# Check if elements of array can be arranged in AP, GP or HP

Given an array arr[] of N integers. The task is to check whether by arranging the elements of the array, is it possible to generate an Arithmetic Progression, Geometric Progression or Harmonic Progression. If possible print “Yes”, with the type of Progression or Else print “No”.

Examples:

Input: arr[] = {2, 16, 4, 8}
Output: Yes, A GP can be formed
Explanation:
Rearrange given array as {2, 4, 8, 16}, forms a Geometric Progression with common ratio 2.

Input: arr[] = {15, 10, 15, 5}
Output: Yes, An AP can be formed
Explanation:
Rearrange given array as {5, 10, 15, 20}, forms Arithmetic Progression with common difference 5.

Input: arr[] = { 1.0/10.0, 1.0/5.0, 1.0/15.0, 1.0/20.0 }
Output: Yes, A HP can be formed
Explanation:
Rearrange given array as { 1.0/5.0, 1.0/10.0, 1.0/15.0, 1.0/20.0 }, forms a Harmonic Progression.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to observe that elements in any of the three progressions A.P., G.P. or H.P. are somewhat related to sorted order. So, we need to first sort the given array.

1. For Arithmetic Progression: Check if the difference between the consecutive elements of the sorted array are same or not. If Yes then given array element forms an Arithmetic Progression.
2. For Geometric Progression: Check if the ratio of the consecutive elements of the sorted array are same or not. If Yes then given array element forms a Geometric Progression.
3. For Harmonic Progression: Check if the difference between the reciprocal of all the consecutive elements of the sorted array are same or not. If Yes then given array element forms a Harmonic Progression.

Below is the implementation of the above approach:

