Check if an array contains only one distinct element

Last Updated : 17 Aug, 2021

Given an array arr[] of size N, the task is to check if the array contains only one distinct element or not. If it contains only one distinct element then print “Yes”, otherwise print “No”.

Examples:

Input: arr[] = {3, 3, 4, 3, 3}
Output: No
Explanation:
There are 2 distinct elements present in the array {3, 4}.
Therefore, the output is No.

Input: arr[] = {9, 9, 9, 9, 9, 9, 9}
Output: Yes
Explanation:
The only distinct element in the array is 9.
Therefore, the output is Yes.

Naive Approach: The idea is to sort the given array and then for each valid index check if the current element and the next element are the same or not. If they are not the same it means the array contains more than one distinct element, therefore print “No”, otherwise print “Yes”.

Time Complexity: O(N*logN)
Auxiliary Space: O(1)

Better Approach: This problem can be solved by using a set data structure. Since in set, no repetitions are allowed. Below are the steps:

Insert elements of the array into the set.
If there is only one distinct element then the size of the set after step 1 will be 1, so print "Yes".
Otherwise, print “No”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach #include <bits/stdc++.h>  using namespace std;   // Function to find if the array  // contains only one distinct element  void uniqueElement(int arr[],int n) {          // Create a set     unordered_set<int> set;          // Traversing the array     for(int i = 0; i < n; i++)     {         set.insert(arr[i]);     }          // Compare and print the result     if(set.size() == 1)     {         cout << "YES" << endl;     }     else     {         cout << "NO" << endl;     } }   // Driver code int main() {     int arr[] = { 9, 9, 9, 9, 9, 9, 9 };          int n = sizeof(arr) / sizeof(arr[0]);          // Function call     uniqueElement(arr,n);     return 0; }   // This code is contributed by rutvik_56

Java

// Java program for the above approach import java.util.*;  public class Main {      // Function to find if the array     // contains only one distinct element     public static void     uniqueElement(int arr[])     {         // Create a set         Set<Integer> set = new HashSet<>();          // Traversing the array         for (int i = 0; i < arr.length; i++) {             set.add(arr[i]);         }          // Compare and print the result         if (set.size() == 1)             System.out.println("Yes");          else             System.out.println("No");     }      // Driver Code     public static void main(String args[])     {         int arr[] = { 9, 9, 9, 9, 9, 9, 9 };          // Function call         uniqueElement(arr);     } }

Python3

# Python3 program for the above approach  # Function to find if the array  # contains only one distinct element  def uniqueElement(arr, n):          # Create a set     s = set(arr)          # Compare and print the result     if(len(s) == 1):         print('YES')     else:         print('NO')  # Driver code if __name__=='__main__':          arr = [ 9, 9, 9, 9, 9, 9, 9 ]     n = len(arr)          # Function call     uniqueElement(arr, n)      # This code is contributed by rutvik_56

// C# program for the above approach using System; using System.Collections.Generic;  class GFG{  // Function to find if the array // contains only one distinct element public static void uniqueElement(int []arr) {          // Create a set     HashSet<int> set = new HashSet<int>();      // Traversing the array     for(int i = 0; i < arr.Length; i++)      {         set.Add(arr[i]);     }      // Compare and print the result     if (set.Count == 1)         Console.WriteLine("Yes");     else         Console.WriteLine("No"); }  // Driver Code public static void Main(String []args) {     int []arr = { 9, 9, 9, 9, 9, 9, 9 };      // Function call     uniqueElement(arr); } }  // This code is contributed by Amit Katiyar

JavaScript

<script>  // Javascript program for the above approach  // Function to find if the array  // contains only one distinct element  function uniqueElement(arr,n) {          // Create a set     var set = new Set();          // Traversing the array     for(var i = 0; i < n; i++)     {         set.add(arr[i]);     }          // Compare and print the result     if(set.size == 1)     {         document.write( "YES");     }     else     {         document.write( "NO");     } }   // Driver code var arr = [9, 9, 9, 9, 9, 9, 9];  var n = arr.length;  // Function call uniqueElement(arr,n);  // This code is contributed by itsok. </script>

Output:

Yes

Time Complexity: O(N)
Auxiliary Space: O(N)

Efficient Approach: This problem can also be solved without using any extra space. Below are the steps:

Assume the first element of the array to be the only unique element in the array and store its value in a variable say X.
Then traverse the array and check if the current element is equal to X or not.
If found to be true, then keep checking for all array elements. If no element is found to be different from X, print "Yes".
Otherwise, if any of the array elements is not equal to X, it means that the array contains more than one unique element. Hence, print "No".

