How To Use For Loops To Create Different Size Towers Java

Minimum number of towers required such that every house is in the range of at to the lowest degree one tower

Given a map of the city and the network range, the task is to determine the minimum number of the tower so that every firm is within range of at to the lowest degree one tower. Each tower must be installed on pinnacle of an existing firm.

Examples:

Input: range : 1        house : i ii three four five Output: 2  Input: range : two        house : seven 2 4 six 5 nine 12 11  Output: 3

All cities can exist covered by inserting 2 towers i.e. at firm 2 and iv.

All cities tin can exist covered by inserting 3 towers i.e. at firm 4, 9, and 12.
Algorithm:-

1. First, sort all the elements.
2. Count only in one case and then traverse till its eye business firm.
3. Afterward this again traverse till tower range.
4. Again echo 1, 2, iii steps till all the houses are covered.

Below is the implementation of above approach:

C++

#include <bits/stdc++.h>

using namespace std;

int number_of_tower( int firm[], int range, int n)

{

sort(firm, house + due north);

int numOfTower = 0;

int i = 0;

while (i < north) {

numOfTower++;

int loc = house[i] + range;

while (i < n && house[i] <= loc)

i++;

--i;

loc = house[i] + range;

while (i < n && firm[i] <= loc)

i++;

}

return numOfTower;

}

int primary()

{

int house[] = { vii, two, 4, six, 5, 9, 12, 11 };

int range = 2;

int n = sizeof (house) / sizeof (firm[0]);

cout << number_of_tower(business firm, range, n);

}

Coffee

import java.util.Arrays;

public class Improve {

static int number_of_tower( int business firm[], int range, int north)

{

Arrays.sort(firm);

int numOfTower = 0 ;

int i = 0 ;

while (i < n) {

numOfTower++;

int loc = house[i] + range;

while (i < n && house[i] <= loc)

i++;

--i;

loc = house[i] + range;

while (i < north && firm[i] <= loc)

i++;

}

render numOfTower;

}

public static void main(String args[])

{

int business firm[] = { 7 , 2 , 4 , six , 5 , ix , 12 , eleven };

int range = 2 ;

int due north = house.length;

Organization.out.println(number_of_tower(house, range, north));

}

}

Python 3

def number_of_tower(house, r, n):

business firm.sort()

numOfTower = 0

i = 0

while (i < northward) :

numOfTower + = 1

loc = house[i] + r

while (i < northward and firm[i] < = loc):

i + = ane

i - = 1

loc = house[i] + r

while (i < n and firm[i] < = loc):

i + = ane

render numOfTower

if __name__ = = "__main__" :

house = [ seven , 2 , 4 , 6 , 5 , 9 , 12 , 11 ]

r = 2

due north = len (business firm)

impress (number_of_tower(house, r, northward))

C#

using Organisation;

public class Better {

static int number_of_tower( int []house, int range, int north)

{

Assortment.Sort(firm);

int numOfTower = 0;

int i = 0;

while (i < n) {

numOfTower++;

int loc = business firm[i] + range;

while (i < n && firm[i] <= loc)

i++;

--i;

loc = house[i] + range;

while (i < due north && firm[i] <= loc)

i++;

}

return numOfTower;

}

public static void Main()

{

int []house = { 7, 2, 4, 6, five, 9, 12, eleven };

int range = two;

int n = house.Length;

Console.WriteLine(number_of_tower(house, range, northward));

}

}

PHP

<?php

function number_of_tower( $house , $range , $north )

{

sort( $house );

$numOfTower = 0;

$i = 0;

while ( $i < $n ) {

$numOfTower ++;

$loc = $house [ $i ] + $range ;

while ( $i < $n && $house [ $i ] <= $loc )

$i ++;

-- $i ;

$loc = $house [ $i ] + $range ;

while ( $i < $n && $house [ $i ] <= $loc )

$i ++;

}

return $numOfTower ;

}

$house = assortment ( 7, 2, 4, half dozen, 5, 9, 12, 11 );

$range = 2;

$n = sizeof( $house ) / sizeof( $house [0]);

echo number_of_tower( $house , $range , $n );

?>

Javascript

<script>

role number_of_tower(business firm,range,due north)

{

house.sort( part (a,b){ render a-b;});

permit numOfTower = 0;

let i = 0;

while (i < north) {

numOfTower++;

permit loc = house[i] + range;

while (i < n && business firm[i] <= loc)

i++;

--i;

loc = house[i] + range;

while (i < n && house[i] <= loc)

i++;

}

return numOfTower;

}

let house=[7, 2, 4, 6, 5, 9, 12, eleven];

let range = 2;

let n = house.length;

document.write(number_of_tower(house, range, n));

</script>

Time Complexity: O(nlogn)
Space Complication: O(1)

How To Use For Loops To Create Different Size Towers Java,

Source: https://www.geeksforgeeks.org/minimum-number-of-towers-required-such-that-every-house-is-in-the-range-of-at-least-one-tower/

Posted by: lewisgoicame.blogspot.com

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