## C++

 `// C++ program to check if a given ` `// array form AP, GP or HP ` `#include ` `using` `namespace` `std; ` ` `  `// Returns true if arr[0..n-1] ` `// can form AP ` `bool` `checkIsAP(``double` `arr[], ``int` `n) ` `{ ` `    ``// Base Case ` `    ``if` `(n == 1) ` `        ``return` `true``; ` ` `  `    ``// Sort array ` `    ``sort(arr, arr + n); ` ` `  `    ``// After sorting, difference ` `    ``// between consecutive elements ` `    ``// must be same. ` `    ``double` `d = arr - arr; ` ` `  `    ``// Traverse the given array and ` `    ``// check if the difference ` `    ``// between ith element and (i-1)th ` `    ``// element is same or not ` `    ``for` `(``int` `i = 2; i < n; i++) { ` `        ``if` `(arr[i] - arr[i - 1] != d) { ` `            ``return` `false``; ` `        ``} ` `    ``} ` ` `  `    ``return` `true``; ` `} ` ` `  `// Returns true if arr[0..n-1] ` `// can form GP ` `bool` `checkIsGP(``double` `arr[], ``int` `n) ` `{ ` `    ``// Base Case ` `    ``if` `(n == 1) ` `        ``return` `true``; ` ` `  `    ``// Sort array ` `    ``sort(arr, arr + n); ` ` `  `    ``// After sorting, common ratio ` `    ``// between consecutive elements ` `    ``// must be same. ` `    ``double` `r = arr / arr; ` ` `  `    ``// Traverse the given array and ` `    ``// check if the common ratio ` `    ``// between ith element and (i-1)th ` `    ``// element is same or not ` `    ``for` `(``int` `i = 2; i < n; i++) { ` `        ``if` `(arr[i] / arr[i - 1] != r) ` `            ``return` `false``; ` `    ``} ` ` `  `    ``return` `true``; ` `} ` ` `  `// Returns true if arr[0..n-1] ` `// can form HP ` `bool` `checkIsHP(``double` `arr[], ``int` `n) ` `{ ` `    ``// Base Case ` `    ``if` `(n == 1) { ` `        ``return` `true``; ` `    ``} ` ` `  `    ``double` `rec[n]; ` ` `  `    ``// Find reciprocal of arr[] ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``rec[i] = ((1 / arr[i])); ` `    ``} ` ` `  `    ``// After finding reciprocal, check if ` `    ``// the reciprocal is in A. P. ` `    ``// To check for A.P. ` `    ``if` `(checkIsAP(rec, n)) ` `        ``return` `true``; ` `    ``else` `        ``return` `false``; ` `} ` ` `  `// Driver's Code ` `int` `main() ` `{ ` `    ``double` `arr[] = { 1.0 / 5.0, 1.0 / 10.0, ` `                     ``1.0 / 15.0, 1.0 / 20.0 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `flag = 0; ` ` `  `    ``// Function to check AP ` `    ``if` `(checkIsAP(arr, n)) { ` `        ``cout << ``"Yes, An AP can be formed"` `             ``<< endl; ` `        ``flag = 1; ` `    ``} ` ` `  `    ``// Function to check GP ` `    ``if` `(checkIsGP(arr, n)) { ` `        ``cout << ``"Yes, A GP can be formed"` `             ``<< endl; ` `        ``flag = 1; ` `    ``} ` ` `  `    ``// Function to check HP ` `    ``if` `(checkIsHP(arr, n)) { ` `        ``cout << ``"Yes, A HP can be formed"` `             ``<< endl; ` `        ``flag = 1; ` `    ``} ` ` `  `    ``else` `if` `(flag == 0) { ` `        ``cout << ``"No"``; ` `    ``} ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to check if a given ` `// array form AP, GP or HP ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `  `  `// Returns true if arr[0..n-1] ` `// can form AP ` `static` `boolean` `checkIsAP(``double` `arr[], ``int` `n) ` `{ ` `    ``// Base Case ` `    ``if` `(n == ``1``) ` `        ``return` `true``; ` `  `  `    ``// Sort array ` `    ``Arrays.sort(arr); ` `  `  `    ``// After sorting, difference ` `    ``// between consecutive elements ` `    ``// must be same. ` `    ``double` `d = arr[``1``] - arr[``0``]; ` `  `  `    ``// Traverse the given array and ` `    ``// check if the difference ` `    ``// between ith element and (i-1)th ` `    ``// element is same or not ` `    ``for` `(``int` `i = ``2``; i < n; i++) { ` `        ``if` `(arr[i] - arr[i - ``1``] != d) { ` `            ``return` `false``; ` `        ``} ` `    ``} ` `  `  `    ``return` `true``; ` `} ` `  `  `// Returns true if arr[0..n-1] ` `// can form GP ` `static` `boolean` `checkIsGP(``double` `arr[], ``int` `n) ` `{ ` `    ``// Base Case ` `    ``if` `(n == ``1``) ` `        ``return` `true``; ` `  `  `    ``// Sort array ` `    ``Arrays.sort(arr); ` `  `  `    ``// After sorting, common ratio ` `    ``// between consecutive elements ` `    ``// must be same. ` `    ``double` `r = arr[``1``] / arr[``0``]; ` `  `  `    ``// Traverse the given array and ` `    ``// check if the common ratio ` `    ``// between ith element and (i-1)th ` `    ``// element is same or not ` `    ``for` `(``int` `i = ``2``; i < n; i++) { ` `        ``if` `(arr[i] / arr[i - ``1``] != r) ` `            ``return` `false``; ` `    ``} ` `  `  `    ``return` `true``; ` `} ` `  `  `// Returns true if arr[0..n-1] ` `// can form HP ` `static` `boolean` `checkIsHP(``double` `arr[], ``int` `n) ` `{ ` `    ``// Base Case ` `    ``if` `(n == ``1``) { ` `        ``return` `true``; ` `    ``} ` `  `  `    ``double` `[]rec = ``new` `double``[n]; ` `  `  `    ``// Find reciprocal of arr[] ` `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `        ``rec[i] = ((``1` `/ arr[i])); ` `    ``} ` `  `  `    ``// After finding reciprocal, check if ` `    ``// the reciprocal is in A. P. ` `    ``// To check for A.P. ` `    ``if` `(checkIsAP(rec, n)) ` `        ``return` `true``; ` `    ``else` `        ``return` `false``; ` `} ` `  `  `// Driver's Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``double` `arr[] = { ``1.0` `/ ``5.0``, ``1.0` `/ ``10.0``, ` `                     ``1.0` `/ ``15.0``, ``1.0` `/ ``20.0` `}; ` `    ``int` `n = arr.length; ` `    ``int` `flag = ``0``; ` `  `  `    ``// Function to check AP ` `    ``if` `(checkIsAP(arr, n)) { ` `        ``System.out.print(``"Yes, An AP can be formed"` `             ``+``"\n"``); ` `        ``flag = ``1``; ` `    ``} ` `  `  `    ``// Function to check GP ` `    ``if` `(checkIsGP(arr, n)) { ` `        ``System.out.print(``"Yes, A GP can be formed"` `             ``+``"\n"``); ` `        ``flag = ``1``; ` `    ``} ` `  `  `    ``// Function to check HP ` `    ``if` `(checkIsHP(arr, n)) { ` `        ``System.out.print(``"Yes, A HP can be formed"` `             ``+``"\n"``); ` `        ``flag = ``1``; ` `    ``} ` `  `  `    ``else` `if` `(flag == ``0``) { ` `        ``System.out.print(``"No"``); ` `    ``} ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python3 program to check if a   ` `# given array form AP, GP or HP  ` ` `  `# Returns true if arr[0..n-1]  ` `# can form AP  ` `def` `checkIsAP(arr, n): ` ` `  `    ``# Base Case  ` `    ``if` `(n ``=``=` `1``):  ` `        ``return` `True` ` `  `    ``# Sort array  ` `    ``arr.sort(); ` ` `  `    ``# After sorting, difference  ` `    ``# between consecutive elements  ` `    ``# must be same.  ` `    ``d ``=` `arr[``1``] ``-` `arr[``0``] ` ` `  `    ``# Traverse the given array and  ` `    ``# check if the difference  ` `    ``# between ith element and (i-1)th  ` `    ``# element is same or not  ` `    ``for` `i ``in` `range``(``2``, n):  ` `        ``if` `(arr[i] ``-` `arr[i ``-` `1``] !``=` `d): ` `            ``return` `False` ` `  `    ``return` `True` ` `  `# Returns true if arr[0..n-1]  ` `# can form GP  ` `def` `checkIsGP(arr, n): ` ` `  `    ``# Base Case  ` `    ``if` `(n ``=``=` `1``):  ` `        ``return` `True` ` `  `    ``# Sort array  ` `    ``arr.sort() ` ` `  `    ``# After sorting, common ratio  ` `    ``# between consecutive elements  ` `    ``# must be same.  ` `    ``r ``=` `arr[``1``] ``/` `arr[``0``] ` ` `  `    ``# Traverse the given array and  ` `    ``# check if the common ratio  ` `    ``# between ith element and (i-1)th  ` `    ``# element is same or not  ` `    ``for` `i ``in` `range``(``2``, n): ` `        ``if` `(arr[i] ``/` `arr[i ``-` `1``] !``=` `r): ` `            ``return` `False` ` `  `    ``return` `True` ` `  `# Returns true if arr[0..n-1]  ` `# can form HP  ` `def` `checkIsHP(arr, n): ` ` `  `    ``# Base Case  ` `    ``if` `(n ``=``=` `1``): ` `        ``return` `True` ` `  `    ``rec ``=` `[] ` ` `  `    ``# Find reciprocal of arr[]  ` `    ``for` `i ``in` `range``(``0``, n):  ` `        ``rec.append((``1` `/` `arr[i])) ` ` `  `    ``# After finding reciprocal, check  ` `    ``# if the reciprocal is in A. P.  ` `    ``# To check for A.P.  ` `    ``if` `(checkIsAP(rec, n)): ` `        ``return` `True` `    ``else``: ` `        ``return` `False` ` `  `# Driver Code  ` `arr ``=` `[ ``1.0` `/` `5.0``, ``1.0` `/` `10.0``,  ` `        ``1.0` `/` `15.0``, ``1.0` `/` `20.0` `] ` `n ``=` `len``(arr) ` `flag ``=` `0` ` `  `# Function to check AP  ` `if` `(checkIsAP(arr, n)): ` `    ``print``(``"Yes, An AP can be formed"``, end ``=` `'\n'``) ` `    ``flag ``=` `1` ` `  `# Function to check GP  ` `if` `(checkIsGP(arr, n)): ` `    ``print``(``"Yes, A GP can be formed"``, end ``=` `'\n'``) ` `    ``flag ``=` `1` `     `  `# Function to check HP  ` `if` `(checkIsHP(arr, n)): ` `    ``print``(``"Yes, A HP can be formed"``, end ``=` `'\n'``) ` `    ``flag ``=` `1` ` `  `elif` `(flag ``=``=` `0``): ` `    ``print``(``"No"``, end ``=` `'\n'``)  ` `     `  `# This code is contributed by Pratik `