Below is the implementation of the above approach:

C++

// C++ program for  // the above approach #include<bits/stdc++.h> using namespace std;   // Function to find if the array // contains only one distinct element void uniqueElement(int arr[], int n) {   // Assume first element to   // be the unique element   int x = arr[0];    int flag = 1;    // Traversing the array   for (int i = 0; i < n; i++)    {      // If current element is not     // equal to X then break the     // loop and print No     if (arr[i] != x)      {       flag = 0;       break;     }   }    // Compare and print the result   if (flag == 1)     cout << "Yes";   else     cout << "No"; }      // Driver Code int main() {   int arr[] = {9, 9, 9,                 9, 9, 9, 9};   int n = sizeof(arr) /            sizeof(arr[0]);    // Function call   uniqueElement(arr, n); }  // This code is contributed by Chitranayal

Java

// Java program for the above approach  import java.util.*;  public class Main {      // Function to find if the array     // contains only one distinct element     public static void     uniqueElement(int arr[])     {         // Assume first element to         // be the unique element         int x = arr[0];          int flag = 1;          // Traversing the array         for (int i = 0; i < arr.length; i++) {              // If current element is not             // equal to X then break the             // loop and print No             if (arr[i] != x) {                 flag = 0;                 break;             }         }          // Compare and print the result         if (flag == 1)             System.out.println("Yes");          else             System.out.println("No");     }      // Driver Code     public static void main(String args[])     {         int arr[] = { 9, 9, 9, 9, 9, 9, 9 };          // Function call         uniqueElement(arr);     } }

Python3

# Python3 program for the above approach   # Function to find if the array  # contains only one distinct element def uniqueElement(arr):      # Assume first element to     # be the unique element     x = arr[0]      flag = 1      # Traversing the array     for i in range(len(arr)):          # If current element is not         # equal to X then break the         # loop and print No         if(arr[i] != x):             flag = 0             break      # Compare and print the result     if(flag == 1):         print("Yes")     else:         print("No")  # Driver Code  # Given array arr[] arr = [ 9, 9, 9, 9, 9, 9, 9 ]  # Function call uniqueElement(arr)  # This code is contributed by Shivam Singh

// C# program for the above approach  using System;  class GFG{  // Function to find if the array // contains only one distinct element public static void uniqueElement(int []arr) {          // Assume first element to     // be the unique element     int x = arr[0];      int flag = 1;      // Traversing the array     for(int i = 0; i < arr.Length; i++)     {                  // If current element is not         // equal to X then break the         // loop and print No         if (arr[i] != x)         {             flag = 0;             break;         }     }      // Compare and print the result     if (flag == 1)         Console.WriteLine("Yes");     else         Console.WriteLine("No"); }  // Driver code static public void Main () {     int []arr = { 9, 9, 9, 9, 9, 9, 9 };      // Function call     uniqueElement(arr); } }  // This code is contributed by AnkitRai01

JavaScript

<script>     // javascript program for the above approach      // Function to find if the array // contains only one distinct element  function uniqueElement(arr) {           // Assume first element to     // be the unique element     var x = arr[0];       var flag = 1;       // Traversing the array     for(var i = 0; i < arr.length; i++)     {                   // If current element is not         // equal to X then break the         // loop and print No         if (arr[i] != x)         {             flag = 0;             break;         }     }       // Compare and print the result     if (flag == 1)         document.write("Yes");     else         document.write("No"); }   // Driver code      var arr = [ 9, 9, 9, 9, 9, 9, 9 ];       // Function call     uniqueElement(arr);     </script>

Output:

Yes

Time Complexity: O(N)
Auxiliary Space: O(1)

Check for Array Contains Duplicate III

jyoti369

Improve

Article Tags :

Practice Tags :

Check if an array contains only one distinct element

Similar Reads