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Check whether Arithmetic Progression can be formed from the given array

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Given an array of n integers. The task is to check whether an arithmetic progression can be formed using all the given elements. If possible print “Yes”, else print “No”.

Examples: 

Input : arr[] = {0, 12, 4, 8}
Output : Yes
Rearrange given array as {0, 4, 8, 12} 
which forms an arithmetic progression.

Input : arr[] = {12, 40, 11, 20}
Output : No

Method 1 (Simple) 
A simple solution is to first find the smallest element, then find second smallest element and find the difference between these two. Let this difference be d. After finding the difference, find third smallest, fourth smallest and so on. After finding every i-th smallest  (from third onward), find the difference between value of current element and value of previous element. If difference is not same as d, return false. If all elements have same difference, return true. Time complexity of this solution is O(n2)

Method 2(Use Sorting) 
The idea is to sort the given array. After sorting, check if differences between consecutive elements are same or not. If all differences are same, Arithmetic Progression is possible. 

Below is the implementation of this approach: 

C++




// C++ program to check if a given array
// can form arithmetic progression
#include<bits/stdc++.h>
using namespace std;
 
// Returns true if a permutation of arr[0..n-1]
// can form arithmetic progression
bool checkIsAP(int arr[], int n)
{
  if (n == 1)
    return true;
 
  // Sort array
  sort(arr, arr + n);
 
  // After sorting, difference between
  // consecutive elements must be same.
  int d = arr[1] - arr[0];
  for (int i=2; i<n; i++)
    if (arr[i] - arr[i-1] != d)
      return false;
 
  return true;
}
 
// Driven Program
int main()
{
  int arr[] = { 20, 15, 5, 0, 10 };
  int n = sizeof(arr)/sizeof(arr[0]);
 
  (checkIsAP(arr, n))? (cout << "Yes" << endl) :
                       (cout << "No" << endl);
 
  return 0;
}

Java



// Java program to check if a given array
// can form arithmetic progression
import java.util.Arrays;
 
class GFG {
         
    // Returns true if a permutation of
    // arr[0..n-1] can form arithmetic
    // progression
    static boolean checkIsAP(int arr[], int n)
    {
        if (n == 1)
            return true;
         
        // Sort array
        Arrays.sort(arr);
         
        // After sorting, difference between
        // consecutive elements must be same.
        int d = arr[1] - arr[0];
        for (int i = 2; i < n; i++)
            if (arr[i] - arr[i-1] != d)
                return false;
         
        return true;
    }
     
    //driver code
    public static void main (String[] args)
    {
        int arr[] = { 20, 15, 5, 0, 10 };
        int n = arr.length;
     
        if(checkIsAP(arr, n))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is contributed by Anant Agarwal.
Python3



# Python3 program to check if a given
# array can form arithmetic progression
 
# Returns true if a permutation of arr[0..n-1]
# can form arithmetic progression
def checkIsAP(arr, n):
    if (n == 1): return True
 
    # Sort array
    arr.sort()
 
    # After sorting, difference between
    # consecutive elements must be same.
    d = arr[1] - arr[0]
    for i in range(2, n):
        if (arr[i] - arr[i-1] != d):
            return False
 
    return True
 
# Driver code
arr = [ 20, 15, 5, 0, 10 ]
n = len(arr)
print("Yes") if(checkIsAP(arr, n)) else print("No")
 
# This code is contributed by Anant Agarwal.
C#



// C# program to check if a given array
// can form arithmetic progression
using System;
 
class GFG {
         
    // Returns true if a permutation of
    // arr[0..n-1] can form arithmetic
    // progression
    static bool checkIsAP(int []arr, int n)
    {
        if (n == 1)
            return true;
         
        // Sort array
        Array.Sort(arr);
         
        // After sorting, difference between
        // consecutive elements must be same.
        int d = arr[1] - arr[0];
        for (int i = 2; i < n; i++)
            if (arr[i] - arr[i - 1] != d)
                return false;
         
        return true;
    }
     
    //Driver Code
    public static void Main ()
    {
        int []arr = {20, 15, 5, 0, 10};
        int n = arr.Length;
     
        if(checkIsAP(arr, n))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by vt_m.
PHP



