Yes i have solved aprox 12-13 different problems.
The links of the separate rank lists of zone1 & zone2 are
Zone 1 : Zone1
Zone 2 : Zone 2
Source : https://campuscommune.tcs.com/channels/codevita-season-8
Anyone who got the mail regarding invitation for a Direct Interview for a
career with TCS.
I have got an email for Call Basis Interview
anyone have taken screenshot of problem statement workbalance of tcs codevita if so please do share it or anyone remember itā¦pls share
Could you pls provide me the ālife guardā solution. I was getting runtime error ,but public test cases were passed!
//solved using binary search to prevent time exceed error you can use brute force also
#include <iostream>
#include <cmath>
using namespace std;
long double xl, yl, xw, yw, f;
long double calculate_it(long double i)
{
return sqrt(((xl - i) * (xl - i)) + (yl * yl)) / f + sqrt((xw - i) * (xw - i) + yw * yw);
}
long double search_it(long double start, long double end)
{
long double mid, current, prev, next;
long double ans = -1;
while (true)
{
mid = (start + end) / 2;
prev = calculate_it(mid - 0.000001);
current = calculate_it(mid);
next = calculate_it(mid + 0.000001);
if (prev > current && next > current)
{
ans = mid;
break;
}
else if (prev > current)
{
start = mid + 0.000001;
}
else
{
end = mid - 0.000001;
}
}
return ans;
}
int main()
{
cin >> xl >> yl >> xw >> yw >> f;
long double count;
if (xl > xw)
{
count = search_it(xw, xl);
}
else
{
count = search_it(xl, xw);
}
printf("%.6lf", (double)count);
return 0;
}
/*
Explanation
(time)
^
| * *
| * *
| * *
| * *
| * <--------(x_min )
|
(0,0)Ā±-------------------------------------->(position)
| |
(x_l) (x_w)
|___________________________|
(x_min is between this point)
we can use binary search by using the following condition :
-
start = min(x_l,x_w) , end = min(x_l,x_w)
-
calculate middle = (start + end )/2
-
find the time taken for position (middle-0.000001) , (middle) ,(middle + 0.000001) by using the formula ;
-
let the calucated time be prev, current and next respectively.
we can find direct our binary search using following condition ;
if( time taken in middle position is smaller than both prev and next )
then this is the required position
else if( time taken in middle position is >time taken n prev postion )
then move right
else
move left
*/
(time_min)