# 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); ` `} ` |

*chevron_right*

*filter_none*

## 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)); ` `} ` `} ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

**Output:**

14

**Time Complexity:** O(N)

## Recommended Posts:

- Minimum cost to cover the given positions in a N*M grid
- Steps required to visit M points in order on a circular ring of N points
- Find if the given number is present in the infinite sequence or not
- Minimum squares to cover a rectangle
- Minimum moves to reach target on a infinite line | Set 2
- Find minimum moves to reach target on an infinite line
- Minimum numbers needed to express every integer below N as a sum
- Find minimum number of Log value needed to calculate Log upto N
- Minimum number of letters needed to make a total of n
- Find minimum operations needed to make an Array beautiful
- Minimum product in a grid of adjacent elements
- Minimum number of integers required to fill the NxM grid
- Count minimum steps to get the given desired array
- Minimum steps to color the tree with given colors
- Find the minimum number of steps to reach M from N

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.