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Forming smallest array with given constraints

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  • Difficulty Level : Basic
  • Last Updated : 17 Aug, 2022
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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 <bits/stdc++.h>
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




<?php
// PHP 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.
function minimumLength($x, $y,$z)
{
    return 1 + abs($x - $y) +
               abs($y - $z);
}
 
    // Driver Code
    $x = 3;
    $y = 1;
    $z = 2;
    echo minimumLength($x, $y, $z);
 
// This code is contributed by anuj_67.
?>

Javascript




<script>
 
// Javascript 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.
function minimumLength(x, y, z)
{
    return 1 + Math.abs(x - y) + Math.abs(y - z);
}
 
// Drivers code
var x = 3, y = 1, z = 2;
document.write( minimumLength(x, y, z));
 
// This code is contributed by noob2000.
</script>

Output

4

Complexity Analysis:

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

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