<?php
// PHP program to check if
// a given array can form
// arithmetic progression
 
// Returns true if a permutation
// of arr[0..n-1] can form
// arithmetic progression
function checkIsAP($arr, $n)
{
    if ($n == 1)
        return true;
     
    // Sort array
    sort($arr);
     
    // After sorting, difference
    // between consecutive elements
    // must be same.
    $d = $arr[1] - $arr[0];
    for ($i = 2; $i < $n; $i++)
        if ($arr[$i] -
            $arr[$i - 1] != $d)
        return false;
     
    return true;
}
 
// Driver Code
$arr = array(20, 15, 5, 0, 10);
$n = count($arr);
 
if(checkIsAP($arr, $n))
echo "Yes";
else
echo "No";
                         
// This code is contributed
// by Sam007
?>
Javascript



<script>
 
// Javascript program to check if a given array
// can form arithmetic progression
 
// Returns true if a permutation of arr[0..n-1]
// can form arithmetic progression
function checkIsAP(arr, n)
{
  if (n == 1)
    return true;
 
  // Sort array
  arr.sort((a, b) => a - b);
 
  // After sorting, difference between
  // consecutive elements must be same.
  let d = arr[1] - arr[0];
  for (let i=2; i<n; i++)
    if (arr[i] - arr[i-1] != d)
      return false;
 
  return true;
}
 
// Driven Program
 
  let arr = [ 20, 15, 5, 0, 10 ];
  let n = arr.length;
 
  (checkIsAP(arr, n))? (document.write("Yes" + "<br>")) :
                       (document.write("No" + "<br>"));
 
  
// This code is contributed by Mayank Tyagi
 
</script>
Output
Yes
Time Complexity: O(n Log n).
Auxiliary Space: O(1)
Method 3(Use Hashing) 
  1. Find out the smallest and second smallest elements
  2. Find difference between the two elements. d = second_smallest – smallest
  3. Store all elements in a hashmap and return “NO” if duplicate element found (can be done together with step 1).
  4. Now start from “second smallest element + d” and one by one check n-2 terms of Arithmetic Progression in hashmap. If any value of progression is missing, return false.
  5. Return “YES” after end of the loop.
Below is the implementation of this method.
C++



// C++ program to check if a given array
// can form arithmetic progression
#include <bits/stdc++.h>
using namespace std;
 
// Returns true if a permutation of arr[0..n-1]
// can form arithmetic progression
bool checkIsAP(int arr[], int n)
{
    unordered_map<int, int> hm;
    int smallest = INT_MAX, second_smallest = INT_MAX;
    for (int i = 0; i < n; i++) {
       
        // Find the smallest and
        // update second smallest
        if (arr[i] < smallest) {
            second_smallest = smallest;
            smallest = arr[i];
        }
       
        // Find second smallest
        else if (arr[i] != smallest
                 && arr[i] < second_smallest)
            second_smallest = arr[i];
       
        // Check if the duplicate element found or not
        if (hm.find(arr[i]) == hm.end())
            hm[arr[i]]++;
       
        // If duplicate found then return false
        else
            return false;
    }
   
    // Find the difference between smallest and second
    // smallest
   
    int diff = second_smallest - smallest;
   
    // As we have used smallest and
    // second smallest,so we
    // should now only check for n-2 elements
    for (int i = 0; i < n - 1; i++) {
        if (hm.find(second_smallest) == hm.end())
            return false;
        second_smallest += diff;
    }
    return true;
}
 
// Driven Program
int main()
{
    int arr[] = { 20, 15, 5, 0, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    (checkIsAP(arr, n)) ? (cout << "Yes" << endl)
                        : (cout << "No" << endl);
 
    return 0;
 
    // This code is contributed by Raman Jha
}
Java



// Java program to check if a given array
// can form arithmetic progression
import java.util.HashMap;
 
class GFG {
 
  // Returns true if a permutation of arr[0..n-1]
  // can form arithmetic progression
  static boolean checkIsAP(int arr[], int n) {
    HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
    int smallest = Integer.MAX_VALUE, second_smallest = Integer.MAX_VALUE;
    for (int i = 0; i < n; i++) {
 
