# Number of groups of magnets formed from N magnets

Last Updated : 19 Sep, 2023

Given N magnets are kept in a row one after another, either with a negative pole on the left and a positive pole on the right (01) or a positive pole on the left and a negative pole on the right (10). Considering the fact that if 2 consecutive magnets have different poles facing each other, they form a group and attract each other, find the total number of groups possible.

Examples:

Input : N = 6, magnets = {10, 10, 10, 01, 10, 10}
Output : 3
Explanation: 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 : 2

Let us consider every pair of consecutive magnets. There are 2 possible cases:

• Both of them have the same configuration. In this case, the connecting ends will have different poles and hence they would belong to the same group.
• Both of them have different configurations. In this case, the connecting ends will have the same pole and hence they would repel each other to form different groups.

So a new group will only be formed in the case when two consecutive magnets have different configurations. To traverse the array of magnets and find the number of consecutive pairs with the different configurations.

Below is the implementation of the above approach:

## C++

 `// C++ program to find number of groups` `// of magnets formed from N magnets`   `#include ` `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

 `// 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`

## Javascript

 ``

## PHP

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Output

```3

```

Complexity Analysis:

• Time Complexity: O(N)
• Auxiliary Space: O(1)

Approach:

• Initialize a stack and a variable called groups to 1.
• Iterate through the magnets from left to right.
• If the stack is empty or the top magnet has the same polarity as the current magnet, push it onto the stack.
• Otherwise, pop all the magnets from the stack until a magnet with the same polarity as the current magnet is found.
• Push the current magnet onto the stack.
• Increment the number of groups.
• After iterating through all the magnets, return the number of groups.

## C++

 `#include ` `using` `namespace` `std;`   `// Function to count number of groups of` `// magnets formed from N magnets` `int` `countGroups(``int` `n, ``int` `magnets[])` `{` `    ``stack<``int``> s;` `    ``int` `groups = 1; ``// Initially, only a single group for the first magnet` `    `  `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``if` `(s.empty() || s.top() == magnets[i]) {` `            ``// If the stack is empty or the top magnet has the same polarity as the current magnet, push it onto the stack` `            ``s.push(magnets[i]);` `        ``} ``else` `{` `            ``// Otherwise, pop all the magnets from the stack until a magnet with the same polarity as the current magnet is found` `            ``while` `(!s.empty() && s.top() != magnets[i]) {` `                ``s.pop();` `            ``}` `            ``s.push(magnets[i]); ``// Push the current magnet onto the stack` `            ``groups++; ``// Increment the number of groups` `        ``}` `    ``}` `    `  `    ``return` `groups;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `n = 6;` `    ``int` `magnets[n] = {10, 10, 10, 01, 10, 10};` `    `  `    ``cout << countGroups(n, magnets);` `    `  `    ``return` `0;` `}`

## Java

 `import` `java.util.*;`   `public` `class` `Main {` `    ``public` `static` `void` `main(String[] args) {` `        ``int` `n = ``6``;` `        ``int``[] magnets = {``10``, ``10``, ``10``, ``01``, ``10``, ``10``};`   `        ``System.out.println(countGroups(n, magnets));` `    ``}`   `    ``public` `static` `int` `countGroups(``int` `n, ``int``[] magnets) {` `        ``Stack stack = ``new` `Stack<>();` `        ``int` `groups = ``1``; ``// Initially, only a single group for the first magnet`   `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``if` `(stack.isEmpty() || stack.peek() == magnets[i]) {` `                ``// If the stack is empty or the top magnet has the same polarity as the current magnet, push it onto the stack` `                ``stack.push(magnets[i]);` `            ``} ``else` `{` `                ``// Otherwise, pop all the magnets from the stack until a magnet with the same polarity as the current magnet is found` `                ``while` `(!stack.isEmpty() && stack.peek() != magnets[i]) {` `                    ``stack.pop();` `                ``}` `                ``stack.push(magnets[i]); ``// Push the current magnet onto the stack` `                ``groups++; ``// Increment the number of groups` `            ``}` `        ``}`   `        ``return` `groups;` `    ``}` `}`

