Category: HackerRank

Arrays Introduction

Problem Statement

An array is a series of elements of the same type placed in contiguous memory locations that can be individually referenced by adding an index to a unique identifier.

Declaration:Movie Fifty Shades Darker (2017)

int arr[10]; //Declares an array named arr of size 10, i.e; you can store 10  integers.

Accessing elements of an array:

Indexing in arrays starts from 0.So the first element is stored at arr[0],the second element at arr[1]...arr[9]

You’ll be an given array of N integers and you have to print the integers in the reverse order.

Input Format

The first line of the input contains N,where N is the number of integers.The next line contains N integers separated by a space.

Constraints

1<=N<=1000

1<=Ai<=10000, where Ai is the ith integer in the array.

Output Format

Print the N integers of the array in the reverse order in a single line separated by a space.

Sample Input

4
1 4 3 2

Sample Output

2 3 4 1

Solution

  
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int n,i,a[max];
    cin>>n;
    for(i=0;i<n;i++)
        cin>>a[i];
    for(i=n-1;i>=0;i--)
        cout<<a[i]<<" ";
    return 0;
}

Input and Output

Problem Statement

For any program that we write the basic things that we need to do is take the input and print the expected output.

In C++ you can take the input using cin and print the output using cout.Unlike C where you need the format specifier in printf and scanf, here you can use cin and cout.

Taking Input:

If you want to input a number: cin>>n ,  where n is the number.
If you want to input a number and a string: cin>>n>>s, where s is the string.

Printing output:

If you want to output a single number: cout<<n
If you want to output a number and a string separated by a new line: cout<<n<<endl<<s (where endl moves the pinter to the new line and then string gets printed.)

Take three numbers as inputs and print the sum of the three numbers.

Input Format

The first line of the input contains three integers A,B and C.

1≤A,B,C≤1000

Output Format

In a single line print the sum of the three numbers.

Sample Input

1 2 7

Sample Output

10

Explanation

Sum of the three numbers 1,2 and 7 is 10.

Solution

  
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    int a,b,c;
    cin>>a>>b>>c;
    cout<<a+b+c;
    return 0;
}

Flipping bits

Problem Statement

You will be given a list of 32 bits unsigned integers. You are required to output the list of the unsigned integers you get by flipping bits in its binary representation (i.e. unset bits must be set, and set bits must be unset).

Input Format

The first line of the input contains the list size T, which is followed by T lines, each line having an integer from the list.

Constraints

1≤T≤100
0≤integer<232

Output Format

Output one line per element from the list with the requested result.

Sample Input

3 
2147483647 
1 
0

Sample Output

2147483648 
4294967294 
4294967295

Explanation

Take 1 for example, as unsigned 32-bits is 00000000000000000000000000000001 and doing the flipping we get 11111111111111111111111111111110 which in turn is 4294967294.

Solution

  
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    int t,n;
    unsigned int a;
    cin>>t;
    while(t--)
    {
        cin>>a;
        cout<<~a<<endl;      
    }
    return 0;
    
}

Funny String

Problem Statement

Suppose you have a string S which has length N and is indexed from 0 to N−1. String R is the reverse of the string S. The string S is funny if the condition |Si−Si−1|=|Ri−Ri−1| is true for every i from 1 to N−1.

(Note: Given a string str, stri denotes the ascii value of the ith character (0-indexed) of str. |x| denotes the absolute value of an integer x)

Input Format

First line of the input will contain an integer T. T testcases follow. Each of the next T lines contains one string S.

Constraints

  • 1<=T<=10
  • 2<=length of S<=10000

Output Format

For each string, print Funny or Not Funny in separate lines.

Sample Input

2
acxz
bcxz

Sample Output

Funny
Not Funny

Explanation

Consider the 1st testcase acxz

here

|c-a| = |x-z| = 2
|x-c| = |c-x| = 21
|z-x| = |a-c| = 2

Hence Funny.

Consider the 2nd testcase bcxz

here

|c-b| != |x-z|

Hence Not Funny.