      // Find the smallest and
      // update second smallest
      if (arr[i] < smallest) {
        second_smallest = smallest;
        smallest = arr[i];
      }
 
      // Find second smallest
      else if (arr[i] != smallest
               && arr[i] < second_smallest)
        second_smallest = arr[i];
 
      // Check if the duplicate element found or not
      if (!hm.containsKey(arr[i])) {
        hm.put(arr[i], 1);
      }
 
      // If duplicate found then return false
      else
        return false;
    }
 
    // Find the difference between smallest and second
    // smallest
    int diff = second_smallest - smallest;
 
    // As we have used smallest and
    // second smallest,so we
    // should now only check for n-2 elements
    for (int i = 0; i < n - 1; i++) {
      if (!hm.containsKey(second_smallest))
        return false;
      second_smallest += diff;
    }
    return true;
  }
 
  // Driven Program
  public static void main(String args[]) {
    int arr[] = { 20, 15, 5, 0, 10 };
    int n = arr.length;
 
    if (checkIsAP(arr, n)) {
      System.out.println("Yes");
    } else {
      System.out.println("No");
    }
    ;
 
  }
}
 
// This code is contributed by gfgking
Python3



# Python3 program to check if a given array
# can form arithmetic progression
 
# Returns true if a permutation of arr[0..n-1]
# can form arithmetic progression
def checkIsAP(arr, n):
     
    hm = {}
    smallest = float('inf')
    second_smallest = float('inf')
     
    for i in range(n):
         
        # Find the smallest and
        # update second smallest
        if (arr[i] < smallest):
            second_smallest = smallest
            smallest = arr[i]
       
        # Find second smallest
        else if (arr[i] != smallest and
              arr[i] < second_smallest):
            second_smallest = arr[i]
       
        # Check if the duplicate element found or not
        if arr[i] not in hm:
            hm[arr[i]] = 1
             
        # If duplicate found then return false
        else:
            return False
   
    # Find the difference between smallest
    # and second smallest
    diff = second_smallest - smallest
   
    # As we have used smallest and
    # second smallest,so we
    # should now only check for n-2 elements
    for i in range(n-1):
        if (second_smallest) not in hm:
            return False
             
        second_smallest += diff
     
    return True
 
# Driver code
arr = [ 20, 15, 5, 0, 10 ]
n = len(arr)
 
if (checkIsAP(arr, n)):
    print("Yes")
else:
    print("No")
     
# This code is contributed by rohitsingh07052
C#



// C# program to check if a given array
// can form arithmetic progression
using System;
using System.Collections.Generic;
 
class GFG
{
 
  // Returns true if a permutation of arr[0..n-1]
  // can form arithmetic progression
  public static bool checkIsAP(int[] arr, int n) {
    Dictionary<int, int> hm = new Dictionary<int, int>();
    int smallest = Int32.MaxValue, second_smallest = Int32.MaxValue;
    for (int i = 0; i < n; i++) {
 
      // Find the smallest and
      // update second smallest
      if (arr[i] < smallest) {
        second_smallest = smallest;
        smallest = arr[i];
      }
 
      // Find second smallest
      else if (arr[i] != smallest
               && arr[i] < second_smallest)
        second_smallest = arr[i];
 
      // Check if the duplicate element found or not
      if (!hm.ContainsKey(arr[i])) {
        hm[arr[i]]= 1;
      }
 
      // If duplicate found then return false
      else
        return false;
    }
 
    // Find the difference between smallest and second
    // smallest
    int diff = second_smallest - smallest;
 
    // As we have used smallest and
    // second smallest,so we
    // should now only check for n-2 elements
    for (int i = 0; i < n - 1; i++) {
      if (!hm.ContainsKey(second_smallest))
        return false;
      second_smallest += diff;
    }
    return true;
  }
 
  // Driven Program
  public static void Main(string[] args) {
    int[] arr = { 20, 15, 5, 0, 10 };
    int n = arr.Length;
 
    if (checkIsAP(arr, n)) {
      Console.WriteLine("Yes");
    } else {
      Console.WriteLine("No");
    }
  }
}
 
// This code is contributed by Pushpesh raj
Javascript



<script>
// Javascript program to check if a given array
// can form arithmetic progression
 