## Python3

 `def` `count_groups(n, magnets):` `    ``s ``=` `[]` `    ``groups ``=` `1`  `# Initially, only a single group for the first magnet` `    `  `    ``for` `i ``in` `range``(n):` `        ``if` `not` `s ``or` `s[``-``1``] ``=``=` `magnets[i]:` `            ``# If the stack is empty or the top magnet has the same polarity as the current magnet, push it onto the stack` `            ``s.append(magnets[i])` `        ``else``:` `            ``# Otherwise, pop all the magnets from the stack until a magnet with the same polarity as the current magnet is found` `            ``while` `s ``and` `s[``-``1``] !``=` `magnets[i]:` `                ``s.pop()` `            ``s.append(magnets[i])  ``# Push the current magnet onto the stack` `            ``groups ``+``=` `1`  `# Increment the number of groups` `    `  `    ``return` `groups`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    ``n ``=` `6` `    ``magnets ``=` `[``10``, ``10``, ``10``, ``1``, ``10``, ``10``]` `    `  `    ``print``(count_groups(n, magnets))`   ` ``# This code is contributed by shivamgupta0987654321`

## C#

 `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG {` `    ``// Function to count number of groups of magnets formed` `    ``// from N magnets` `    ``static` `int` `CountGroups(``int` `n, ``int``[] magnets)` `    ``{` `        ``Stack<``int``> s = ``new` `Stack<``int``>();` `        ``int` `groups = 1; ``// Initially, only a single group` `                        ``// for the first magnet`   `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``if` `(s.Count == 0 || s.Peek() == magnets[i]) {` `                ``// If the stack is empty or the top magnet` `                ``// has the same polarity as the current` `                ``// magnet, push it onto the stack` `                ``s.Push(magnets[i]);` `            ``}` `            ``else` `{` `                ``// Otherwise, pop all the magnets from the` `                ``// stack until a magnet with the same` `                ``// polarity as the current magnet is found` `                ``while` `(s.Count > 0` `                       ``&& s.Peek() != magnets[i]) {` `                    ``s.Pop();` `                ``}` `                ``s.Push(magnets[i]); ``// Push the current` `                                    ``// magnet onto the stack` `                ``groups++; ``// Increment the number of groups` `            ``}` `        ``}`   `        ``return` `groups;` `    ``}`   `    ``static` `void` `Main(``string``[] args)` `    ``{` `        ``int` `n = 6;` `        ``int``[] magnets = { 10, 10, 10, 01, 10, 10 };`   `        ``Console.WriteLine(CountGroups(n, magnets));` `    ``}` `}`   `// This code is contributed by shivamgupta0987654321`

## Javascript

 `function` `countGroups(n, magnets) {` `    ``let stack = [];` `    ``let groups = 1;` `    ``for` `(let i = 0; i < n; i++) {` `        ``if` `(stack.length === 0 || stack[stack.length - 1] === magnets[i]) {` `            ``// If the stack is empty or the top magnet has the ` `            ``// same polarity as the current magnet push it onto the stack` `            ``stack.push(magnets[i]);` `        ``} ``else` `{` `            ``// Otherwise, pop all the magnets from the stack until ` `            ``// magnet with the same polarity as the current magnet is found` `            ``while` `(stack.length > 0 && stack[stack.length - 1] !== magnets[i]) {` `                ``stack.pop();` `            ``}` `            ``stack.push(magnets[i]);` `            ``groups++;` `        ``}` `    ``}` `    ``return` `groups;` `}`   `const n = 6;` `const magnets = [10, 10, 10, 1, 10, 10];`   `console.log(countGroups(n, magnets));`

Output

```3

```

Time Complexity: O(n), where n is the number of magnets in the row, as we are iterating through each magnet only once.
Space Complexity: O(n), as in the worst case scenario (when all magnets have different polarities), we might need to store all n magnets in the stack.

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