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Count maximum number of giants that can be destroyed before they reach the city

  • Last Updated : 03 Aug, 2021

Given two arrays dist[] and speed[] consisting of N integers, where dist[i] is the initial distance of the ith giant from the city and speed[i] is the speed of the ith giant. One giant can be eliminated every minute. Therefore, at any time t, at most t giants can be killed. If more giants are able to reach the city in time t, then the game is over. The task is to find the maximum number of giants that can be eliminated before losing, or N if all of the giants can be eliminated before they reach the city.

Examples:

Input: dist[] = [1, 3, 4], speed[] = [1, 1, 1]
Output: 3
Explanation: At the start of minute 0, the distances of the  giants are [1, 3, 4]. The first giant is eliminated.
At the start of 1st minute, the distances of the giants are [X, 2, 3]. No giant is eliminated.
At the start of 2nd minute, the distances of the giants are [X, 1, 2]. The second giant is eliminated.
At the start of 3rd minute, the distances of the giants are [X, X, 1]. The third  giant is eliminated.
All 3  giants can be eliminated.

Input: dist[] = [1, 1, 2, 3], speed[] = [1, 1, 1, 1]
Output: 1

Approach: The idea is to use the greedy approach to solve the problem. Find the time for each giant to come into the city and try to destroy the giant with the least possible time of approaching. Follow the steps below to solve the problem:



  • Initialize a vector timezone[] to store the times.
  • Iterate over the range [0, N] using the variable i and perform the following steps:
    • Push the value of dist[i]/speed[i] into the vector timezone[].
  • Sort the vector timezone[] in ascending order.
  • Initialize the variables curr_time as 0 to store the current time and killcount as 0 to store the number of giants killed.
  • Iterate over the range [0, N] using the variable i and perform the following steps:
    • If timezone[i] is less than curr_time, then break the loop.
    • Else, increase the count of curr_time and killcount by 1.
  • After performing the above steps, return the value of killcount as the 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 count the maximum number
// of giants that can be destroyed
int eliminateMaximum(int dist[], int N,
                     int speed[])
{
    // Make a vector of time and
    // store the time corresponding
    // to its distance and speed
    vector<double> timezone;
 
    for (int i = 0; i < N; i++) {
 
        timezone.push_back((double)dist[i]
                           / (double)speed[i]);
    }
 
    // Sort time in ascending order
    sort(timezone.begin(), timezone.end());
 
    // Stores the time at each instant
    int Curr_time = 0;
 
    // Stores count of giants killed
    int killcount = 0;
 
    for (auto i : timezone) {
        if (i <= Curr_time) {
 
            // Game is lost
            break;
        }
        else {
            Curr_time++;
            killcount++;
        }
    }
    return killcount;
}
 
// Driver Code
int main()
{
    // Given input
    int dist[] = { 1, 3, 4 };
    int N = sizeof(dist) / sizeof(dist[0]);
 
    int speed[] = { 1, 1, 1 };
    int M = sizeof(speed) / sizeof(speed[0]);
 
    // function call
    cout << eliminateMaximum(dist, N, speed);
}


Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to count the maximum number
// of giants that can be destroyed
static int eliminateMaximum(int dist[], int N,
                     int speed[])
{
   
    // Make a vector of time and
    // store the time corresponding
    // to its distance and speed
    Vector<Double> timezone = new Vector<Double>();
 
    for (int i = 0; i < N; i++) {
 
        timezone.add((double)dist[i]
                           / (double)speed[i]);
    }
 
    // Sort time in ascending order
    Collections.sort(timezone);
 
    // Stores the time at each instant
    int Curr_time = 0;
 
    // Stores count of giants killed
    int killcount = 0;
 
    for (Double i : timezone) {
        if (i <= Curr_time) {
 
            // Game is lost
            break;
        }
        else {
            Curr_time++;
            killcount++;
        }
    }
    return killcount;
}
 
// Driver Code
public static void main(String[] args)
{
    // Given input
    int dist[] = { 1, 3, 4 };
    int N = dist.length;
 
    int speed[] = { 1, 1, 1 };
    int M = speed.length;
 
    // function call
    System.out.print(eliminateMaximum(dist, N, speed));
}
}
 
// This code is contributed by shikhasingrajput


Python3




# Python program for the above approach
 
# Function to count the maximum number
# of giants that can be destroyed
def eliminateMaximum(dist, N, speed):
 
    # Make a vector of time and
    # store the time corresponding
    # to its distance and speed
    timezone = []
 
    for i in range(N):
        timezone.append(dist[i] / speed[i])
 
 
    # Sort time in ascending order
    timezone.sort()
 
    # Stores the time at each instant
    Curr_time = 0
 
    # Stores count of giants killed
    killcount = 0
 
    for i in timezone:
        if (i <= Curr_time) :
            # Game is lost
            break
 
        else:
            Curr_time += 1
            killcount += 1
         
    return killcount
 
# Driver Code
 
# Given input
dist = [1, 3, 4]
N = len(dist)
 
speed = [1, 1, 1]
M = len(speed)
 
# function call
print(eliminateMaximum(dist, N, speed))
 
# This code is contributed by gfgking


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
// Function to count the maximum number
// of giants that can be destroyed
static int eliminateMaximum(int []dist, int N,
                     int []speed)
{
    // Make a vector of time and
    // store the time corresponding
    // to its distance and speed
    List<double> timezone = new List<double>();
 
    for (int i = 0; i < N; i++) {
 
        timezone.Add((double)dist[i]/speed[i]);
    }
 
    // Sort time in ascending order
    timezone.Sort();
 
    // Stores the time at each instant
    int Curr_time = 0;
 
    // Stores count of giants killed
    int killcount = 0;
 
    foreach(double i in timezone) {
        if (i <= Curr_time) {
 
            // Game is lost
            break;
        }
        else {
            Curr_time++;
            killcount++;
        }
    }
    return killcount;
}
 
// Driver Code
public static void Main()
{
    // Given input
    int []dist = { 1, 3, 4 };
    int N = dist.Length;
 
    int []speed = { 1, 1, 1 };
    int M = speed.Length;
 
    // function call
    Console.Write(eliminateMaximum(dist, N, speed));
}
}
 
// This code is contributed by ipg2016107.


Javascript




<script>
 
        // JavaScript program for the above approach
 
        // Function to count the maximum number
        // of giants that can be destroyed
        function eliminateMaximum(dist, N, speed)
        {
         
            // Make a vector of time and
            // store the time corresponding
            // to its distance and speed
            let timezone = [];
 
            for (let i = 0; i < N; i++) {
 
                timezone.push(dist[i] / speed[i]);
            }
 
            // Sort time in ascending order
            timezone.sort(function (a, b) { return a - b; });
 
            // Stores the time at each instant
            let Curr_time = 0;
 
            // Stores count of giants killed
            let killcount = 0;
 
            for (let i of timezone) {
                if (i <= Curr_time) {
 
                    // Game is lost
                    break;
                }
                else {
                    Curr_time++;
                    killcount++;
                }
            }
            return killcount;
        }
 
        // Driver Code
 
        // Given input
        let dist = [1, 3, 4];
        let N = dist.length;
 
        let speed = [1, 1, 1];
        let M = speed.length;
 
        // function call
        document.write(eliminateMaximum(dist, N, speed));
 
// This code is contributed by Potta Lokesh
    </script>


 
 

Output: 

3

 

 

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

 

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