// Returns true if a permutation of arr[0..n-1]
// can form arithmetic progression
function checkIsAP(arr, n)
{
    var hm = new Map();
    var smallest = 1000000000, second_smallest = 1000000000;
    for (var i = 0; i < n; i++) {
       
        // Find the smallest and
        // update second smallest
        if (arr[i] < smallest) {
            second_smallest = smallest;
            smallest = arr[i];
        }
       
        // Find second smallest
        else if (arr[i] != smallest
                 && arr[i] < second_smallest)
            second_smallest = arr[i];
       
        // Check if the duplicate element found or not
        if (!hm.has(arr[i]))
        {
            hm.set(arr[i], 1);
        }
       
        // If duplicate found then return false
        else
            return false;
    }
   
    // Find the difference between smallest and second
    // smallest
   
    var diff = second_smallest - smallest;
   
    // As we have used smallest and
    // second smallest,so we
    // should now only check for n-2 elements
    for (var i = 0; i < n - 1; i++) {
        if (!hm.has(second_smallest))
            return false;
        second_smallest += diff;
    }
    return true;
}
 
// Driven Program
var arr = [20, 15, 5, 0, 10 ];
var n = arr.length;
(checkIsAP(arr, n)) ? (document.write( "Yes"))
                    : (document.write( "No" ));
 
// This code is contributed by famously.
</script>
Output
Yes
Time Complexity: O(n) 
Auxiliary Space: O(n)
Thanks to Chenna Rao for suggesting this method,
Method 4(Using counting sort) 
We can reduce space required in method 3 if given array can be modified. 
  1. Find smallest and second smallest elements.
  2. Find d = second_smallest – smallest
  3. Subtract smallest element from all elements.
  4. Now if given array represent AP, all elements should be of form i*d where i varies from 0 to n-1.
  5. One by one divide all reduced elements with d. If any element is not divisible by d, return false.
  6. Now if array represents AP, it must be a permutation of numbers from 0 to n-1. We can easily check this using counting sort.
C++



// C++ program to check if a given array
// can form arithmetic progression
#include <bits/stdc++.h>
using namespace std;
 
 
// Checking if array is permutation
// of 0 to n-1 using counting sort
bool countingsort(int arr[], int n)
{
    int count[n] = { 0 };
  
    // Counting the frequency
    for (int i = 0; i < n; i++) {
        count[arr[i]]++;
    }
  
    // Check if each frequency is 1 only
    for (int i = 0; i <= n-1; i++) {
        if (count[i] != 1)
            return false;
    }
  
    return true;
}
 
// Returns true if a permutation of arr[0..n-1]
// can form arithmetic progression
bool checkIsAP(int arr[], int n)
{
    int smallest = INT_MAX, second_smallest = INT_MAX;
    for (int i = 0; i < n; i++) {
     
        // Find the smallest and
        // update second smallest
        if (arr[i] < smallest) {
            second_smallest = smallest;
            smallest = arr[i];
        }
     
        // Find second smallest
        else if (arr[i] != smallest
                && arr[i] < second_smallest)
            second_smallest = arr[i];
    }
 
    // Find the difference between smallest and second
    // smallest
    int diff = second_smallest - smallest;
 
    for (int i = 0; i < n; i++) {
        arr[i]=arr[i]-smallest;
    }
     
    for(int i=0;i<n;i++)
    {
        if(arr[i]%diff!=0)
        {
            return false;
        }
        else
        {
            arr[i]=arr[i]/diff;
        }
    }
     
    // If array represents AP, it must be a
    // permutation of numbers from 0 to n-1.
    // Check this using counting sort.
    if(countingsort(arr,n))
    return true;
    else
    return false;
}
 
// Driven Program
int main()
{
    int arr[] = { 20, 15, 5, 0, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    (checkIsAP(arr, n)) ? (cout << "Yes" << endl)
                        : (cout << "No" << endl);
 
    return 0;
 
    // This code is contributed by Pushpesh Raj
}
Java



// Java program to check if a given array
// can form arithmetic progression
import java.io.*;
 
class GFG {
 
  // Checking if array is permutation
  // of 0 to n-1 using counting sort
  static boolean countingsort(int arr[], int n)
  {
    int[] count = new int[n];
    for(int i = 0; i < n; i++)
      count[i] = 0;
 
