        switch (c){
            case '(':
            case '{':
            case '[':
                  st.push(c);
                   break;
            case '}':
                if(st.empty() || st.top()!='{' )
                    return "NO";
                st.pop();
                break;
            case ']':
                if(st.empty() || st.top()!='[')
                    return "NO";
                st.pop();
                break;
            case ')':
                if(st.empty() || st.top()!='(')
                    return "NO";
                st.pop();
                break;
            default: break;
        }
    }
    return st.empty() ? "YES":"NO";
}

int main(){
    int t;
    cin >> t;
    for(int a0 = 0; a0 < t; a0++){
        string s;
        cin >> s;
        cout << isBalanced(s) << endl;
    }
    return 0;
}