## C#

 `// C# program to check if a given ` `// array form AP, GP or HP ` `using` `System; ` ` `  `class` `GFG{ ` `   `  `// Returns true if arr[0..n-1] ` `// can form AP ` `static` `bool` `checkIsAP(``double` `[]arr, ``int` `n) ` `{ ` `    ``// Base Case ` `    ``if` `(n == 1) ` `        ``return` `true``; ` `   `  `    ``// Sort array ` `    ``Array.Sort(arr); ` `   `  `    ``// After sorting, difference ` `    ``// between consecutive elements ` `    ``// must be same. ` `    ``double` `d = arr - arr; ` `   `  `    ``// Traverse the given array and ` `    ``// check if the difference ` `    ``// between ith element and (i-1)th ` `    ``// element is same or not ` `    ``for` `(``int` `i = 2; i < n; i++) { ` `        ``if` `(arr[i] - arr[i - 1] != d) { ` `            ``return` `false``; ` `        ``} ` `    ``} ` `   `  `    ``return` `true``; ` `} ` `   `  `// Returns true if arr[0..n-1] ` `// can form GP ` `static` `bool` `checkIsGP(``double` `[]arr, ``int` `n) ` `{ ` `    ``// Base Case ` `    ``if` `(n == 1) ` `        ``return` `true``; ` `   `  `    ``// Sort array ` `    ``Array.Sort(arr); ` `   `  `    ``// After sorting, common ratio ` `    ``// between consecutive elements ` `    ``// must be same. ` `    ``double` `r = arr / arr; ` `   `  `    ``// Traverse the given array and ` `    ``// check if the common ratio ` `    ``// between ith element and (i-1)th ` `    ``// element is same or not ` `    ``for` `(``int` `i = 2; i < n; i++) { ` `        ``if` `(arr[i] / arr[i - 1] != r) ` `            ``return` `false``; ` `    ``} ` `   `  `    ``return` `true``; ` `} ` `   `  `// Returns true if arr[0..n-1] ` `// can form HP ` `static` `bool` `checkIsHP(``double` `[]arr, ``int` `n) ` `{ ` `    ``// Base Case ` `    ``if` `(n == 1) { ` `        ``return` `true``; ` `    ``} ` `   `  `    ``double` `[]rec = ``new` `double``[n]; ` `   `  `    ``// Find reciprocal of []arr ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``rec[i] = ((1 / arr[i])); ` `    ``} ` `   `  `    ``// After finding reciprocal, check if ` `    ``// the reciprocal is in A. P. ` `    ``// To check for A.P. ` `    ``if` `(checkIsAP(rec, n)) ` `        ``return` `true``; ` `    ``else` `        ``return` `false``; ` `} ` `   `  `// Driver's Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``double` `[]arr = { 1.0 / 5.0, 1.0 / 10.0, ` `                     ``1.0 / 15.0, 1.0 / 20.0 }; ` `    ``int` `n = arr.Length; ` `    ``int` `flag = 0; ` `   `  `    ``// Function to check AP ` `    ``if` `(checkIsAP(arr, n)) { ` `        ``Console.Write(``"Yes, An AP can be formed"` `             ``+``"\n"``); ` `        ``flag = 1; ` `    ``} ` `   `  `    ``// Function to check GP ` `    ``if` `(checkIsGP(arr, n)) { ` `        ``Console.Write(``"Yes, A GP can be formed"` `             ``+``"\n"``); ` `        ``flag = 1; ` `    ``} ` `   `  `    ``// Function to check HP ` `    ``if` `(checkIsHP(arr, n)) { ` `        ``Console.Write(``"Yes, A HP can be formed"` `             ``+``"\n"``); ` `        ``flag = 1; ` `    ``} ` `   `  `    ``else` `if` `(flag == 0) { ` `        ``Console.Write(``"No"``); ` `    ``} ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```Yes, A HP can be formed
```

Time Complexity: O(N*log N)

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Improved By : princiraj1992, PratikBasu