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Expected number of cluster of cars formed on infinite Road

  • Difficulty Level : Easy
  • Last Updated : 19 Jan, 2022

Given an array speed[] of size N denoting the speed of N cars moving on an infinitely long single lane road (i.e. no overtaking is allowed) from left to right. Initially, the rightmost car is at the first position and whenever a car with higher speed reaches a car with lower speed they start moving with the speed to the slower car and form a cluster. The task is to find the total number of clusters that will form.

Note: A single car is also considered a cluster.

Examples:

Input: speed[] = {1, 4, 5, 2, 17, 4 }
Output: 3
Explanation: After a certain time, car with speed 17 will reach car with speed 4(car5) and will form a cluster. 
And similarly, cars with speed 4 and 5 will start moving with speed 2 on reaching car3.
The clusters will be (car0), (car1, car2, car3), (car4, car5).

Input: speed[] = {2, 3, 4, 7, 7}
Output: 5
Explanation: Each car will form a cluster.

 

Approach: The solution is based on greedy approach. Here, a car with less speed forms a cluster with all the cars behind it and having speeds greater than itself. Follow the steps below to solve the problem:

  1. Start iterating from the last index of array speed[].
  2. Store the speed of the car in a variable, say currentCar.
  3. A cluster is formed till the speed of cars behind the currentCar is greater than itself. Therefore increase the value of clusters as soon as a car with less speed is found behind it or when the array is out of bounds.

Below is the implementation of the above approach.

C++




// C++ code to implement above approach
#include <bits/stdc++.h>
#include <vector>
using namespace std;
 
// Function to count total number of clusters
int countClusters(vector<int>& speed)
{
    int N = speed.size();
     
    // For number of clusters
    int cluster = 0;
 
    for (int i = N - 1; i >= 0; i--) {
        int currentCar = speed[i];
         
        // comparing with previous car
        while (i != 0 and speed[i - 1]
               > currentCar) {
            i--;
        }
        cluster++;
    }
    return cluster;
}
 
// Driver code
int main()
{
    vector<int> speed = { 1, 4, 5, 2, 17, 4 };
    int clusters = countClusters(speed);
    cout << clusters;
    return 0;
}

Java




// Java program to implement
// the above approach
class GFG
{
 
  // Function to count total number of clusters
  static int countClusters(int []speed)
  {
    int N = speed.length;
 
    // For number of clusters
    int cluster = 0;
 
    for (int i = N - 1; i >= 0; i--) {
      int currentCar = speed[i];
 
      // comparing with previous car
      while (i != 0 && speed[i - 1]
             > currentCar) {
        i--;
      }
      cluster++;
    }
    return cluster;
  }
 
  // Driver Code
  public static void main(String args[])
  {
    int []speed = { 1, 4, 5, 2, 17, 4 };
    int clusters = countClusters(speed);
    System.out.println(clusters);
 
  }
}
 
// This code is contributed by Saurabh Jaiswal

Python3




# Python code to implement above approach
 
# Function to count total number of clusters
def countClusters(speed):
    N = len(speed)
    clusters = 0
    i = N-1
    while(i >= 0):
        currentCar = speed[i]
        while(i != 0 and speed[i-1] > currentCar):
            i = i-1
        clusters = clusters+1
        i = i-1
    return clusters
 
# Driver code
if __name__ == '__main__':
    speed = [1, 4, 5, 2, 17, 4]
    clusters = countClusters(speed)
    print(clusters)

C#




// C# program to implement
// the above approach
using System;
class GFG
{
 
  // Function to count total number of clusters
  static int countClusters(int []speed)
  {
    int N = speed.Length;
 
    // For number of clusters
    int cluster = 0;
 
    for (int i = N - 1; i >= 0; i--) {
      int currentCar = speed[i];
 
      // comparing with previous car
      while (i != 0 && speed[i - 1]
             > currentCar) {
        i--;
      }
      cluster++;
    }
    return cluster;
  }
 
  // Driver Code
  public static void Main()
  {
    int []speed = { 1, 4, 5, 2, 17, 4 };
    int clusters = countClusters(speed);
    Console.Write(clusters);
 
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
       // JavaScript code for the above approach
 
       // Function to count total number of clusters
       function countClusters(speed) {
           let N = speed.length;
 
           // For number of clusters
           let cluster = 0;
 
           for (let i = N - 1; i >= 0; i--) {
               let currentCar = speed[i];
 
               // comparing with previous car
               while (i != 0 && speed[i - 1]
                   > currentCar) {
                   i--;
               }
               cluster++;
           }
           return cluster;
       }
 
       // Driver code
 
       let speed = [1, 4, 5, 2, 17, 4];
       let clusters = countClusters(speed);
       document.write(clusters);
 
 // This code is contributed by Potta Lokesh
   </script>

 
 

Output
3

 

Time Complexity: O(N)
Auxiliary Space: O(1)

 


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