Solution

  
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */ 
    int t,count;
    string s,r;
    cin>>t;
    while(t--)
    {
        count=0;
        r.clear();
        s.clear();
        cin>>s;
        r=s;
        reverse(r.begin(),r.end());
        for(int i=1;i<s.length();i++)
        {
            if(abs(s[i]-s[i-1])==abs(r[i]-r[i-1]))
            {
                count++;
            }
        }
        if(count==(s.length()-1))
            cout<<"Funny"<<endl;
        else cout<<"Not Funny"<<endl;
    }
    return 0;
}

Pangrams

Problem Statement

Roy wanted to increase his typing speed for programming contests. So, his friend advised him to type the sentence “The quick brown fox jumps over the lazy dog” repeatedly, because it is a pangram. (Pangrams are sentences constructed by using every letter of the alphabet at least once.)

After typing the sentence several times, Roy became bored with it. So he started to look for other pangrams.

Given a sentence s, tell Roy if it is a pangram or not.

Input Format Input consists of a line containing s.

Constraints
Length of s can be at most 103 (1≤|s|≤103) and it may contain spaces, lower case and upper case letters. Lower case and upper case instances of a letter are considered the same.

Output Format Output a line containing pangram if s is a pangram, otherwise output not pangram.

Sample Input #1

We promptly judged antique ivory buckles for the next prize

Sample Output #1

pangram

Sample Input #2

We promptly judged antique ivory buckles for the prize

Sample Output #2

not pangram

Explanation
In the first test case, the answer is pangram because the sentence contains all the letters of the English alphabet.

Solution in Python: 

  
# Enter your code here. Read input from STDIN. Print output to STDOUT
print 'pangram' if set(raw_input().lower().replace(' ', '')) == set('abcdefghijklmnopqrstuvwxyz') else 'not pangram'

Manasa and Stones

Problem Statement

Manasa is out on a hike with friends. She finds a trail of stones with numbers on them. She starts following the trail and notices that two consecutive stones have a difference of either a or b. Legend has it that there is a treasure trove at the end of the trail and if Manasa can guess the value of the last stone, the treasure would be hers. Given that the number on the first stone was 0, find all the possible values for the number on the last stone.

Note: The numbers on the stones are in increasing order.

Input Format
The first line contains an integer T, i.e. the number of test cases. T test cases follow; each has 3 lines. The first line contains n (the number of stones). The second line contains a, and the third line contains b.

Output Format
Space-separated list of numbers which are the possible values of the last stone in increasing order.

Constraints
1≤T≤10
1≤n,a,b≤103

Sample Input

2
3 
1
2
4
10
100

Sample Output

2 3 4 
30 120 210 300

Explanation

All possible series for the first test case are given below:

  1. 0,1,2
  2. 0,1,3
  3. 0,2,3
  4. 0,2,4

Hence the answer 2 3 4.

Series with different number of final steps for second test case are the following:

  1. 0, 10, 20, 30
  2. 0, 10, 20, 120
  3. 0, 10, 110, 120
  4. 0, 10, 110, 210
  5. 0, 100, 110, 120
  6. 0, 100, 110, 210
  7. 0, 100, 200, 210
  8. 0, 100, 200, 300

Hence the answer 30 120 210 300.

Solution

  
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    int t,n,a,b,i;
    cin>>t;
    while(t--)
    {
        cin>>n>>a>>b;
        if(b<a){
            int temp=a;
            a=b;
            b=temp;
            
        }
        //cout<<a<<" "<<b<<endl;
        if(a==b)
        {
            cout<<(n-1)*a<<endl;
            
        }
        else 
        {
            for(i=0;i<=n-1;i++)
            {
                cout<<(n-1-i)*a+(i)*b<<" ";
            
            }
        cout<<endl;
        }   
    }
    return 0;
}

Cavity Map

Problem Statement

You are given a square map of size n×n. Each cell of the map has a value denoting its depth. We will call a cell of the map a cavity if and only if this cell is not on the border of the map and each cell adjacent to it has strictly smaller depth. Two cells are adjacent if they have a common side (edge).

You need to find all the cavities on the map and depict them with the uppercase character X.

Input Format

The first line contains an integer, n, denoting the size of the map. Each of the following n lines contains n positive digits without spaces. Each digit (1-9) denotes the depth of the appropriate area.

Constraints
1≤n≤100

Output Format

Output n lines, denoting the resulting map. Each cavity should be replaced with character X.

