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Convert given array to Arithmetic Progression by adding an element
• Difficulty Level : Easy
• Last Updated : 09 Jan, 2019

Given an array arr[], the task is to find an element that can be added to the array in order to convert it to Arithmetic Progression. If it’s impossible to convert the given array into an AP then print -1.

Examples:

Input: arr[] = {3, 7}
Output: 11
3, 7 and 11 is a finite AP sequence.

Input: a[] = {4, 6, 8, 15}
Output: -1

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

Approach:

• Sort the array and start traversing the array element by element and note the difference between the two consecutive elements.
• If the difference for all the elements is same then print last element + common difference.
• If the difference is different for at most one pair (arr[i – 1], arr[i]) and diff = 2 * common difference for all other elements then print arr[i] – common difference.
• Else print -1.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach`` ` `#include``using` `namespace` `std;`` ` `// Function to return the number to be``// added``int` `getNumToAdd(``int` `arr[], ``int` `n)``{``    ``sort(arr,arr+n);``    ``int` `d = arr[1] - arr[0];``    ``int` `numToAdd = -1;``    ``bool` `numAdded = ``false``;`` ` `    ``for` `(``int` `i = 2; i < n; i++) {``        ``int` `diff = arr[i] - arr[i - 1];`` ` `        ``// If difference of the current ``        ``// consecutive elements is``        ``// different from the common``        ``// difference``        ``if` `(diff != d) {`` ` `            ``// If number has already been``            ``// chosen then it's not possible ``            ``// to add another number``            ``if` `(numAdded)``                ``return` `-1;`` ` `            ``// If the current different is``            ``// twice the common difference ``            ``// then a number can be added midway``            ``// from current and previous element``            ``if` `(diff == 2 * d) {``                ``numToAdd = arr[i] - d;`` ` `                ``// Number has been chosen``                ``numAdded = ``true``;``            ``}`` ` `            ``// It's not possible to maintain``            ``// the common difference``            ``else``                ``return` `-1;``        ``}``    ``}`` ` `    ``// Return last element + common difference``    ``// if no element is chosen and the array``    ``// is already in AP``    ``if` `(numToAdd == -1)``        ``return` `(arr[n - 1] + d);`` ` `    ``// Else return the chosen number``    ``return` `numToAdd;``}`` ` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 3, 5, 7, 11, 13, 15 };``    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);``    ``cout << getNumToAdd(arr, n);``}`` ` ` ` `// This code is contributed``// by ihritik`

## Java

 `// Java implementation of the approach``import` `java.util.*;``public` `class` `GFG {`` ` `    ``// Function to return the number to be``    ``// added``    ``static` `int` `getNumToAdd(``int` `arr[], ``int` `n)``    ``{``        ``Arrays.sort(arr);``        ``int` `d = arr[``1``] - arr[``0``];``        ``int` `numToAdd = -``1``;``        ``boolean` `numAdded = ``false``;`` ` `        ``for` `(``int` `i = ``2``; i < n; i++) {``            ``int` `diff = arr[i] - arr[i - ``1``];`` ` `            ``// If difference of the current ``            ``// consecutive elements is``            ``// different from the common``            ``// difference``            ``if` `(diff != d) {`` ` `                ``// If number has already been``                ``// chosen then it's not possible ``                ``// to add another number``                ``if` `(numAdded)``                    ``return` `-``1``;`` ` `                ``// If the current different is``                ``// twice the common difference ``                ``// then a number can be added midway``                ``// from current and previous element``                ``if` `(diff == ``2` `* d) {``                    ``numToAdd = arr[i] - d;`` ` `                    ``// Number has been chosen``                    ``numAdded = ``true``;``                ``}`` ` `                ``// It's not possible to maintain``                ``// the common difference``                ``else``                    ``return` `-``1``;``            ``}``        ``}`` ` `        ``// Return last element + common difference``        ``// if no element is chosen and the array``        ``// is already in AP``        ``if` `(numToAdd == -``1``)``            ``return` `(arr[n - ``1``] + d);`` ` `        ``// Else return the chosen number``        ``return` `numToAdd;``    ``}`` ` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``1``, ``3``, ``5``, ``7``, ``11``, ``13``, ``15` `};``        ``int` `n = arr.length;``        ``System.out.println(getNumToAdd(arr, n));``    ``}``}`

