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Maximize count of planes that can be stopped per second with help of given initial position and speed

  • Difficulty Level : Easy
  • Last Updated : 27 Sep, 2021

Given two arrays A[] and B[] consisting of N integers where A[i] represents the initial position of the ith plane and B[i] is the speed at which the plane is landing, the task is to print the number of the plane that can be stopped from landing by shooting an aircraft at every second.

Examples:

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Input: A[] = {1, 3, 5, 4, 8}, B[] = {1, 2, 2, 1, 2}
Output: 4
Explanation: 

  1. At second 1, shoot the plane at index 0, the positions of the planes are now A[] = {_, 1, 3, 3, 6}.
  2. At second 2, shoot the plane at index 1, the positions of the planes are now A[] = {_, _, 1, 2, 4}.
  3. At second 3, shoot the plane at index 2, the positions of the planes are now A[] = {_, _, _, 1, 2}.
  4. At second 4, shoot the plane at index 4 or 5, the positions of the planes are now A[] = {_, _, _, _, _}.

Therefore, a total of 4 planes can be stopped from landing.



Input: A[] = {2, 8, 2, 3, 1}, B[] = {1, 4, 2, 2, 2}
Output: 2

Approach: The given problem can be solved based on the following observations:

  • It can be observed that the count of planes that can be stopped is the set of distinct times needed to land for each plane, as at the same time, only one plane can be destroyed.
  • The time needed for a plane to land is A[i] / B[i] can be calculated using the formula: Time = Distance / Velocity

Follow the steps below to solve the problem:

  • Initialize a Hash-Set, say S that stores the times needed for plane landing.
  • Iterate over a range [0, N] using the variable i and perform the following tasks:
    • Initialize a variable, say t as 1 if the value of A[i]%B[i] is greater than 0. Otherwise, update the variable t as 0.
    • Add the value of A[i]/B[i] into the variable, say t.
    • Add the value of t into the HashSet S.
  • After performing the above steps, return the size of the HashSet as the resultant answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find maximum number
// of planes that can be stopped
// from landing
int maxPlanes(int A[], int B[], int N)
{
   
    // Stores the times needed for
    // landing for each plane
    set<int> St;
 
    // Iterate over the arrays
    for (int i = 0; i < N; i++) {
 
        // Stores the time needed for
        // landing of current plane
        int t = (A[i] % B[i] > 0) ? 1 : 0;
 
        // Update the value of t
        t += (A[i] / B[i]) + t;
 
        // Append the t in set St
        St.insert(t);
    }
 
    // Return the answer
    return St.size();
}
 
 // Driver Code
int main() {
 
    int A[] = { 1, 3, 5, 4, 8 };
    int B[] = { 1, 2, 2, 1, 2 };
      int N = sizeof(A)/sizeof(A[0]);
   
    cout << maxPlanes(A, B, N);
             
    return 0;
}
 
// This code is contributed by Dharanendra L V.

Java




// Java program for the above approach
 
import java.util.*;
 
class GFG {
 
    // Function to find maximum number
    // of planes that can be stopped
    // from landing
    static int maxPlanes(int[] A, int[] B)
    {
        // Stores the times needed for
        // landing for each plane
        Set<Integer> St = new HashSet<>();
 
        // Iterate over the arrays
        for (int i = 0; i < A.length; i++) {
 
            // Stores the time needed for
            // landing of current plane
            int t = (A[i] % B[i] > 0) ? 1 : 0;
 
            // Update the value of t
            t += (A[i] / B[i]) + t;
 
            // Append the t in set St
            St.add(t);
        }
 
        // Return the answer
        return St.size();
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] A = { 1, 3, 5, 4, 8 };
        int[] B = { 1, 2, 2, 1, 2 };
        System.out.println(
            maxPlanes(A, B));
    }
}

Python3




# Python program for the above approach
 
# Function to find maximum number
# of planes that can be stopped
# from landing
def maxPlanes(A, B, N):
     
    # Stores the times needed for
    # landing for each plane
    St = set()
     
    # Iterate over the arrays
    for i in range(N):
         
        # Stores the time needed for
        # landing of current plane
        t = 1 if (A[i] % B[i] > 0) else 0
         
        # Update the value of t
        t += (A[i] // B[i]) + t
         
        # Append the t in set St
        St.add(t)
         
    # Return the answer
    return len(St)
 
# Driver Code
A = [ 1, 3, 5, 4, 8 ]
B = [ 1, 2, 2, 1, 2 ]
N = len(A)
print(maxPlanes(A, B, N))
 
# This code is contributed by shivanisinghss2110

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
public class GFG
{
   
    // Function to find maximum number
    // of planes that can be stopped
    // from landing
    static int maxPlanes(int[] A, int[] B)
    {
       
        // Stores the times needed for
        // landing for each plane
         HashSet<int> St = new HashSet<int>();
 
        // Iterate over the arrays
        for (int i = 0; i < A.Length; i++) {
 
            // Stores the time needed for
            // landing of current plane
            int t = (A[i] % B[i] > 0) ? 1 : 0;
 
            // Update the value of t
            t += (A[i] / B[i]) + t;
 
            // Append the t in set St
            St.Add(t);
        }
 
        // Return the answer
        return St.Count;
    }
   
  // Driver code
    static public void Main (){
 
       int[] A = { 1, 3, 5, 4, 8 };
        int[] B = { 1, 2, 2, 1, 2 };
        Console.WriteLine(
            maxPlanes(A, B));
    }
}
 
// This code is contributed by Potta Lokesh

Javascript




<script>
 
    // Function to find maximum number
    // of planes that can be stopped
    // from landing
    function maxPlanes(A, B)
    {
     
        // Stores the times needed for
        // landing for each plane
        let St = new Set();
 
        // Iterate over the arrays
        for (let i = 0; i < A.length; i++) {
 
            // Stores the time needed for
            // landing of current plane
            let t = (A[i] % B[i] > 0) ? 1 : 0;
 
            // Update the value of t
            t += Math.floor(A[i] / B[i]) + t;
 
            // Append the t in set St
            St.add(t);
        }
 
        // Return the answer
        return St.size;
    }
 
// Driver Code
        let A = [ 1, 3, 5, 4, 8 ];
        let B = [ 1, 2, 2, 1, 2 ];
        document.write(
            maxPlanes(A, B));
     
    // This code is contributed by sanjoy_62.
</script>
Output: 
3

 

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




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