# Minimum steps needed to cover a sequence of points on an infinite grid

Given an infinite grid, initial cell position (x, y) and a sequence of other cell position which needs to be covered in the given order. The task is to find the minimum number of steps needed to travel to all those cells.

**Note:** Movement can be done in any of the eight possible directions from a given cell i.e from cell (x, y) you can move to any of the following eight positions:(x-1, y+1), (x-1, y), (x-1, y-1), (x, y-1), (x+1, y-1), (x+1, y), (x+1, y+1), (x, y+1) is possible.

**Examples:**

Input:points[] = [(0, 0), (1, 1), (1, 2)]

Output:2

Move from (0, 0) to (1, 1) in 1 step(diagonal) and

then from (1, 1) to (1, 2) in 1 step (rightwards)

Input:points[] = [{4, 6}, {1, 2}, {4, 5}, {10, 12}]

Output:14

Move from(4, 6)-> (3, 5) -> (2, 4) -> (1, 3) ->

(1, 2)-> (2, 3) -> (3, 4) ->

(4, 5)-> (5, 6) -> (6, 7) ->

(7, 8) -> (8, 9) -> (9, 10) -> (10, 11) ->(10, 12)

**Approach:** Since all the given points are to be covered in the specified order. Find the minimum number of steps required to reach from a starting point to next point, then the sum of all such minimum steps for covering all the points would be the answer. One way to reach from a point (x1, y1) to (x2, y2) is to move abs(x2-x1) steps in the horizontal direction and abs(y2-y1) steps in the vertical direction, but this is not the shortest path to reach (x2, y2). The best way would be to cover the maximum possible distance in a diagonal direction and remaining in horizontal or vertical direction.

If we look closely this just reduces to the maximum of **abs(x2-x1)** and **abs(y2-y1)**. Traverse for all points and summation of all diagonal distance will be the answer.

Below is the implementation of the above approach:

## C++

`// C++ program to cover a sequence of points ` `// in minimum steps in a given order. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// cell structure denoted as point ` `struct` `point { ` ` ` `int` `x, y; ` `}; ` ` ` `// function to give minimum steps to ` `// move from point p1 to p2 ` `int` `shortestPath(point p1, point p2) ` `{ ` ` ` `// dx is total horizontal ` ` ` `// distance to be covered ` ` ` `int` `dx = ` `abs` `(p1.x - p2.x); ` ` ` ` ` `// dy is total vertical ` ` ` `// distance to be covered ` ` ` `int` `dy = ` `abs` `(p1.y - p2.y); ` ` ` ` ` `// required answer is ` ` ` `// maximum of these two ` ` ` `return` `max(dx, dy); ` `} ` ` ` `// Function to return the minimum steps ` `int` `coverPoints(point sequence[], ` `int` `size) ` `{ ` ` ` `int` `stepCount = 0; ` ` ` ` ` `// finding steps for each ` ` ` `// consecutive point in the sequence ` ` ` `for` `(` `int` `i = 0; i < size - 1; i++) { ` ` ` `stepCount += shortestPath(sequence[i], ` ` ` `sequence[i + 1]); ` ` ` `} ` ` ` ` ` `return` `stepCount; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `// arr stores sequence of points ` ` ` `// that are to be visited ` ` ` `point arr[] = { { 4, 6 }, { 1, 2 }, { 4, 5 }, { 10, 12 } }; ` ` ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` `cout << coverPoints(arr, n); ` `} ` |

