# Steps required to visit M points in order on a circular ring of N points

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 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); ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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## 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 ` `?> ` |

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**Output:**

4

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