    // Counting the frequency
    for (int i = 0; i < n; i++) {
      count[arr[i]]++;
    }
 
    // Check if each frequency is 1 only
    for (int i = 0; i <= n-1; i++) {
      if (count[i] != 1)
        return false;
    }
 
    return true;
  }
 
  // Returns true if a permutation of arr[0..n-1]
  // can form arithmetic progression
  static boolean checkIsAP(int arr[], int n)
  {
    int smallest = Integer.MAX_VALUE, second_smallest = Integer.MAX_VALUE ;
    for (int i = 0; i < n; i++) {
 
      // Find the smallest and
      // update second smallest
      if (arr[i] < smallest) {
        second_smallest = smallest;
        smallest = arr[i];
      }
 
      // Find second smallest
      else if (arr[i] != smallest
               && arr[i] < second_smallest)
        second_smallest = arr[i];
    }
 
    // Find the difference between smallest and second
    // smallest
    int diff = second_smallest - smallest;
 
    for (int i = 0; i < n; i++) {
      arr[i] = arr[i] - smallest;
    }
 
    for(int i = 0; i < n; i++)
    {
      if(arr[i] % diff != 0)
      {
        return false;
      }
      else
      {
        arr[i] = arr[i]/diff;
      }
    }
 
    // If array represents AP, it must be a
    // permutation of numbers from 0 to n-1.
    // Check this using counting sort.
    if(countingsort(arr,n))
      return true;
    else
      return false;
  }
 
  // Driven Program
  public static void main (String[] args)
  {
    int arr[] = { 20, 15, 5, 0, 10 };
    int n = arr.length;
 
    if(checkIsAP(arr, n))
      System.out.println("Yes");
    else System.out.println("No");
 
  }
}
 
// This code is contributed by Utkarsh
Python3



# Python program to check if a given array
# can form arithmetic progression
import sys
 
# Checking if array is permutation
# of 0 to n-1 using counting sort
def countingsort( arr,  n):
    count = [0]*n;
  
    # Counting the frequency
    for i in range(0, n):
        count[arr[i]] += 1;
  
    # Check if each frequency is 1 only
    for i in range(0, n - 1):
        if (count[i] != 1):
            return False;
  
    return True;
 
# Returns true if a permutation of arr[0..n-1]
# can form arithmetic progression
def checkIsAP( arr,  n):
    smallest = sys.maxsize;
    second_smallest = sys.maxsize;
    for i in range(0,n):
     
        # Find the smallest and
        # update second smallest
        if (arr[i] < smallest) :
            second_smallest = smallest;
            smallest = arr[i];
     
     
        # Find second smallest
        elif (arr[i] != smallest and arr[i] < second_smallest):
            second_smallest = arr[i];
 
    # Find the difference between smallest and second
    # smallest
    diff = second_smallest - smallest;
 
    for i in range(0,n):
        arr[i]=arr[i]-smallest;
    
     
    for i in range(0,n):
        if(arr[i]%diff!=0):
            return False;
        else:
            arr[i]=(int)(arr[i]/diff);
     
    # If array represents AP, it must be a
    # permutation of numbers from 0 to n-1.
    # Check this using counting sort.
    if(countingsort(arr,n)):
        return True;
    else:
        return False;
 
# Driven Program
arr = [ 20, 15, 5, 0, 10 ];
n = len(arr);
 
if(checkIsAP(arr, n)):
    print("Yes");
else:
    print("NO");
     
    # This code is contributed by ratiagrawal.
Javascript



// Javascript program to check if a given array
// can form arithmetic progression
 
// Checking if array is permutation
// of 0 to n-1 using counting sort
function countingsort( arr, n)
{
    let count=new Array(n).fill(0);
  
    // Counting the frequency
    for (let i = 0; i < n; i++) {
        count[arr[i]]++;
    }
  
    // Check if each frequency is 1 only
    for (let i = 0; i <= n-1; i++) {
        if (count[i] != 1)
            return false;
    }
  
    return true;
}
 
// Returns true if a permutation of arr[0..n-1]
// can form arithmetic progression
function checkIsAP( arr,  n)
{
    let smallest = Number.MAX_SAFE_INTEGER, second_smallest = Number.MAX_SAFE_INTEGER;
    for (let i = 0; i < n; i++) {
     
