Find the final radiations of each Radiated Stations in C++

C++Server Side ProgrammingProgramming

Suppose there are N stations in the straight line. Each of them has same non-negative power of radiation power. Every station can increase the radiation power of its neighboring stations in the following way.

Suppose the station i with radiation power R, will increase (i – 1)th station’s radiation power, by R-1, (i - 2)th station’s radiation power by R-2, and will increase (i + 1)th station’s radiation power, by R-1, (i + 2)th station’s radiation power by R-2. So on. So for example, if the array is like Arr = [1, 2, 3], then the output will be 3, 4, 4. The new radiation will be [1 + (2 – 1) + (3 - 2), 2 + (1 – 1) + (3 - 1), 3 + (2 – 1)] = [3, 4, 4]

The idea is simple. For each station i increases the radiation of neighboring stations as mentioned above, up to when the effective radiation becomes negative.

Example

 Live Demo

#include <iostream>
using namespace std;
class pump {
   public:
      int petrol;
      int distance;
};
int findStartIndex(pump pumpQueue[], int n) {
   int start_point = 0;
   int end_point = 1;
   int curr_petrol = pumpQueue[start_point].petrol - pumpQueue[start_point].distance;
   while (end_point != start_point || curr_petrol < 0) {
      while (curr_petrol < 0 && start_point != end_point) {
         curr_petrol -= pumpQueue[start_point].petrol - pumpQueue[start_point].distance;
         start_point = (start_point + 1) % n;
         if (start_point == 0)
            return -1;
      }
      curr_petrol += pumpQueue[end_point].petrol - pumpQueue[end_point].distance;
      end_point = (end_point + 1) % n;
   }
   return start_point;
}
int main() {
   pump PumpArray[] = {{4, 6}, {6, 5}, {7, 3}, {4, 5}};
   int n = sizeof(PumpArray)/sizeof(PumpArray[0]);
   int start = findStartIndex(PumpArray, n);
   if(start == -1)
      cout<<"No solution";
   else
      cout<<"Index of first petrol pump : "<<start;
}

Output

Index of first petrol pump : 1
raja
Published on 17-Dec-2019 16:24:55
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