# Forming smallest array with given constraints

• Difficulty Level : Basic
• Last Updated : 17 Aug, 2022

Given three integers x, y and z (can be negative). The task is to find the length of the smallest array that can be made such that absolute difference between adjacent elements is less than or equal to 1, the first element of the array is x, having one integer y and last element z.

Examples:

Input : x = 5, y = 7, z = 11
Output :
The smallest starts with 5, having 7, ends
with 11 and having absolute difference 1
is { 5, 6, 7, 8, 9, 10, 11 }.

Input : x = 3, y = 1, z = 2
Output :
The array would become { 3, 2, 1, 2 }

The idea is to consider the number line since the difference between adjacent elements is 1 so to move from X to Y all numbers between x and y have to be covered and to end the array with integer z, so move from element y to element z.

So, the length of the smallest array that can be formed will be the number of points that will be covered from x to z i.e.

`1 + abs(x - y) + abs(y - z) (1 is added to count point x).`

Implementation:

## C++

 `// C++ program to find the length of``// smallest array begin with x, having y,``// ends with z and having absolute difference``// between adjacent elements <= 1.``#include ``using` `namespace` `std;` `// Return the size of smallest``// array with given constraint.``int` `minimumLength(``int` `x, ``int` `y, ``int` `z)``{``    ``return` `1 + ``abs``(x - y) + ``abs``(y - z);``}` `// Drivers code``int` `main()``{``    ``int` `x = 3, y = 1, z = 2;``    ``cout << minimumLength(x, y, z);` `    ``return` `0;``}`

## Java

 `// Java program to find the length``// of smallest array begin with x,``// having y, ends with z and having``// absolute difference between``// adjacent elements <= 1.``import` `java.io.*;` `class` `GFG {``    ` `    ``// Return the size of smallest``    ``// array with given constraint.``    ``static` `int` `minimumLength(``int` `x,``                      ``int` `y, ``int` `z)``    ``{``        ``return` `1` `+ Math.abs(x - y)``                 ``+ Math.abs(y - z);``    ``}``    ` `    ``// Drivers code` `    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `x = ``3``, y = ``1``, z = ``2``;``        ``System.out.println(``              ``minimumLength(x, y, z));``    ``}``}` `// This code is contributed by anuj_67.`

## Python 3

 `# Python 3 program to find``# the length of smallest``# array begin with x, having``# y, ends with z and having``# absolute difference between``# adjacent elements <= 1.` `# Return the size of smallest``# array with given constraint.``def` `minimumLength(x, y, z):` `    ``return` `(``1` `+` `abs``(x ``-` `y)``              ``+` `abs``(y ``-` `z))` `# Drivers code``x ``=` `3``y ``=` `1``z ``=` `2``print``(minimumLength(x, y, z))` `# This code is contributed``# by Smitha`

## C#

 `// C# program to find the length``// of smallest array begin with x,``// having y, ends with z and having``// absolute difference between``// adjacent elements <= 1.``using` `System;``class` `GFG {``    ` `    ``// Return the size of smallest``    ``// array with given constraint.``    ``static` `int` `minimumLength(``int` `x,``                             ``int` `y,``                             ``int` `z)``    ``{``        ``return` `1 + Math.Abs(x - y)``                 ``+ Math.Abs(y - z);``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int` `x = 3, y = 1, z = 2;``        ``Console.WriteLine(minimumLength(x, y, z));``    ``}``}` `// This code is contributed by anuj_67.`

## PHP

 ``

## Javascript

 ``

Output

`4`

Complexity Analysis:

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

My Personal Notes arrow_drop_up