        // Find the smallest and
        // update second smallest
        if (arr[i] < smallest) {
            second_smallest = smallest;
            smallest = arr[i];
        }
     
        // Find second smallest
        else if (arr[i] != smallest
                && arr[i] < second_smallest)
            second_smallest = arr[i];
    }
 
    // Find the difference between smallest and second
    // smallest
    let diff = second_smallest - smallest;
 
    for (let i = 0; i < n; i++) {
        arr[i]=arr[i]-smallest;
    }
     
    for(let i=0;i<n;i++)
    {
        if(arr[i]%diff!=0)
        {
            return false;
        }
        else
        {
            arr[i]=arr[i]/diff;
        }
    }
     
    // If array represents AP, it must be a
    // permutation of numbers from 0 to n-1.
    // Check this using counting sort.
    if(countingsort(arr,n))
        return true;
    else
        return false;
}
 
// Driven Program
let arr = [20, 15, 5, 0, 10 ];
let n = arr.length;
 
(checkIsAP(arr, n)) ? (console.log("Yes\n"))
                    : (console.log("No\n"));
  
 // // This code was contributed by poojaagrawal2.
C#



using System;
 
 
    class GFG
    {
        // Checking if array is permutation
  // of 0 to n-1 using counting sort
        static bool CountingSort(int[] arr, int n)
        {
          // Counting the frequency
            int[] count = new int[n];
 
            for (int i = 0; i < n; i++)
            {
                count[arr[i]]++;
            }
 // Check if each frequency is 1 only
            for (int i = 0; i <= n - 1; i++)
            {
                if (count[i] != 1)
                {
                    return false;
                }
            }
 
            return true;
        }// Returns true if a permutation of arr[0..n-1]
  // can form arithmetic progression
 
        static bool CheckIsAP(int[] arr, int n)
        {// Find the smallest and
      // update second smallest
            int smallest = int.MaxValue;
            int secondSmallest = int.MaxValue;
 
            for (int i = 0; i < n; i++)
            {
                if (arr[i] < smallest)
                {
                    secondSmallest = smallest;
                    smallest = arr[i];
                }
                else if (arr[i] != smallest && arr[i] < secondSmallest)
                {
                    secondSmallest = arr[i];
                }
            }
 
            int diff = secondSmallest - smallest;
 
            for (int i = 0; i < n; i++)
            {
                arr[i] = arr[i] - smallest;
            }
 
            for (int i = 0; i < n; i++)
            {
                if (arr[i] % diff != 0)
                {
                    return false;
                }
                else
                {
                    arr[i] = arr[i] / diff;
                }
            }
// If array represents AP, it must be a
    // permutation of numbers from 0 to n-1.
    // Check this using counting sort.
            if (CountingSort(arr, n))
            {
                return true;
            }
            else
            {
                return false;
            }
        }
// Driven Program
        static void Main(string[] args)
        {
            int[] arr = new int[] { 20, 15, 5, 0, 10 };
            int n = arr.Length;
 
            Console.WriteLine(CheckIsAP(arr, n) ? "Yes" : "No");
        }
    }
Output
Yes
Time Complexity – O(n) 
Auxiliary Space – O(n)
Method 5( Hashing with Single Pass)
The basic idea is to find the common difference of the AP by finding out the maximum and the minimum element of the array. After that start from the maximum value and keep on decreasing the value by the common difference alongside checking that whether this new value is present in the hashmap or not . If at any point the value isn’t present in the hashset , break the loop . The ideal situation after the loop breaking is that all n elements have been covered and if yes , then return true else return false. 
C++



// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
bool checkIsAP(int arr[], int n)
{
    unordered_set<int> st;
    int maxi = INT_MIN;
    int mini = INT_MAX;
    for (int i=0;i<n;i++) {
        maxi = max(arr[i], maxi);
        mini = min(arr[i], mini);
        st.insert(arr[i]);
    }
    // FINDING THE COMMON DIFFERENCE
    int diff = (maxi - mini) / (n - 1);
    int count = 0;
 
