由 N 个磁体形成的磁体组的数量
给定 N 个磁体，它们一个接一个地排成一行，要么负极在左边，正极在右边(01)，要么正极在左边，负极在右边(10)。考虑到这样一个事实，如果两个连续的磁铁有不同的磁极彼此面对，它们形成一个组并相互吸引，找到可能的组的总数。
示例:
Input : N = 6
magnets = {10, 10, 10, 01, 10, 10}
Output : 3
The groups are formed by the following magnets:
1, 2, 3
4
5, 6
Input : N = 5
magnets = {10, 10, 10, 10, 10, 01}
Output : 1
让我们考虑每对连续的磁铁。有两种可能的情况:
- 两者配置相同。在这种情况下，连接端将具有不同的极点，因此它们将属于同一组。
- 两者都有不同的配置。在这种情况下，连接端将具有相同的极，因此它们将相互排斥以形成不同的组。
因此，只有在两个连续的磁体具有不同配置的情况下，才会形成新的组。遍历磁体阵列，找出具有不同配置的连续对的数量。
下面是上述方法的实现:
C++
// C++ program to find number of groups
// of magnets formed from N magnets
#include <bits/stdc++.h>
using namespace std;
// Function to count number of groups of
// magnets formed from N magnets
int countGroups(int n, string m[])
{
// Intinially only a single group
// for the first magnet
int count = 1;
for (int i = 1; i < n; i++)
// Different configuration increases
// number of groups by 1
if (m[i] != m[i - 1])
count++;
return count;
}
// Driver Code
int main()
{
int n = 6;
string m[n] = { "10", "10", "10", "01", "10", "10" };
cout << countGroups(n, m);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find the maximum number
// of elements that can be added to a set
// such that it is the absolute difference // of magnets formed from N magnets
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG{
// Function to count number of groups of
// magnets formed from N magnets
static int countGroups(int n, String m[])
{
// Intinially only a single group
// for the first magnet
int count = 1;
for (int i = 1; i < n; i++)
// Different configuration increases
// number of groups by 1
if (m[i] != m[i - 1])
count++;
return count;
}
// Driver Code
public static void main(String args[])
{
int n = 6;
String []m = { "10", "10", "10", "01", "10", "10" };
System.out.println( countGroups(n, m));
}
}
Python 3
# Python 3 program to find number
# of groups of magnets formed
# from N magnets
# Function to count number of
# groups of magnets formed
# from N magnets
def countGroups(n, m):
# Intinially only a single
# group for the first magnet
count = 1
for i in range(1, n):
# Different configuration increases
# number of groups by 1
if (m[i] != m[i - 1]):
count += 1
return count
# Driver Code
if __name__ == "__main__":
n = 6
m = [ "10", "10", "10",
"01", "10", "10" ]
print(countGroups(n, m))
# This code is contributed
# by ChitraNayal
C
// C# program to find number of groups
// of magnets formed from N magnets
using System;
class GFG {
// Function to count number of groups of
// magnets formed from N magnets
static int countGroups(int n, String []m)
{
// Intinially only a single group
// for the first magnet
int count = 1;
for (int i = 1; i < n; i++)
// Different configuration increases
// number of groups by 1
if (m[i] != m[i - 1])
count++;
return count;
}
// Driver Code
public static void Main()
{
int n = 6;
String [] m = {"10", "10", "10",
"01", "10", "10"};
Console.WriteLine(countGroups(n, m));
}
}
// This code is contributed by ANKITRAI1
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find number of groups
// of magnets formed from N magnets
// Function to count number of groups
// of magnets formed from N magnets
function countGroups($n, $m)
{
// Intinially only a single group
// for the first magnet
$count = 1;
for ($i = 1; $i < $n; $i++)
// Different configuration increases
// number of groups by 1
if ($m[$i] != $m[$i - 1])
$count++;
return $count;
}
// Driver Code
$n = 6;
$m = array( "10", "10", "10",
"01", "10", "10" );
echo(countGroups($n, $m));
// This code is contributed
// by Shivi_Aggarwal
?>
java 描述语言
<script>
// Javascript program to find the maximum number
// of elements that can be added to a set
// such that it is the absolute difference
// of magnets formed from N magnets
// Function to count number of groups of
// magnets formed from N magnets
function countGroups(n,m) {
// Intinially only a single group
// for the first magnet
let count = 1;
for (let i = 1; i < n; i++)
// Different configuration increases
// number of groups by 1
if (m[i] != m[i - 1])
count++;
return count;
}
// Driver Code
let n = 6;
let m=[ "10", "10", "10", "01", "10", "10"];
document.write( countGroups(n, m));
// This code is contributed by rag2127
</script>
Output:
3
时间复杂度: O(N)
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