# Find whether it is possible to make array elements same using one external number

Given an Array, three operations can be performed using any external number x.

1. Add x to an element once
2. Subtract x from an element once
3. Perform no operation on the element
1. Count of unique elements is 1. Answer is YES with x = 0
2. Count of unique elements is 2. Answer is YES with x = Difference of two unique elements.
3. Count of unique elements is 3.
• If difference between mid and max is same as difference between mid and min, answer is YES with x = difference between mid and max or mid and min.

In Python, we can quickly find unique elements using set in Python.

## C++

 `// C++ Program to find an element X``// that can be used to operate on an array and``// get equal elements` `#include``using` `namespace` `std;` `// Prints "YES" and an element x if we can``// equalize array using x. Else prints "NO"``void` `canEqualise(``int` `array[], ``int` `n)``{``    ``// We all the unique elements (using set``    ``// function).``    ``set<``int``> s;``    ``for``(``int` `i=0;i

## Java

 `// Java Program to find an element X``// that can be used to operate on an array and``// get equal elements` `// Importing generic java libraries``import` `java.util.*;` `public` `class` `GFG {` `    ``// Prints "YES" and an element x if we can``    ``// equalize array using x. Else prints "NO"``    ``static` `void` `canEqualise(``int` `array[], ``int` `n)``    ``{``        ``// We all the unique elements (using set``        ``// function).``        ``Set s = ``new` `HashSet();``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``s.add(array[i]);``        ``}` `        ``// if there are only 1 or 2 unique elements,``        ``// then we can add or subtract x from one of them``        ``// to get the other element``        ``if` `(s.size() == ``1``)``            ``System.out.println(``"YES 0"``);``        ``else` `if` `(s.size() == ``2``) {``            ``int` `x = s.stream().findFirst().get();``            ``s.remove(x);``            ``int` `y = s.stream().findFirst().get();``            ``s.remove(y);``            ``System.out.println(``"YES "` `+ (y - x));``        ``}` `        ``// If count of unique elements is three, then``        ``// difference between the middle and minimum``        ``// should be same as difference between maximum``        ``// and middle``        ``else` `if` `(s.size() == ``3``) {``            ``int` `x = s.stream().findFirst().get();``            ``s.remove(x);``            ``int` `y = s.stream().findFirst().get();``            ``s.remove(y);``            ``int` `z = s.stream().findFirst().get();``            ``s.remove(z);` `            ``if` `((z - y) == (y - x))``                ``System.out.println(``"YES "` `+ (z - y));``            ``else``                ``System.out.println(``"NO"``);``        ``}` `        ``// if there are more than three unique elements,``        ``// then we cannot add or subtract the same value``        ``// from all the elements.``        ``else``            ``System.out.println(``"NO"``);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `array[] = { ``55``, ``52``, ``52``, ``49``, ``52` `};``        ``int` `n = array.length;``        ``canEqualise(array, n);``    ``}``}` `// This code is contributed by Aarti_Rathi`

## Python

 `# Program in python 2.x to find an element X``# that can be used to operate on an array and``# get equal elements``  ` `# Prints "YES" and an element x if we can``# equalize array using x. Else prints "NO"``def` `canEqualise(array):``  ` `    ``# We all the unique elements (using set``    ``# function). Then we sort unique elements.``    ``uniques ``=` `sorted``(``set``(array))``  ` `    ``# if there are only 1 or 2 unique elements,``    ``# then we can add or subtract x from one of them``    ``# to get the other element``    ``if` `len``(uniques) ``=``=` `1``:``        ``print``(``"YES "` `+` `"0"``)``    ``elif` `len``(uniques) ``=``=` `2``:``        ``print``(``"YES "` `+` `str``(uniques[``1``] ``-` `uniques[``0``]))``  ` `    ``# If count of unique elements is three, then``    ``# difference between the middle and minimum``    ``# should be same as difference between maximum``    ``# and middle``    ``elif` `len``(uniques) ``=``=` `3``:``        ``if` `uniques[``2``] ``-` `uniques[``1``] ``=``=` `uniques[``1``] ``-` `uniques[``0``]:``            ``X ``=` `uniques[``2``] ``-` `uniques[``1``]``            ``print``(``"YES "` `+` `str``(X))``        ``else``:``            ``print``(``"NO"``)``  ` `    ``# if there are more than three unique elements, then``    ``# we cannot add or subtract the same value from all``    ``# the elements.``    ``else``:``        ``print``(``"NO"``)``  ` `# Driver code``array ``=` `[``55``, ``52``, ``52``, ``49``, ``52``]``canEqualise(array)`

