There are aliens in n buildings (minimum of 1 in each) and you have to kill all of them minimum number of bombings. Buildings are numbered as 1 – n. Aliens in a bombed building gets injured at the first bombing and die at the second. When a building is bombed for the first time, aliens in that building try to escape to the nearest building (for first building the nearest one is the second one and for nth building it is n-1). Calculate the minimum number of bombs required to kill all the aliens and the order of bombings.
Example:
Input: 3
Output: 4
2 1 3 2
Explanation: Minimum number of bombs required are 4.
First bomb the 2nd building, aliens
will move to 1st or 3rd to save
themselves. Then bomb at 1st building,
if some aliens have moved from 2nd
building to 1st they will be killed and
the 1st building aliens will be injured,
and they will move to the 2nd building
as it is nearest to them. Now, bomb at
the 3rd building to kill aliens who
moved from the 2nd building to 3rd and
injure 3rd building aliens so they move
to 2nd building as it is nearest to them.
Now, bomb at the 2nd building again and
all aliens who moved from 1st or 3rd
building will be killed.
Input: 2
Output: 3
2 1 2
We can make a constructive way to kill all aliens. Since everyone either moves to the left or to the right, we have to make sure that all the even positions are attacked, once at the start to injure aliens and the other time at the end. When we attack aliens at the even positions first time they move to the odd position buildings, so attack them at odd to kill all the previous even positions and injure the odd position aliens. The odd position aliens get injured and will move to the even position, so attack them at the even at the end to kill them.
The number of ways will be n/2 + n/2 + n/2 which is n + n/2.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void print( int n)
{
cout << n + n / 2 << endl;
for ( int i = 2; i <= n; i += 2)
cout << i << " " ;
for ( int i = 1; i <= n; i += 2)
cout << i << " " ;
for ( int i = 2; i <= n; i += 2)
cout << i << " " ;
}
int main()
{
int n = 3;
print(n);
return 0;
}
|
Java
class GFG {
static void print( int n)
{
System.out.println(n + n / 2 );
for ( int i = 2 ; i <= n; i += 2 )
System.out.print( i + " " );
for ( int i = 1 ; i <= n; i += 2 )
System.out.print(i + " " );
for ( int i = 2 ; i <= n; i += 2 )
System.out.print( i + " " );
}
public static void main (String[] args)
{
int n = 3 ;
print(n);
}
}
|
Python3
def bomb_required(n):
print (n + n / / 2 )
for i in range ( 2 , n + 1 , 2 ):
print (i, end = " " )
for i in range ( 1 , n + 1 , 2 ):
print (i, end = " " )
for i in range ( 2 , n, 2 ):
print (i, end = " " )
bomb_required( 3 )
|
C#
using System;
class GFG {
static void print( int n)
{
Console.WriteLine(n + n / 2);
for ( int i = 2; i <= n; i += 2)
Console.Write( i + " " );
for ( int i = 1; i <= n; i += 2)
Console.Write(i + " " );
for ( int i = 2; i <= n; i += 2)
Console.Write( i + " " );
}
public static void Main ()
{
int n = 3;
print(n);
}
}
|
PHP
<?php
function p_rint( $n )
{
echo floor ( $n + $n / 2), "\n" ;
for ( $i = 2; $i <= $n ; $i += 2)
echo $i , " " ;
for ( $i = 1; $i <= $n ; $i += 2)
echo $i , " " ;
for ( $i = 2; $i <= $n ; $i += 2)
echo $i , " " ;
}
$n = 3;
p_rint( $n );
?>
|
Javascript
<script>
function print(n) {
document.write(n + parseInt(n / 2) + "<br/>" );
for (i = 2; i <= n; i += 2)
document.write(i + " " );
for (i = 1; i <= n; i += 2)
document.write(i + " " );
for (i = 2; i <= n; i += 2)
document.write(i + " " );
}
var n = 3;
print(n);
</script>
|
Output:
4
2 1 3 2
Time complexity: O(n)
Auxiliary Space: O(1)
Last Updated :
18 Sep, 2023
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