Hello dear readers. We again and again delight you with a new selection of interesting questions and tasks from interviews to leading IT companies!By the way, the answers to the problems from the previous issue have already been published !
Issues will appear every week - stay tuned! The column is supported by the recruitment agency Spice IT .This week we collected tasks from interviews at the Indian company MakeMyTrip.Questions
1. 10 Coins PuzzleYou are blindfolded and 10 coins are place in front of you on table. You are allowed to touch the coins, but can't tell which way up they are by feel. You are told that there are 5 coins head up, and 5 coins tails up but not which ones are which.
Can you make two piles of coins each with the same number of heads up? You can flip the coins any number of times.
Transfer10 . , , . , 5 , («») 5 , («») , — .
, ? .
2. Newspaper PuzzleA newspaper made of 16 large sheets of paper folded in half. The newspaper has 64 pages altogether. The first sheet contains pages 1, 2, 63, 64.
If we pick up a sheet containing page number 45. What are the other pages that this sheet contains?
Transfer16 , . 64 . 1, 2, 63, 64.
*
, 45. ?
Tasks
1. Transpose of MatrixWrite a program to find transpose of a square matrix mat[][] of size N*N. Transpose of a matrix is obtained by changing rows to columns and columns to rows.
Input:
The first line of input contains an integer T, denoting the number of testcases. Then T test cases follow. Each test case contains an integer N, denoting the size of the square matrix. Then in the next line are N*N space separated values of the matrix.
Output:
For each test case output will be the space separated values of the transpose of the matrix
Constraints:
1 <= T <= 15
1 <= N <= 20
-103 <= mat[i][j] <= 103
Example:
Input:
2
4
1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4
2
1 2 -9 -2
Output:
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
1 -9 2 -2
Explanation:
Testcase 1: The matrix after rotation will be: 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4.
Transfer, mat[][] N*N. .
:
T, . T . N, . N*N .
:
:
1 < = T <= 15
1 < = N <= 20
-103 < = mat[i] [j] < = 103
:
:
2
4
1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4
2
1 2 -9 -2
:
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
1 -9 2 -2
:
1: : 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4.
2. Trailing zeroes in factorialFor an integer n find number of trailing zeroes in n! .
Input:
The first line contains an integer ' T ' denoting the total number of test cases. In each test cases, it contains an integer ' N '.
Output:
In each seperate line output the answer to the problem.
Constraints: Example: Input: Output:
1 <= T <= 100
1 <= N <= 1000
1
9
1
Transfern n!.
:
'T', . 'N'.
:
.
:
1 < = T <= 100
1 < = N < = 1000
:
:
1
9
:
1
3. Steps by KnightGiven a square chessboard of N x N size, the position of Knight and position of a target is given. We need to find out minimum steps a Knight will take to reach the target position.
Input:
The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case contains an integer n denoting the size of the square chessboard. The next line contains the X-Y coordinates of the knight. The next line contains the X-Y coordinates of the target.
Output:
Print the minimum steps the Knight will take to reach the target position.
Constraints:
1<=T<=100
1<=N<=20
1<=knight_pos,targer_pos<=N
Example:
Input:
2
6
4 5
1 1
20
5 7
15 20
Output:
3
9
N x N . , , .
:
T, . T . n, . X-Y . X-Y .
:
, , .
:
1<=T<=100
1<=N<=20
1<=knight_pos,targer_pos<=N
:
:
2
6
4 5
1 1
20
5 7
15 20
:
3
9
1: .
:
2 . .
:
— , — .
1:
2:
1.
1:
2:
.
245- 46. , P, 65-p .
,
65-45 = 20
65-46 = 19
, — 19, 20, 45, 46.
1O(1) :
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
public static void main (String[] args)
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
try{
int testCases = Integer.parseInt(br.readLine());
while(testCases-->0){
int size = Integer.parseInt(br.readLine());
int[] arr = Arrays.stream(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
int num = 0;
int i=0;
int total = size*size;
while(num<total && i<size){
if(num >= size*(size-1)) {
System.out.print(arr[num] + " ");
i++;
num = i;
}
else {
System.out.print(arr[num] + " ");
num+=size;
}
}
System.out.println();
}
}
catch (Exception e){
e.printStackTrace();
}
}
}
2import math
for _ in range(int(input())):
n=math.factorial(int(input()))
print(len(str(n))-len(str(n).rstrip("0")))
3#include<bits/stdc++.h>
using namespace std;
void bfs(bool visited[20][20],int n,int x,int y,int X,int Y,int& count){
queue<pair<int,int>>q;
visited[x-1][y-1]=true;
q.push(make_pair(x,y));
q.push(make_pair(0,0));
while(q.size()>1){
x=q.front().first;
y=q.front().second;
if(q.front().first==0 && q.front().second==0){
q.pop();
count++;
q.push(make_pair(0,0));
continue;
}
if(q.front().first==X && q.front().second==Y){
return;
}
if(x>2 && y>1 && visited[x-3][y-2]==false){
visited[x-3][y-2]=true;
q.push(make_pair(x-2,y-1));
}
if(x>2 && y<n && visited[x-3][y]==false){
visited[x-3][y]=true;
q.push(make_pair(x-2,y+1));
}
if(x>1 && y<n-1 && visited[x-2][y+1]==false){
visited[x-2][y+1]=true;
q.push(make_pair(x-1,y+2));
}
if(x<n && y<n-1 && visited[x][y+1]==false){
visited[x][y+1]=true;
q.push(make_pair(x+1,y+2));
}
if(x<n-1 && y<n && visited[x+1][y]==false){
visited[x+1][y]=true;
q.push(make_pair(x+2,y+1));
}
if(x<n-1 && y>1 && visited[x+1][y-2]==false){
visited[x+1][y-2]=true;
q.push(make_pair(x+2,y-1));
}
if(x<n && y>2 && visited[x][y-3]==false){
visited[x][y-3]=true;
q.push(make_pair(x+1,y-2));
}
if(x>1 && y>2 && visited[x-2][y-3]==false){
visited[x-2][y-3]=true;
q.push(make_pair(x-1,y-2));
}
q.pop();
}
}
int main()
{
int t;
cin>>t;
while(t--){
int n;
cin>>n;
int x,y,X,Y;
cin>>x>>y>>X>>Y;
int count=0;
bool visited[20][20]={false};
bfs(visited,n,x,y,X,Y,count);
cout<<count<<'\n';
}
return 0;
}