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Edisi # 35: Pelatihan TI - masalah saat ini dan tantangan dari perusahaan terkemuka

Hai Hai! Kami sekali lagi mempersiapkan untuk Anda sejumlah pertanyaan dan tugas menarik dari wawancara di perusahaan IT terkemuka!


Masalah akan muncul setiap minggu - tetap disini! Kolom ini didukung oleh agen perekrutan Spice IT .

Minggu ini kami mengumpulkan tugas dari wawancara di perusahaan Amerika PayPal. Omong-omong, jawaban untuk masalah dari masalah sebelumnya telah diterbitkan .

Pertanyaan


1. Tahanan Tahanan dan Polisi
Polisi memutuskan untuk menghukum Tahanan dan memintanya untuk membuat pernyataan. Tahanan harus membuat pernyataan sedemikian rupa sehingga ia akan hidup.
Jika pernyataan itu dianggap benar oleh Polisi, Tahanan akan digantung sampai mati dan jika pernyataan itu dinyatakan salah, Tahanan akan ditembak mati.

Transfer
. , .
, , , . ?

2. Teka-teki Batang Korek Api
Bagaimana cara membuat 4 segitiga sama sisi dengan 6 batang korek api yang identik?

Transfer
4 6 ?

Tugas


1. Persimpangan dua array
Given two arrays A and B respectively of size N and M. The task is to print the count of elements in the intersection (or common elements) of the two arrays.
For this question, intersection of two arrays can be defined as the set containing distinct common elements between the two arrays.

Input:
The first line of input contains an integer T denoting the number of test cases. The first line of each test case is N and M, N is the size of array A and M is size of array B. The second line of each test case contains N input A[i].
The third line of each test case contains M input B[i].

Output:
Print the count of intersecting elements.

Constraints:
1 ≤ T ≤ 100
1 ≤ N, M ≤ 105
1 ≤ A[i], B[i] ≤ 105

Contoh:
Input: Output: Penjelasan: Testcase 1: 89 adalah satu-satunya elemen di persimpangan dua array. Testcase 2: 3 4 5 dan 6 adalah elemen-elemen di persimpangan dua array. Testcase 3: Bukan elemen yang umum sehingga hasilnya akan -1. Testcase 4: 10 adalah satu-satunya elemen yang ada di persimpangan dua array.
4
5 3
89 24 75 11 23
89 2 4
6 5
1 2 3 4 5 6
3 4 5 6 7
4 4
10 10 10 10
20 20 20 20
3 3
10 10 10
10 10 10

1
4
0
1






Transfer
A B N M. , ( ) .
, .

:
T, . — N M, N- A, M- B. N A[i].
M B[i].

:
.

:
1 ≤ T ≤ 100
1 ≤ N, M ≤ 105
1 ≤ A[i], B[i] ≤ 105

:
:
4
5 3
89 24 75 11 23
89 2 4
6 5
1 2 3 4 5 6
3 4 5 6 7
4 4
10 10 10 10
20 20 20 20
3 3
10 10 10
10 10 10
:
1
4
0
1

:
1: 89 — .
2: 3 4 5 6 — .
3: , 0.
4: 10 — , .

2. Gabungan String Zig-Zag di Baris 'n'
Diberikan string dan jumlah baris 'n'. Cetak string yang dibentuk dengan menyatukan n baris ketika string input ditulis dengan cara Zig-Zag.

Contoh: Input: Baris input pertama terdiri dari jumlah kasus uji. Deskripsi kasus uji T adalah sebagai berikut: Baris pertama dari setiap kasus uji berisi string, dan baris kedua memiliki 'n' jumlah baris. Output: Di setiap baris terpisah cetak string setelah menyatukan n baris dalam bentuk zig zag. Kendala: Contoh: Input: Output:
Input:
str = "ABCDEFGH"
n = 2
Output: "ACEGBDFH"
Explanation: Let us write input string in Zig-Zag fashion in 2 rows.
A___C___E___G
__B___D___F___H
Now concatenate the two rows and ignore spaces
in every row. We get "ACEGBDFH"

Input:
str = "SPICEITRECRUITMENT"
n = 3
Output: SEEINPCIRCUTETITRM
Explanation: Let us write input string in Zig-Zag fashion in 3 rows.
S_______E_______E_______I_______N
__P___C___I___R___C___U___T___E___T
____I_______T_______R_______M
Now concatenate the two rows and ignore spaces
in every row. We get "SEEINPCIRCUTETITRM"









1 ≤ T ≤ 70
1 ≤ N ≤ size of string



2
qrrc
3
rfkqyuqfjkxy
2

qrcr
rkyqjxfqufky

Transfer
‘n'. , n , .