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## Java

`// Java program to cover a ` `// sequence of points in ` `// minimum steps in a given order. ` `import` `java.io.*; ` `import` `java.util.*; ` `import` `java.lang.*; ` ` ` `// class denoted as point ` `class` `point ` `{ ` ` ` `int` `x, y; ` ` ` `point(` `int` `a, ` `int` `b) ` ` ` `{ ` ` ` `x = a; ` ` ` `y = b; ` ` ` ` ` `} ` `} ` ` ` `class` `GFG ` `{ ` `// function to give minimum ` `// steps to move from point ` `// p1 to p2 ` `static` `int` `shortestPath(point p1, ` ` ` `point p2) ` `{ ` ` ` `// dx is total horizontal ` ` ` `// distance to be covered ` ` ` `int` `dx = Math.abs(p1.x - p2.x); ` ` ` ` ` `// dy is total vertical ` ` ` `// distance to be covered ` ` ` `int` `dy = Math.abs(p1.y - p2.y); ` ` ` ` ` `// required answer is ` ` ` `// maximum of these two ` ` ` `return` `Math.max(dx, dy); ` `} ` ` ` `// Function to return ` `// the minimum steps ` `static` `int` `coverPoints(point sequence[], ` ` ` `int` `size) ` `{ ` ` ` `int` `stepCount = ` `0` `; ` ` ` ` ` `// finding steps for ` ` ` `// each consecutive ` ` ` `// point in the sequence ` ` ` `for` `(` `int` `i = ` `0` `; i < size - ` `1` `; i++) ` ` ` `{ ` ` ` `stepCount += shortestPath(sequence[i], ` ` ` `sequence[i + ` `1` `]); ` ` ` `} ` ` ` ` ` `return` `stepCount; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` ` ` `// arr stores sequence of points ` ` ` `// that are to be visited ` ` ` `point arr[] = ` `new` `point[` `4` `]; ` ` ` `arr[` `0` `] = ` `new` `point(` `4` `, ` `6` `); ` ` ` `arr[` `1` `] = ` `new` `point(` `1` `, ` `2` `); ` ` ` `arr[` `2` `] = ` `new` `point(` `4` `, ` `5` `); ` ` ` `arr[` `3` `] = ` `new` `point(` `10` `, ` `12` `); ` ` ` ` ` `int` `n = arr.length; ` ` ` `System.out.print(coverPoints(arr, n)); ` `} ` `} ` |

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## C#

`// C# program to cover a ` `// sequence of points in ` `// minimum steps in a given order. ` ` ` `using` `System; ` `// class denoted as point ` `public` `class` `point ` `{ ` ` ` `public` `int` `x, y; ` ` ` `public` `point(` `int` `a, ` `int` `b) ` ` ` `{ ` ` ` `x = a; ` ` ` `y = b; ` ` ` ` ` `} ` `} ` ` ` `public` `class` `GFG ` `{ ` ` ` `// function to give minimum ` ` ` `// steps to move from point ` ` ` `// p1 to p2[] ` ` ` `static` `int` `shortestPath(point p1, ` ` ` `point p2) ` ` ` `{ ` ` ` `// dx is total horizontal ` ` ` `// distance to be covered ` ` ` `int` `dx = Math.Abs(p1.x - p2.x); ` ` ` ` ` `// dy is total vertical ` ` ` `// distance to be covered ` ` ` `int` `dy = Math.Abs(p1.y - p2.y); ` ` ` ` ` `// required answer is ` ` ` `// maximum of these two ` ` ` `return` `Math.Max(dx, dy); ` ` ` `} ` ` ` ` ` `// Function to return ` ` ` `// the minimum steps ` ` ` `static` `int` `coverPoints(point []sequence, ` ` ` `int` `size) ` ` ` `{ ` ` ` `int` `stepCount = 0; ` ` ` ` ` `// finding steps for ` ` ` `// each consecutive ` ` ` `// point in the sequence ` ` ` `for` `(` `int` `i = 0; i < size - 1; i++) ` ` ` `{ ` ` ` `stepCount += shortestPath(sequence[i], ` ` ` `sequence[i + 1]); ` ` ` `} ` ` ` ` ` `return` `stepCount; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `// arr stores sequence of points ` ` ` `// that are to be visited ` ` ` `point []arr = ` `new` `point[4]; ` ` ` `arr[0] = ` `new` `point(4, 6); ` ` ` `arr[1] = ` `new` `point(1, 2); ` ` ` `arr[2] = ` `new` `point(4, 5); ` ` ` `arr[3] = ` `new` `point(10, 12); ` ` ` ` ` `int` `n = arr.Length; ` ` ` `Console.WriteLine(coverPoints(arr, n)); ` ` ` `} ` `} ` `// This code is contributed by Rajput-Ji ` |

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**Output:**

14

**Time Complexity:** O(N)

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