## C#

 `using` `System;``using` `System.Collections.Generic;` `public` `static` `class` `GFG {``  ``// C# Program to find an element X``  ``// that can be used to operate on an array and``  ``// get equal elements` `  ``// Prints "YES" and an element x if we can``  ``// equalize array using x. Else prints "NO"``  ``public` `static` `void` `canEqualise(``int``[] array, ``int` `n)``  ``{``    ``// We all the unique elements (using set``    ``// function).``    ``HashSet<``int``> s = ``new` `HashSet<``int``>();``    ``for` `(``int` `i = 0; i < n; i++) {``      ``s.Add(array[i]);``    ``}` `    ``// if there are only 1 or 2 unique elements,``    ``// then we can add or subtract x from one of them``    ``// to get the other element``    ``if` `(s.Count == 1) {``      ``Console.Write(``"YES "``);``      ``Console.Write(``"0"``);``    ``}``    ``else` `if` `(s.Count == 2) {``      ``int` `x = 0;``      ``int` `y = 0;``      ``int` `m = 0;``      ``Console.Write(``"YES "``);``      ``foreach``(``var` `i ``in` `s)``      ``{``        ``if` `(m == 0) {``          ``x = i;``        ``}``        ``else` `{``          ``y = i;``        ``}``        ``m++;``      ``}``      ``Console.Write((y - x));``    ``}` `    ``// If count of unique elements is three, then``    ``// difference between the middle and minimum``    ``// should be same as difference between maximum``    ``// and middle``    ``else` `if` `(s.Count == 3) {``      ``int` `x = 0;``      ``int` `y = 0;``      ``int` `z = 0;``      ``int` `m = 0;``      ``foreach``(``var` `i ``in` `s)``      ``{``        ``if` `(m == 0) {``          ``x = i;``        ``}``        ``else` `if` `(m == 1) {``          ``y = i;``        ``}``        ``else` `{``          ``z = i;``        ``}``        ``m++;``      ``}` `      ``if` `((z - y) == (y - x)) {``        ``Console.Write(``"YES "``);``        ``Console.Write((z - y));``      ``}``      ``else` `{``        ``Console.Write(``"NO"``);``      ``}``    ``}` `    ``// if there are more than three unique elements,``    ``// then we cannot add or subtract the same value``    ``// from all the elements.``    ``else` `{``      ``Console.Write(``"NO"``);``    ``}``  ``}` `  ``// Driver code``  ``public` `static` `void` `Main()``  ``{``    ``int``[] array = { 55, 52, 52, 49, 52 };``    ``int` `n = array.Length;``    ``canEqualise(array, n);``  ``}``}``  ``// This code is contributed by Aarti_Rathi`

## Javascript

 `// JavaScript program to find an element X``// that can be used to operate on an array and``// get equal elements``function` `canEqualise(array, n) {``    ``// We find all the unique elements (using Set function)``    ``let s = ``new` `Set(array);`  `    ``// If there are only 1 or 2 unique elements,``    ``// then we can add or subtract x from one of them``    ``// to get the other element``    ``if` `(s.size === 1)``        ``console.log(``"YES 0"``);``    ``else` `if` `(s.size === 2) {``        ``let x = s.values().next().value;``        ``s.``delete``(x);``        ``let y = s.values().next().value;``        ``s.``delete``(y);``        ``console.log(`YES \${y-x}`);``    ``}` `    ``// If count of unique elements is three, then``    ``// difference between the middle and minimum``    ``// should be same as difference between maximum``    ``// and middle``    ``else` `if` `(s.size === 3) {``        ``let x = s.values().next().value;``        ``s.``delete``(x);``        ``let y = s.values().next().value;``        ``s.``delete``(y);``        ``let z = s.values().next().value;``        ``s.``delete``(z);` `        ``if` `((z - y) === (y - x))``            ``console.log(`YES \${z-y}`);``        ``else``            ``console.log(``"NO"``);``    ``}` `    ``// If there are more than three unique elements, then``    ``// we cannot add or subtract the same value from all``    ``// the elements.``    ``else``        ``console.log(``"NO"``);``}` `// Driver code``let array = [55, 52, 52, 49, 52];``let n = array.length;``canEqualise(array, n);`

OUTPUT

`YES 3`

Time Complexity: O(n logn)//we are running a for loop for n times and inside the loop the set insert operation is taking place which takes log n time to compute
Auxiliary space: O(n). //since we are using a set to store all the elements of the array

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