Maximum number of bomb blasts that may occur before the thief gets caught

• Difficulty Level : Expert
• Last Updated : 13 Jul, 2021

Given an array, arr[] of M integers, where the ith element represents the time after which the ith bomb will blast after dropping it, and three integers N, X, and Y representing the number of adjacent continuous cells on the X-coordinate, and the initial cell positions of a thief and police. The task is to find the maximum number of bomb blasts that may occur before the thief gets caught if, at every second, the thief can either drop a bomb or move to the left or right of an existing cell not visited by the police.

Examples:

Input: arr[] = {1, 4}, N = 7, X = 3, Y = 6
Output: 2
Explanation:
One possible way is:

1. At t = 0: Thief drops the bomb of activating time equal to 4. Meanwhile, the police move one cell towards the thief. Thereafter, the positions of the thief and police are 3 and 5 respectively.
2. At t = 1: The police move one cell towards the thief and the thief moves one cell to its left. Thereafter, the positions of the thief and police are 2 and 4 respectively.
3. At t = 2: The police move one cell towards the thief and the thief moves one cell to its left. Thereafter, the positions of the thief and police are 1 and 3 respectively.
4. At t = 3: The police move one cell towards the thief and the thief drops the bomb of activating time equal to 1. Thereafter, the positions of the thief and police are 1 and 2 respectively.
5. At t = 4: The bombs dropped at time (t= 3, and t = 0) blasts. Now the thief cannot move to any cell, and it does not have any bombs left. The police move one cell towards the thief, finally catching it at cell 1.

Therefore, the maximum bomb blasts that occurred before the thief got caught is 2.

Input: arr[] = {5, 1}, N = 7, X = 3, Y = 6
Output: 1

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

1. If both police and thief move optimally, then at every second the police will move towards the thief. Therefore, the maximum time the thief has before getting caught is the distance between their positions.
2. It can be observed that the best choice is to drop the bomb with more activating time first than the less activating time. If a bomb with less time is dropped first and then dropping the bomb with more activating time may exceed the time that the thief has before getting caught.

Follow the steps below to solve the problem:

• Sort the array arr[] in descending order.
• Initialize two variables, say count and time with value 0 to store the maximum count of the bomb blast that may occur and the time passed.
• Find the absolute difference between X and Y and store it in a variable, say maxSec.
• Iterate in the range[0, M-1], using the variable i, and do the following steps:
• If the sum of the current element and the time is less than or equal to the maxSec then increment count and time by 1.
• After the above step, update the count as count = min(count, abs(X-Y)-1).
• Finally, after completing the above steps, print the value of count as the answer.

Below is the implementation of the above approach:

C++

 // C++ program for the above approach#include using namespace std; // Function to find the maximum number// of bomb that can be blasted before// the thief gets caughtint findMaxBomb(int N, int M, int X, int Y, int arr[]){    // Sort the array arr[] in    // descending order    sort(arr, arr + M, greater());     // Stores the maxtime the thief    // has before getting caught    int maxSec;     // If Y is less than X    if (Y < X) {        maxSec = N - Y;    }    // Otherwise    else {        maxSec = Y - 1;    }     // Stores the current    // second    int time = 1;     // Stores the count of    // bomb blasts    int count = 0;     // Traverse the array arr[]    for (int i = 0; i < M; i++) {         // If arr[i]+time is less        // than or equal to the        // maxSec        if (arr[i] + time <= maxSec) {            // Increment time and            // count by 1            time++;            count++;        }    }     // Update count    count = min(count, abs(X - Y) - 1);     // Return the value of count    return count;} // Driver Codeint main(){    // Given Input    int N = 7, X = 3, Y = 6;    int arr[] = { 1, 4 };    int M = sizeof(arr) / sizeof(arr);     // Function Call    cout << findMaxBomb(N, M, X, Y, arr);    return 0;}

