Given an integer ‘n’, consider a circular ring containing ‘n’ points numbered from ‘1’ to ‘n’ such that you can move in the following way :

1 -> 2 -> 3 -> ….. -> n -> 1 -> 2 -> 3 -> ……

Also, given an array of integers (of size ‘m’), the task is to find the number of steps it’ll take to get to every point in the array in order starting at ‘1’**Examples :**

Input:n = 3, m = 3, arr[] = {2, 1, 2}Output:4 The sequence followed is 1->2->3->1->2Input:n = 2, m = 1, arr[] = {2}Output:1 The sequence followed is 1->2

**Approach:** Let’s denote the current position by **cur** and the next position by **nxt**. This gives us 2 cases:

- If
**cur**is smaller than**nxt**, you can move to it in**nxt – cur**steps. - Otherwise, you first need to reach the point n in
**n – cur**steps, and then you can move to nxt in**cur**steps.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to count the steps required` `int` `findSteps(` `int` `n, ` `int` `m, ` `int` `a[])` `{` ` ` `// Start at 1` ` ` `int` `cur = 1;` ` ` `// Initialize steps` ` ` `int` `steps = 0;` ` ` `for` `(` `int` `i = 0; i < m; i++) {` ` ` `// If nxt is greater than cur` ` ` `if` `(a[i] >= cur)` ` ` `steps += (a[i] - cur);` ` ` `else` ` ` `steps += (n - cur + a[i]);` ` ` `// Now we are at a[i]` ` ` `cur = a[i];` ` ` `}` ` ` `return` `steps;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 3, m = 3;` ` ` `int` `a[] = { 2, 1, 2 };` ` ` `cout << findSteps(n, m, a);` `}` |

## Java

`// Java implementation of the approach` `class` `GFG` `{` ` ` `// Function to count the steps required` `static` `int` `findSteps(` `int` `n, ` `int` `m,` ` ` `int` `a[])` `{` ` ` `// Start at 1` ` ` `int` `cur = ` `1` `;` ` ` `// Initialize steps` ` ` `int` `steps = ` `0` `;` ` ` `for` `(` `int` `i = ` `0` `; i < m; i++)` ` ` `{` ` ` `// If nxt is greater than cur` ` ` `if` `(a[i] >= cur)` ` ` `steps += (a[i] - cur);` ` ` `else` ` ` `steps += (n - cur + a[i]);` ` ` `// Now we are at a[i]` ` ` `cur = a[i];` ` ` `}` ` ` `return` `steps;` `}` `// Driver code` `public` `static` `void` `main(String []args)` `{` ` ` `int` `n = ` `3` `, m = ` `3` `;` ` ` `int` `a[] = { ` `2` `, ` `1` `, ` `2` `};` ` ` `System.out.println(findSteps(n, m, a));` `}` `}` `// This code is contributed by ihritik` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG` `{` `// Function to count the` `// steps required` `static` `int` `findSteps(` `int` `n,` ` ` `int` `m, ` `int` `[]a)` `{` ` ` `// Start at 1` ` ` `int` `cur = 1;` ` ` `// Initialize steps` ` ` `int` `steps = 0;` ` ` `for` `(` `int` `i = 0; i < m; i++)` ` ` `{` ` ` `// If nxt is greater than cur` ` ` `if` `(a[i] >= cur)` ` ` `steps += (a[i] - cur);` ` ` `else` ` ` `steps += (n - cur + a[i]);` ` ` `// Now we are at a[i]` ` ` `cur = a[i];` ` ` `}` ` ` `return` `steps;` `}` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `int` `n = 3, m = 3;` ` ` `int` `[]a = { 2, 1, 2 };` ` ` `Console.WriteLine(findSteps(n, m, a));` `}` `}` `// This code is contributed by ihritik` |

## Python3

`# Python3 implementation of the approach` `# Function to count the steps required` `def` `findSteps(n, m, a):` ` ` `# Start at 1` ` ` `cur ` `=` `1` ` ` `# Initialize steps` ` ` `steps ` `=` `0` ` ` `for` `i ` `in` `range` `(` `0` `, m):` ` ` `# If nxt is greater than cur` ` ` `if` `(a[i] >` `=` `cur):` ` ` `steps ` `+` `=` `(a[i] ` `-` `cur)` ` ` `else` `:` ` ` `steps ` `+` `=` `(n ` `-` `cur ` `+` `a[i])` ` ` `# Now we are at a[i]` ` ` `cur ` `=` `a[i]` ` ` ` ` `return` `steps` `# Driver code` `n ` `=` `3` `m ` `=` `3` `a ` `=` `[` `2` `, ` `1` `, ` `2` `]` `print` `(findSteps(n, m, a))` `# This code is contributed by ihritik` |

## PHP

`<?php` `// PHP implementation of the approach` `// Function to count the steps required` `function` `findSteps(` `$n` `, ` `$m` `, ` `$a` `)` `{` ` ` `// Start at 1` ` ` `$cur` `= 1;` ` ` `// Initialize steps` ` ` `$steps` `= 0;` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$m` `; ` `$i` `++)` ` ` `{` ` ` `// If nxt is greater than cur` ` ` `if` `(` `$a` `[` `$i` `] >= ` `$cur` `)` ` ` `$steps` `+= (` `$a` `[` `$i` `] - ` `$cur` `);` ` ` `else` ` ` `$steps` `+= (` `$n` `- ` `$cur` `+ ` `$a` `[` `$i` `]);` ` ` `// Now we are at a[i]` ` ` `$cur` `= ` `$a` `[` `$i` `];` ` ` `}` ` ` `return` `$steps` `;` `}` `// Driver code` `$n` `= 3;` `$m` `= 3;` `$a` `= ` `array` `(2, 1, 2 );` `echo` `findSteps(` `$n` `, ` `$m` `, ` `$a` `);` `// This code is contributed by ihritik` `?>` |

## Javascript

`<script>` `// Javascript implementation of the approach` `// Function to count the steps required` `function` `findSteps(n, m, a)` `{` ` ` `// Start at 1` ` ` `var` `cur = 1;` ` ` `// Initialize steps` ` ` `var` `steps = 0;` ` ` `for` `(` `var` `i = 0; i < m; i++) {` ` ` `// If nxt is greater than cur` ` ` `if` `(a[i] >= cur)` ` ` `steps += (a[i] - cur);` ` ` `else` ` ` `steps += (n - cur + a[i]);` ` ` `// Now we are at a[i]` ` ` `cur = a[i];` ` ` `}` ` ` `return` `steps;` `}` `// Driver code` `var` `n = 3, m = 3;` `var` `a = [ 2, 1, 2 ];` `document.write( findSteps(n, m, a));` `</script>` |

**Output:**

4

**Time Complexity:** O(M)

**Auxiliary Space:** O(1)

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the **Essential Maths for CP Course** at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**