Sample Input

4
1112
1912
1892
1234

Sample Output

1112
1X12
18X2
1234

Solution

  
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner sc = new Scanner(System.in);
        String line;
        int n,i,j;
        n = sc.nextInt();
        char [][]CharArray = new char[n][n];
        for(i=0;i<n;i++){
            line = sc.next();
            CharArray[i]= line.toCharArray();
        }
        for(i=1;i<n-1;i++){
            for(j=1;j<n-1;j++){
                if(CharArray[i][j] > CharArray[i][j-1] & CharArray[i][j] > CharArray[i][j+1] & CharArray[i][j] > CharArray[i-1][j] & CharArray[i][j] > CharArray[i+1][j])
                    CharArray[i][j] = 'X';
            }
        }
        for(i=0;i<n;i++){
            for(j=0;j<n;j++){
                System.out.print(CharArray[i][j]);
            }
            System.out.println();
        }
    }
}

Chocolate Feast

Problem Statement

Little Bob loves chocolate, and he goes to a store with $N in his pocket. The price of each chocolate is $C. The store offers a discount: for every M wrappers he gives to the store, he gets one chocolate for free. How many chocolates does Bob get to eat?

Input Format:
The first line contains the number of test cases, T.
T lines follow, each of which contains three integers, N, C, and M.

Output Format:
Print the total number of chocolates Bob eats.

Constraints:
1≤T≤1000
2≤N≤105
1≤C≤N
2≤M≤N

Sample input

3
10 2 5
12 4 4
6 2 2

Sample Output

6
3
5

Explanation
In the first case, he can buy 5 chocolates with $10 and exchange the 5 wrappers to get one more chocolate. Thus, the total number of chocolates is 6.

In the second case, he can buy 3 chocolates for $12. However, it takes 4 wrappers to get one more chocolate. He can’t avail the offer and hence the total number of chocolates remains 3.

In the third case, he can buy 3 chocolates for $6. Now he can exchange 2 of the 3 wrappers and get 1 additional piece of chocolate. Now he can use his 1 unused wrapper and the 1 wrapper of the new piece of chocolate to get one more piece of chocolate. So the total is 5.

Solution

  
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int t,n,c,m,total,brought;
    cin>>t;
    while(t--){
        cin>>n>>c>>m;
        int ans=0;
        // Computer answer
        brought = n/c;
        total = brought +(brought-1)/(m-1);
        cout<<total<<endl;
    }
    return 0;
}

Cut the sticks

Problem Statement

You are given N sticks, where the length of each stick is a positive integer. A cut operation is performed on the sticks such that all of them are reduced by the length of the smallest stick.

Suppose we have six sticks of the following lengths:

5 4 4 2 2 8

Then, in one cut operation we make a cut of length 2 from each of the six sticks. For the next cut operation four sticks are left (of non-zero length), whose lengths are the following:

3 2 2 6

The above step is repeated until no sticks are left.

Given the length of N sticks, print the number of sticks that are left before each subsequent cut operations.

Note: For each cut operation, you have to recalcuate the length of smallest sticks (excluding zero-length sticks).

Input Format
The first line contains a single integer N.
The next line contains N integers: a0, a1,…aN-1 separated by space, where ai represents the length of ith stick.

Output Format
For each operation, print the number of sticks that are cut, on separate lines.

Constraints
1 ≤ N ≤ 1000
1 ≤ ai ≤ 1000

Sample Input #00

6
5 4 4 2 2 8

Sample Output #00

6
4
2
1

Sample Input #01

8
1 2 3 4 3 3 2 1

Sample Output #01

8
6
4
1

Explanation

Sample Case #00 :

sticks-length        length-of-cut   sticks-cut
5 4 4 2 2 8             2               6
3 2 2 _ _ 6             2               4
1 _ _ _ _ 4             1               2
_ _ _ _ _ 3             3               1
_ _ _ _ _ _           DONE            DONE

Sample Case #01

sticks-length         length-of-cut   sticks-cut
1 2 3 4 3 3 2 1         1               8
_ 1 2 3 2 2 1 _         1               6
_ _ 1 2 1 1 _ _         1               4
_ _ _ 1 _ _ _ _         1               1
_ _ _ _ _ _ _ _       DONE            DONE

