Reverse Gear | Code Vita 2015 | round 1

Reverse Gear

Problem Description:

        A futuristic company is building an autonomous car. The scientists at the company are training the car to perform Reverse parking. To park, the car needs to be able to move in backward as well as forward direction. The car is programmed to move backwards B meters and forwards again, say F meters, in a straight line. The car does this repeatedly until it is able to park or collides with other objects. The car covers 1 meter in T units of time. There is a wall after distance D from car's initial position in the backward direction.

The car is currently not without defects and hence often hits the wall. The scientists are devising a strategy to prevent this from happening. Your task is to help the scientists by providing them with exact information on amount of time available before the car hits the wall.

Input Format:

First line contains total number of test cases, denoted by N
Next N lines, contain a tuple containing 4 values delimited by space
F B T D, where

1. F denotes forward displacement in meters
2. B denotes backward displacement in meters
3. T denotes time taken to cover 1 meter
4. D denotes distance from Car's starting position and the wall in backward direction

Output Format:

For each test case print time taken by the Car to hit the wall

Constraints:

First move will always be in backward direction

1 <= N <= 100

backward displacement > forward displacement i.e. (B > F)

forward displacement (F) > 0

backward displacement (B) > 0

time (T) > 0 

distance (D) > 0

All input values must be positive integers only

Sample Input and Output

SNo.	Input	Output
1	2 6 9 3 18 3 7 5 20	162 220

Program:

#include <stdio.h>
int main() {
int f,b,d,t,sdist,sdisp,step,rd,td,tot,tt,test;
scanf("%d",&test);
while(test)
{
scanf("%d %d %d %d",&f,&b,&t,&d);
sdist=b+f;
sdisp=b-f;
step=(d-b)/sdisp;
if((d-b)%sdisp!=0)
step=step+1;
rd=d-(step*sdisp);
tot=(step*sdist)+rd;
tt=tot*t;
printf("%d\n",tt);
test--;
}
return 0;
}

Output:

2
6 9 3 18
3 7 5 20

162
220

You can also run it on an online IDE: https://ide.geeksforgeeks.org/S994F0QfMs
Your feedback and if you have any doubts it is welcomed! Do contact me or comment it below! Cheers!

Related Links: consecutive prime sum

Bride Hunting -Code Vita 2018 | round 1 | TCS Code Vita Solution 2018 | Bride Hunting TCS Code Vita solution

Bride Hunting

Problem Description

Sam is an eligible bachelor. He decides to settle down in life and start a family. He goes bride hunting.

He wants to marry a girl who has at least one of the 8 qualities mentioned below:-

1) The girl should be rich.

2) The girl should be an Engineer/Doctor.

3) The girl should be beautiful.

4) The girl should be of height 5.3".

5) The girl should be working in an MNC.

6) The girl should be an extrovert.

7) The girl should not have spectacles.

8) The girl should be kind and honest.

He is in search of a bride who has some or all of the 8 qualities mentioned above. On bride hunting, he may find more than one contenders to be his wife.

In that case, he wants to choose a girl whose house is closest to his house. Find a bride for Sam who has maximum qualities. If in case, there are more than one contenders who are at equal distance from Sam’'s house; then

print "“Polygamy not allowed”".

In case there is no suitable girl who fits the criteria then print “"No suitable girl found"”

Given a Matrix N*M, Sam's house is at (1, 1). It is denoted by 1. In the same matrix, the location of a marriageable Girl is also denoted by 1. Hence 1 at location (1, 1) should not be considered as the location of a marriageable Girl’s location.

The qualities of that girl, as per Sam’'s criteria, have to be decoded from the number of non-zero neighbors (max 8-way) she has. Similar to the condition above, 1 at location (1, 1) should not be considered as the quality of a Girl. See Example section to get a better understanding.

Find Sam, a suitable Bride and print the row and column of the bride, and find out the number of qualities that the Bride possesses.

NOTE: - Distance is calculated in number of hops in any direction i.e. (Left, Right, Up, Down and Diagonal)

Constraints

2 <= N,M <= 10^2

Input Format

First Line contains the row (N) and column (M) of the houses.

Next N lines contain the data about girls and their qualities.

Output

It will contain the row and column of the bride, and the number of qualities that Bride possess separated by a colon (i.e. :).

Explanation

Example 1

Input:

2 9

1 0 1 1 0 1 1 1 1

0 0 0 1 0 1 0 0 1

Output:

1:7:3

Explanation:

The girl and qualities are present at (1,3),(1,4),(1,6),(1,7),(1,8),(1,9),(2,4),(2,6),(2,9).

The girl present at (1,3) has 2 qualities (i.e. (1,4)and (2,4)).

The girl present at (1,4) has 2 qualities.

The Bride present at (1,6) has 2 qualities.

The Bride present at (1,7) has 3 qualities.

The Bride present at (1,8) has 3 qualities.

The Bride present at (1,9) has 2 qualities.

The Bride present at (2,4) has 2 qualities.

The Bride present at (2,6) has 2 qualities.

The Bride present at (2,9) has 2 qualities.

As we see, there are two contenders who have maximum qualities, one is at (1,7) and another at (1,8).

The girl who is closest to Sam's house is at (1,7). Hence, she is the bride.

Hence, the output will be 1:7:3.

