Check whether Arithmetic Progression can be formed from the given array

Difficulty Level : Medium
Last Updated : 01 Aug, 2021

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 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).

Method 3(Use Hashing)

Find out the smallest and second smallest elements
Find different between the two elements. d = second_smallest – smallest
Store all elements in a hashmap and return “NO” if duplicate element found (can be done together with step 1).
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.
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 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
}

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 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]
       
        # 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

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 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.

Find smallest and second smallest elements.
Find d = second_smallest – smallest
Subtract smallest element from all elements.
Now if given array represent AP, all elements should be of form i*d where i varies from 0 to n-1.
One by one divide all reduced elements with d. If any element is not divisible by d, return false.
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.

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.

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;
    }
}

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.
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