:
:
str = "ABCDEFGH"
n = 2
: "ACEGBDFH"
: 2 .
A___C___E___G
__B___D___F___H
. "ACEGBDFH"

: SEEINPCIRCUTETITRM
str = "SPICEITRECRUITMENT"
n = 3
:
: 3 .
S_______E_______E_______I_______N
__P___C___I___R___C___U___T___E___T
____I_______T_______R_______M
. "SEEINPCIRCUTETITRM"

:
. t :
, 'n' — .

:
n .

:
1 ≤ T ≤ 70
1 ≤ N ≤

:
:
2
qrrc
3
rfkqyuqfjkxy
2
:
qrcr
rkyqjxfqufky

3. Selesaikan Sudoku
Given an incomplete Sudoku configuration in terms of a 9 x 9 2-D square matrix (mat[][]). The task to print a solved Sudoku. For simplicity you may assume that there will be only one unique solution.

Sample Sudoku for you to get the logic for its solution:


Input:
The first line of input contains an integer T denoting the no of test cases. Then T test cases follow. Each test case contains 9*9 space separated values of the matrix mat[][] representing an incomplete Sudoku state where a 0 represents empty block.

Output:
For each test case, in a new line, print the space separated values of the solution of the the sudoku.

Constraints:
1 <= T <= 10
0 <= mat[] <= 9

Example:
Input:
1
3 0 6 5 0 8 4 0 0
5 2 0 0 0 0 0 0 0
0 8 7 0 0 0 0 3 1
0 0 3 0 1 0 0 8 0
9 0 0 8 6 3 0 0 5
0 5 0 0 9 0 6 0 0
1 3 0 0 0 0 2 5 0
0 0 0 0 0 0 0 7 4
0 0 5 2 0 6 3 0 0

Output:
3 1 6 5 7 8 4 9 2 5 2 9 1 3 4 7 6 8 4 8 7 6 2 9 5 3 1 2 6 3 4 1 5 9 8 7 9 7 4 8 6 3 1 2 5 8 5 1 7 9 2 6 4 3 1 3 8 9 4 7 2 5 6 6 9 2 3 5 1 8 7 4 7 4 5 2 8 6 3 1 9

Explanation:
Testcase 1: The solved sudoku is:
3 1 6 5 7 8 4 9 2
5 2 9 1 3 4 7 6 8
4 8 7 6 2 9 5 3 1
2 6 3 4 1 5 9 8 7
9 7 4 8 6 3 1 2 5
8 5 1 7 9 2 6 4 3
1 3 8 9 4 7 2 5 6
6 9 2 3 5 1 8 7 4
7 4 5 2 8 6 3 1 9

9 x 9 2D (mat [] []). . , .

, :


:
T, . T . 9 * 9 mat [] [], , 0 .

:
.

:
1 <= T <= 10
0 <= mat[] <= 9

:
:
1
3 0 6 5 0 8 4 0 0
5 2 0 0 0 0 0 0 0
0 8 7 0 0 0 0 3 1
0 0 3 0 1 0 0 8 0
9 0 0 8 6 3 0 0 5
0 5 0 0 9 0 6 0 0
1 3 0 0 0 0 2 5 0
0 0 0 0 0 0 0 7 4
0 0 5 2 0 6 3 0 0

:
3 1 6 5 7 8 4 9 2 5 2 9 1 3 4 7 6 8 4 8 7 6 2 9 5 3 1 2 6 3 4 1 5 9 8 7 9 7 4 8 6 3 1 2 5 8 5 1 7 9 2 6 4 3 1 3 8 9 4 7 2 5 6 6 9 2 3 5 1 8 7 4 7 4 5 2 8 6 3 1 9

:
1: :
3 1 6 5 7 8 4 9 2
5 2 9 1 3 4 7 6 8
4 8 7 6 2 9 5 3 1
2 6 3 4 1 5 9 8 7
9 7 4 8 6 3 1 2 5
8 5 1 7 9 2 6 4 3
1 3 8 9 4 7 2 5 6
6 9 2 3 5 1 8 7 4
7 4 5 2 8 6 3 1 9



1
— , — ’
, , , .
, , , .

2
.