Java

 // Java program for the above approachimport java.util.Arrays;import java.util.Collections; public class GFG {     // Function to find the maximum number    // of bomb that can be blasted before    // the thief gets caught    static int findMaxBomb(int N, int M, int X, int Y,                           Integer arr[])    {        // Sort the array arr[] in        // descending order        Arrays.sort(arr, Collections.reverseOrder());         // Stores the maxtime the thief        // has before getting caught        int maxSec;         // If Y is less than X        if (Y < X) {            maxSec = N - Y;        }        // Otherwise        else {            maxSec = Y - 1;        }         // Stores the current        // second        int time = 1;         // Stores the count of        // bomb blasts        int count = 0;         // Traverse the array arr[]        for (int i = 0; i < M; i++) {             // If arr[i]+time is less            // than or equal to the            // maxSec            if (arr[i] + time <= maxSec) {                // Increment time and                // count by 1                time++;                count++;            }        }         // Update count        count = Math.min(count, Math.abs(X - Y) - 1);         // Return the value of count        return count;    }     // Driver code    public static void main(String[] args)    {        // Given Input        int N = 7, X = 3, Y = 6;        Integer arr[] = { 1, 4 };        int M = arr.length;         // Function Call        System.out.println(findMaxBomb(N, M, X, Y, arr));    }} // This code is contributed by abhinavjain194

Python3

 # Python3 program for the above approach # Function to find the maximum number# of bomb that can be blasted before# the thief gets caughtdef findMaxBomb(N, M, X, Y, arr):         # Sort the array arr[] in    # descending order    arr.sort(reverse = True)     # Stores the maxtime the thief    # has before getting caught    maxSec = 0     # If Y is less than X    if (Y < X):        maxSec = N - Y     # Otherwise    else:        maxSec = Y - 1     # Stores the current    # second    time = 1     # Stores the count of    # bomb blasts    count = 0     # Traverse the array arr[]    for i in range(M):                 # If arr[i]+time is less        # than or equal to the        # maxSec        if (arr[i] + time <= maxSec):                         # Increment time and            # count by 1            time += 1            count += 1     # Update count    count = min(count, abs(X - Y) - 1)     # Return the value of count    return count # Driver Codeif __name__ == '__main__':         # Given Input    N = 7    X = 3    Y = 6    arr = [ 1, 4 ]    M = len(arr)     # Function Call    print(findMaxBomb(N, M, X, Y, arr)) # This code is contributed by ipg2016107

C#

 // C# program for the above approachusing System; class GFG{ // Function to find the maximum number// of bomb that can be blasted before// the thief gets caughtstatic int findMaxBomb(int N, int M, int X,                       int Y, int[] arr){         // Sort the array arr[] in    // descending order    Array.Sort(arr);     // Reverse array    Array.Reverse(arr);     // Stores the maxtime the thief    // has before getting caught    int maxSec;     // If Y is less than X    if (Y < X)    {        maxSec = N - Y;    }         // Otherwise    else    {        maxSec = Y - 1;    }     // Stores the current    // second    int time = 1;     // Stores the count of    // bomb blasts    int count = 0;     // Traverse the array arr[]    for(int i = 0; i < M; i++)    {                 // If arr[i]+time is less        // than or equal to the        // maxSec        if (arr[i] + time <= maxSec)        {                         // Increment time and            // count by 1            time++;            count++;        }    }     // Update count    count = Math.Min(count, Math.Abs(X - Y) - 1);     // Return the value of count    return count;} // Driver Codepublic static void Main(String[] args){         // Given Input    int N = 7, X = 3, Y = 6;    int[] arr = { 1, 4 };    int M = arr.Length;     // Function Call    Console.WriteLine(findMaxBomb(N, M, X, Y, arr));}} // This code is contributed by target_2

Javascript


Output
2

Time Complexity: O(M*log(M)), where M is the size of the array arr[].
Auxiliary Space: O(1)

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