Solution

  
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#define max 1000
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    int l,i,j,num[max],min=1000,count=0;
    cin>>l;
    for(i=0;i<l;i++){
        cin>>num[i];
        if(num[i]<min)
            min=num[i];
    }
    sort(num,num+l);
    for(j=0;j<l;j++)
    {
        count=0;
        if(num[l-1]==0)
            break;
        for(i=0;i<l;i++){
            if(num[i]>=min){
                num[i]-=min;
                count++;
            }
        }
        cout<<count<<endl;
        min=1000;
        for(i=0;i<l;i++){
            if(num[i]!=0 & num[i]<min)
                min=num[i];
        }    
    }
    return 0;
}

Service Lane

Problem Statement

Calvin is driving his favorite vehicle on the 101 freeway. He notices that the check engine light of his vehicle is on, and he wants to service it immediately to avoid any risks. Luckily, a service lane runs parallel to the highway. The length of the service lane is N units. The service lane consists of N segments of equal length and different width.

Calvin can enter to and exit from any segment. Let’s call the entry segment as index i and the exit segment as index j. Assume that the exit segment lies after the entry segment (i≤j) and 0≤i. Calvin has to pass through all segments from index i to index j (both inclusive).

Paradise Highway

Calvin has three types of vehicles – bike, car, and truck – represented by 1, 2 and 3, respectively. These numbers also denote the width of the vehicle.

You are given an array width of length N, where width[k] represents the width of the kth segment of the service lane. It is guaranteed that while servicing he can pass through at most 1000 segments, including the entry and exit segments.

  • If width[k]=1, only the bike can pass through the kth segment.
  • If width[k]=2, the bike and the car can pass through the kth segment.
  • If width[k]=3, all three vehicles can pass through the kth segment.

Given the entry and exit point of Calvin’s vehicle in the service lane, output the type of the largest vehicle which can pass through the service lane (including the entry and exit segments).

Input Format

The first line of input contains two integers, N and T, where N denotes the length of the freeway and T the number of test cases. The next line has N space-separated integers which represent the width array.

T test cases follow. Each test case contains two integers, i and j, where i is the index of the segment through which Calvin enters the service lane and j is the index of the lane segment through which he exits.

Constraints
2≤N≤100000
1≤T≤1000
0≤i<j<N
2≤j−i+1≤min(N,1000)
1≤width[k]≤3,where 0≤k<N

Output Format

For each test case, print the number that represents the largest vehicle type that can pass through the service lane.

Note: Calvin has to pass through all segments from index i to index j (both inclusive).

Sample Input

8 5
2 3 1 2 3 2 3 3
0 3
4 6
6 7
3 5
0 7

Sample Output

1
2
3
2
1

Explanation

Below is the representation of the lane:

   |HIGHWAY|Lane|    ->    Width

0: |       |--|            2
1: |       |---|           3
2: |       |-|             1
3: |       |--|            2
4: |       |---|           3
5: |       |--|            2
6: |       |---|           3
7: |       |---|           3
  1. (0, 3): Because width[2] = 1, only the bike can pass through it.
  2. (4, 6): Here the largest allowed vehicle which can pass through the 5th segment is the car and for the 4th and 6th segment it’s the truck. Hence the largest vehicle allowed in these segments is a car.
  3. (6, 7): In this example, the vehicle enters at the 6th segment and exits at the 7th segment. Both segments allow even trucks to pass through them. Hence the answer is 3.
  4. (3, 5): width[3] = width[5] = 2. While the 4th segment allows the truck, the 3rd and 5th allow up to a car. So 2 will be the answer here.
  5. (0, 7): The bike is the only vehicle which can pass through the 2nd segment, which limits the strength of the whole lane to 1.

Solution

  
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#define max 100000
using namespace std;

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    int l,t,i,a,b,min,width[max];
    cin>>l>>t;
    for(i=0;i>width[i];
    while(t--){
        cin>>a>>b;
        min = 3;
        for(i=a;i<=b;i++)
        {
           if(width[i]<min)
               min=width[i];
        }
        cout<<min<<endl;
    }
    return 0;
}