    // CHECK IF PRESENT IN THE HASHSET OR NOT
    while (st.find(maxi)!=st.end()) {
        count++;
        maxi = maxi - diff;
    }
    if (count == n)
        return true;
 
    return false;
}
 
// Driver Code
int main()
{
    int arr[] = { 0, 12, 4, 8 };
    int n = 4;
    cout << boolalpha << checkIsAP(arr, n);
    return 0;
}
 
// This code is contributed by Rohit Pradhan
Java



/*package whatever //do not write package name here */
 
import java.io.*;
import java.util.*;
 
class GFG {
    public static void main(String[] args)
    {
        int[] arr = { 0, 12, 4, 8 };
        int n = arr.length;
        System.out.println(checkIsAP(arr, n));
    }
 
    static boolean checkIsAP(int arr[], int n)
    {
        HashSet<Integer> set = new HashSet<Integer>();
        int max = Integer.MIN_VALUE;
        int min = Integer.MAX_VALUE;
        for (int i : arr) {
            max = Math.max(i, max);
            min = Math.min(i, min);
            set.add(i);
        }
        // FINDING THE COMMON DIFFERENCE
        int diff = (max - min) / (n - 1);
        int count = 0;
 
        // CHECK IF PRESENT IN THE HASHSET OR NOT
        while (set.contains(max)) {
            count++;
            max = max - diff;
        }
        if (count == arr.length)
            return true;
 
        return false;
    }
}
Python3



import sys
 
def checkIsAP(arr, n):
 
    Set = set()
    Max = -sys.maxsize - 1
    Min = sys.maxsize
    for i in arr:
        Max = max(i, Max)
        Min = min(i, Min)
        Set.add(i)
     
    # FINDING THE COMMON DIFFERENCE
    diff = (Max - Min) // (n - 1)
    count = 0
 
    # CHECK IF PRESENT IN THE HASHSET OR NOT
    while (Max in Set):
        count += 1
        Max = Max - diff
 
    if (count == len(arr)):
        return True
 
    return False
 
# driver code
arr = [ 0, 12, 4, 8 ]
n = len(arr)
print(checkIsAP(arr, n))
 
# This code is contributed by shinjanpatra
C#



using System;
using System.Collections.Generic;
 
public class GFG
{
 
  // C# program for above approach
  static bool checkIsAP(int[] arr, int n)
  {
    HashSet<int> st = new HashSet<int>();
    int maxi = int.MinValue;
    int mini = int.MaxValue;
    for (int i = 0; i < n; i++) {
      maxi = Math.Max(arr[i], maxi);
      mini = Math.Min(arr[i], mini);
      st.Add(arr[i]);
    }
    // FINDING THE COMMON DIFFERENCE
    int diff = (maxi - mini) / (n - 1);
    int count = 0;
 
    // CHECK IF PRESENT IN THE HASHSET OR NOT
    while (st.Contains(maxi)) {
      count++;
      maxi = maxi - diff;
    }
    if (count == n) {
      return true;
    }
 
    return false;
  }
 
  // Driver Code
  internal static void Main()
  {
    int[] arr = { 0, 12, 4, 8 };
    int n = 4;
    Console.Write(checkIsAP(arr, n));
  }
 
  // This code is contributed by Aarti_Rathi
}
Javascript



<script>
 
function checkIsAP(arr, n){
 
    set = new Set()
    let Max = Number.MIN_VALUE
    let Min = Number.MAX_VALUE
    for(let i of arr){
        Max = Math.max(i, Max)
        Min = Math.min(i, Min)
        set.add(i)
    }
     
    // FINDING THE COMMON DIFFERENCE
    let diff = Math.floor((Max - Min) / (n - 1))
    let count = 0
 
    // CHECK IF PRESENT IN THE HASHSET OR NOT
    while (set.has(Max)){
        count += 1
        Max = Max - diff
    }
 
    if (count == arr.length)
        return true
 
    return false
}
 
// driver code
let arr = [ 0, 12, 4, 8 ]
let n = arr.length
document.write(checkIsAP(arr, n))
 
// This code is contributed by shinjanpatra
 
</script>
Output
true
Time Complexity – O(n) 
Auxiliary Space – O(n)
This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Last Updated :
15 Mar, 2023
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