Example 2

Input:

6 6

1 0 0 0 0 0

0 0 0 0 0 0

0 0 1 1 1 0

0 0 1 1 1 0

0 0 1 1 1 0

0 0 0 0 0 0

Output:

4:4:8

Explanation:

The bride and qualities are present at (3,3),(3,4),(3,5),(4,3),(4,4),(4,5),(5,3),(5,4),(5,5)

The Bride present at (3,3) has 3 qualities (i.e. (3,4),(4,3) and (4,4)).

The Bride present at (3,4) has 5 qualities.

The Bride present at (3,5) has 3 qualities.

The Bride present at (4,3) has 5 qualities.

The Bride present at (4,4) has 8 qualities.

The Bride present at (4,5) has 5 qualities.

The Bride present at (5,3) has 3 qualities.

The Bride present at (5,4) has 5 qualities.

The Bride present at (5,5) has 3 qualities.

As we see, the girl present in (4,4) has maximum number of Qualities. Hence, she is the bride.

Hence, the output will be 4:4:8.

Solution in C:

 
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
// Function to check if given indices are valid in the matrix
int isValid(int row, int col, int N, int M) {
return (row >= 0 && row < N && col >= 0 && col < M);
}
// Function to count the number of qualities a girl possesses
int countQualities(int matrix[][100], int row, int col, int N, int M) {
int count = 0;
for (int i = row - 1; i <= row + 1; i++) {
for (int j = col - 1; j <= col + 1; j++) {
if (isValid(i, j, N, M) && !(i == row && j == col) && matrix[i][j] == 1) {
count++;
}
}
}
return count;
}
// Function to find the suitable bride for Sam
void findBride(int matrix[][100], int N, int M) {
int minDistance = N + M; // Initialize to maximum possible distance
int maxQualities = 0;
int brideRow = -1, brideCol = -1;
// Loop through the matrix to find the bride with maximum qualities
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
if (matrix[i][j] == 1) {
int qualities = countQualities(matrix, i, j, N, M);
if (qualities > maxQualities) {
maxQualities = qualities;
brideRow = i;
brideCol = j;
} else if (qualities == maxQualities) {
// If multiple brides have the same number of qualities,
// find the one with minimum distance to Sam's house
int distance = abs(0 - i) + abs(0 - j);
if (distance < minDistance) {
minDistance = distance;
brideRow = i;
brideCol = j;
} else if (distance == minDistance) {
printf("Polygamy not allowed\n");
return;
}
}
}
}
}
// If no suitable bride found
if (brideRow == -1 || brideCol == -1) {
printf("No suitable girl found\n");
} else {
printf("%d:%d:%d\n", brideRow + 1, brideCol + 1, maxQualities);
}
}
int main() {
int N, M;
scanf("%d %d", &N, &M);
int matrix[100][100];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
scanf("%d", &matrix[i][j]);
}
}
findBride(matrix, N, M);
return 0;
}
 

Alternate Solution (Brute Force):

#include<stdio.h>
int main()
{
int n,m,i,g[50][50],j,p,q,max=0,cnt=0,k=1,c=0,u=1,x[30],y[30],t1,min=0, sc[50],e,f,ct=0,a[50],count=0,t2=0,t=0;
scanf("%d %d",&n,&m);
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
scanf("%d",&g[i][j]);
} }
g[1][1]=0;
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{ cnt=0;
if(g[i][j]==1)
{
t++;
for(p=i-1;p<=i+1;p++)
{
for(q=j-1;q<=j+1;q++)
{
if(g[p][q]==1)
{ cnt++;
} } }
cnt=cnt-1;
a[k]=cnt;
k++;
} } }
for(k=1;k<=t;k++)
{ if(a[k]>max) max=a[k];
}
if(max==0)
{ printf("No suitable girl found"); goto x; }
for(k=1;k<=t;k++)
{ if(a[k]==max)
c++; }
for(k=1;k<=t;k++)
{ t2=0;
if(a[k]==max)
{ for(i=1;i<=n;i++)
{ for(j=1;j<=m;j++)
{ if(g[i][j]==1) t2++;
if(t2==k)
{ x[u]=i; y[u]=j; u++;
} } } } }
t1=u-1;
if(c==1)
printf("%d:%d:%d",x[1],y[1],max);
else
{ for(u=1;u<=t1;u++)
{ e=x[u]-1;
f=y[u]-1;
if(e>=f)
{ sc[u]=e;
}
else sc[u]=f;
}
min=sc[1];
for(u=1;u<=t1;u++)
{ if(sc[u]<min) min=sc[u];
}
for(u=1;u<=t1;u++)
{ if(sc[u]==min) count++;
}
if(count>1)
printf("Polygamy not allowed");
if(count==1)
{ for(u=1;u<=t1;u++)
{ if(sc[u]==min)
printf("%d:%d:%d",x[u],y[u],max);
} } }
x: return 0;
}

OUTPUT:

6 6
1 0 0 0 0 0
0 0 0 0 0 0
0 0 1 1 1 0
0 0 1 1 1 0
0 0 1 1 1 0
0 0 0 0 0 0

4:4:8

Hey guys you can run this program in the online ide in this link: 

https://ide.geeksforgeeks.org/online-c-compiler/e9bd0a6a-12a8-459f-82dc-ebc592e1d16e

Your feedback are welcomed and can comment or post your doubts in the comment Cheers!

Related Links: 

https://codepiggy.blogspot.com/2024/03/super-market-problem-tcs-code-vita-2023.html