## Python3

 `# Python 3 implementation of the approach`` ` `# Function to return the number ``# to be added``def` `getNumToAdd(arr, n):``    ``arr.sort(reverse ``=` `False``)``    ``d ``=` `arr[``1``] ``-` `arr[``0``]``    ``numToAdd ``=` `-``1``    ``numAdded ``=` `False`` ` `    ``for` `i ``in` `range``(``2``, n, ``1``):``        ``diff ``=` `arr[i] ``-` `arr[i ``-` `1``]`` ` `        ``# If difference of the current consecutive ``        ``# elements is different from the common``        ``# difference``        ``if` `(diff !``=` `d):``             ` `            ``# If number has already been chosen ``            ``# then it's not possible to add ``            ``# another number``            ``if` `(numAdded):``                ``return` `-``1`` ` `            ``# If the current different is twice ``            ``# the common difference then a ``            ``# number can be added midway from ``            ``# current and previous element``            ``if` `(diff ``=``=` `2` `*` `d):``                ``numToAdd ``=` `arr[i] ``-` `d`` ` `                ``# Number has been chosen``                ``numAdded ``=` `True``         ` `            ``# It's not possible to maintain``            ``# the common difference``            ``else``:``                ``return` `-``1``     ` `    ``# Return last element + common difference``    ``# if no element is chosen and the array``    ``# is already in AP``    ``if` `(numToAdd ``=``=` `-``1``):``        ``return` `(arr[n ``-` `1``] ``+` `d)`` ` `    ``# Else return the chosen number``    ``return` `numToAdd`` ` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``1``, ``3``, ``5``, ``7``, ``11``, ``13``, ``15``]``    ``n ``=` `len``(arr)``    ``print``(getNumToAdd(arr, n))`` ` `# This code is contributed``# mohit kumar 29`

## C#

 `// C# implementation of the approach`` ` `using` `System;``public` `class` `GFG {`` ` `    ``// Function to return the number to be``    ``// added``    ``static` `int` `getNumToAdd(``int` `[]arr, ``int` `n)``    ``{``        ``Array.Sort(arr);``        ``int` `d = arr[1] - arr[0];``        ``int` `numToAdd = -1;``        ``bool` `numAdded = ``false``;`` ` `        ``for` `(``int` `i = 2; i < n; i++) {``            ``int` `diff = arr[i] - arr[i - 1];`` ` `            ``// If difference of the current ``            ``// consecutive elements is``            ``// different from the common``            ``// difference``            ``if` `(diff != d) {`` ` `                ``// If number has already been``                ``// chosen then it's not possible ``                ``// to add another number``                ``if` `(numAdded)``                    ``return` `-1;`` ` `                ``// If the current different is``                ``// twice the common difference ``                ``// then a number can be added midway``                ``// from current and previous element``                ``if` `(diff == 2 * d) {``                    ``numToAdd = arr[i] - d;`` ` `                    ``// Number has been chosen``                    ``numAdded = ``true``;``                ``}`` ` `                ``// It's not possible to maintain``                ``// the common difference``                ``else``                    ``return` `-1;``            ``}``        ``}`` ` `        ``// Return last element + common difference``        ``// if no element is chosen and the array``        ``// is already in AP``        ``if` `(numToAdd == -1)``            ``return` `(arr[n - 1] + d);`` ` `        ``// Else return the chosen number``        ``return` `numToAdd;``    ``}`` ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = { 1, 3, 5, 7, 11, 13, 15 };``        ``int` `n = arr.Length;``        ``Console.WriteLine(getNumToAdd(arr, n));``    ``}``}`` ` `// This code is contributed``// by ihritik`

## PHP

 ``
Output:
```9
```

Time Complexity : O(n Log n)

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