1
t=0
try:
    t=int(input().strip())
except:
    pass
if __name__ == "__main__":
    while t:
        t-=1
        n, m = map(int, input().strip().split())
        a = list(map(int, input().strip().split()))
        b = list(map(int, input().strip().split()))
        a.sort()
        count=0;x=0
        for i in a:
            if i==x:
                pass
            else:
                if i in b:
                    count+=1
                    x=i
        print(count)

2
#include<bits/stdc++.h>
using namespace std;
int main()
 {
	//code
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int t;
	cin>>t;
	while(t--)
	{
	    string str;
	    cin>>str;
	    int row;
	    cin>>row;
	    vector<char> v[row];
	    int flag=1,k=0;
	    while(1)
	    {
	        if(flag==1 && k<str.length())
	        {
	            for(int i=0;i<row && k<str.length();i++)
	             { 
	                 v[i].push_back(str[k]);
	                 k++;
	             }
	           flag=0;
	        }
	        if(flag==0 && k<str.length())
	        {
	            for(int i=row-2;i>0 && k<str.length();i--)
	            {
	                v[i].push_back(str[k]);
	                k++;
	            }
	            flag=1;
	        }
	        if(k==str.length())
	         break;
	    }
	    string res="";
	    for(int i=0;i<row;i++)
	    {
	         for(char ch:v[i])
	         {
	             res+=ch;
	         }
	         //cout<<endl;
	    }
	    cout<<res<<endl;
	}
	return 0;
}

3
#include<bits/stdc++.h>
using namespace std;
#define ll long long
int grid[9][9];
int row[9][10];
int col[9][10];
int box[3][3][10];
bool flag;
vector<pair<int,int> >V;
void call(int I)
{
    if(V.size()==I)
    {
        flag=false;
        for(int i=0;i<9;i++)
        {
            for(int j=0;j<9;j++)
            cout<<grid[i][j]<<" ";
        }
        cout<<endl;
        return;
    }
    int x=V[I].first,y=V[I].second;
    int boxx,boxy;
    if(x<3)
    boxx=0;
    else if(x<6)
    boxx=1;
    else
    boxx=2;
    
    if(y<3)
    boxy=0;
    else if(y<6)
    boxy=1;
    else
    boxy=2;
    
    
    for(int i=1;i<=9 && flag;i++)
    {
        if(row[x][i]==0 && col[y][i]==0 && box[boxx][boxy][i]==0)
        {
            row[x][i]=1; col[y][i]=1; box[boxx][boxy][i]=1;
            grid[x][y]=i;
            call(I+1);
            row[x][i]=0; col[y][i]=0; box[boxx][boxy][i]=0;
            grid[x][y]=0;
        }
    }
    return;
}
int main()
 {
	int t;
	cin>>t;
	while(t--)
	{
	    for(int i=0;i<9;i++)
	    {
	        for(int j=0;j<9;j++)
	        grid[i][j]=0;
	    }
	    for(int i=0;i<9;i++)
	    {
	        for(int j=1;j<=9;j++)
	        {
	            row[i][j]=0;
	            col[i][j]=0;
	        }
	    }
	    for(int i=0;i<3;i++)
	    {
	        for(int j=0;j<3;j++)
	            for(int k=1;k<=9;k++)
	            box[i][j][k]=0;
	    }
	    flag=true;
	    V.clear();
	    for(int i=0;i<9;i++)
	    {
	        for(int j=0;j<9;j++)
	        {
	            cin>>grid[i][j];
	            if(grid[i][j]==0)
	            {
	                V.push_back(make_pair(i,j));
	                continue;
	            }
	            row[i][grid[i][j]]=1;
	            col[j][grid[i][j]]=1;
	            if(i<3)
	            {
	               if(j<3)
	               {
	                    box[0][0][grid[i][j]]=1;   
	               }
	               else if(j<6)
	               {
	                   box[0][1][grid[i][j]]=1;
	               }
	               else
	               {
	                   box[0][2][grid[i][j]]=1;
	               }
	               
	            }
	            else if(i<6)
	            {
	                if(j<3)
	               {
	                    box[1][0][grid[i][j]]=1;   
	               }
	               else if(j<6)
	               {
	                   box[1][1][grid[i][j]]=1;
	               }
	               else
	               {
	                   box[1][2][grid[i][j]]=1;
	               }
	            }
	            else
	            {
	                if(j<3)
	               {
	                    box[2][0][grid[i][j]]=1;   
	               }
	               else if(j<6)
	               {
	                   box[2][1][grid[i][j]]=1;
	               }
	               else
	               {
	                   box[2][2][grid[i][j]]=1;
	               }
	            }
	        }
	    }
	    call(0);
	}
	return 0;
}

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