Check if an array contains all elements of a given range

An array containing positive elements is given. ‘A’ and ‘B’ are two numbers defining a range. Write a function to check if the array contains all elements in the given range.

Examples :

Input : arr[] = {1 4 5 2 7 8 3}
           A : 2, B : 5
Output : Yes

Input : arr[] = {1 4 5 2 7 8 3}
          A : 2, B : 6
Output : No

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Method 1 : (Intuitive)
The most intuitive approach is to sort the array and check from the element greater than ‘A’ to the element greater than ‘B’. If these elements are in continuous order, all elements in the range exists in the array.

Time complexity : O(n log n)
Auxiliary space : O(1)

Method 2 : (Hashing)
We can maintain a count array or a hash table which stores the count of all numbers in the array that are in the range A…B. Then we can simply check if every number occurred at least once.

Time complexity : O(n)
Auxiliary space : O(max_element)

Method 3 : (Best)
Do a linear traversal of the array. If an element is found such that |arr[i]| >= A and |arr[i]|

C++

#include <iostream> 
using namespace std; 
   
// Function to check the array for elements in 
// given range 
bool check_elements(int arr[], int n, int A, int B) 
{ 
    // Range is the no. of elements that are 
    // to be checked 
    int range = B - A; 
   
    // Traversing the array 
    for (int i = 0; i < n; i++) { 
   
        // If an element is in range 
        if (abs(arr[i]) >= A && abs(arr[i]) <= B) { 
   
            // Negating at index ‘element – A’ 
            int z = abs(arr[i]) - A; 
            if (arr[z] > 0) { 
                arr[z] = arr[z] * -1; 
            } 
        } 
    } 
   
    // Checking whether elements in range 0-range 
    // are negative 
    int count=0; 
    for (int i = 0; i <= range && i<n; i++) { 
   
        // Element from range is missing from array 
        if (arr[i] > 0) 
            return false; 
        else
            count++; 
    } 
     if(count!= (range+1)) 
        return false; 
    // All range elements are present 
    return true; 
} 
   
// Driver code 
int main() 
{ 
    // Defining Array and size 
    int arr[] = { 1, 4, 5, 2, 7, 8, 3 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
   
    // A is lower limit and B is the upper limit 
    // of range 
    int A = 2, B = 5; 
   
    // True denotes all elements were present 
    if (check_elements(arr, n, A, B)) 
        cout << "Yes"; 
   
    // False denotes any element was not present 
    else
        cout << "No"; 
  
    return 0; 
}

Java

// JAVA Code for Check if an array contains 
// all elements of a given range 
import java.util.*; 
  
class GFG { 
      
    // Function to check the array for elements in 
    // given range 
     public static boolean check_elements(int arr[], int n, 
                                             int A, int B) 
    { 
        // Range is the no. of elements that are 
        // to be checked 
        int range = B - A; 
        
        // Traversing the array 
        for (int i = 0; i < n; i++) { 
        
            // If an element is in range 
            if (Math.abs(arr[i]) >= A &&  
                           Math.abs(arr[i]) <= B) { 
        
                  
                int z = Math.abs(arr[i]) - A; 
                if (arr[z] > 0) { 
                    arr[z] = arr[z] * -1; 
                } 
            } 
        } 
        
        // Checking whether elements in range 0-range 
        // are negative 
        int count=0; 
  
        for (int i = 0; i <= range && i<n; i++) { 
        
            // Element from range is missing from array 
            if (arr[i] > 0) 
                return false; 
            else
                count++; 
        } 
  
        if(count!= (range+1)) 
            return false; 
  
        // All range elements are present 
        return true; 
    } 
      
    /* Driver program to test above function */
    public static void main(String[] args)  
    { 
        // Defining Array and size 
        int arr[] = { 1, 4, 5, 2, 7, 8, 3 }; 
        int n = arr.length; 
        
        // A is lower limit and B is the upper limit 
        // of range 
        int A = 2, B = 5; 
        
        // True denotes all elements were present 
        if (check_elements(arr, n, A, B)) 
            System.out.println("Yes"); 
        
        // False denotes any element was not present 
        else
            System.out.println("No"); 
    } 
} 
// This code is contributed by Arnav Kr. Mandal. 

Python3

# Function to check the array for  
# elements in given range  
def check_elements(arr, n, A, B) : 
      
    # Range is the no. of elements  
    # that are to be checked  
    rangeV = B - A  
      
    # Traversing the array  
    for i in range(0, n):  
      
        # If an element is in range  
        if (abs(arr[i]) >= A and 
            abs(arr[i]) <= B) :  
      
            # Negating at index ‘element – A’  
            z = abs(arr[i]) - A  
            if (arr[z] > 0) :  
                arr[z] = arr[z] * -1
              
    # Checking whether elements in  
    # range 0-range are negative  
    count = 0
    for i in range(0, rangeV + 1):  
        if i >= n: 
            break
              
        # Element from range is 
        # missing from array 
        if (arr[i] > 0):  
            return False
        else: 
            count = count + 1
      
    if(count != (rangeV + 1)):  
        return False
          
    # All range elements are present  
    return True
  
# Driver code  
  
# Defining Array and size  
arr = [ 1, 4, 5, 2, 7, 8, 3 ]  
n = len(arr)  
      
# A is lower limit and B is  
# the upper limit of range  
A = 2
B = 5
      
# True denotes all elements  
# were present  
if (check_elements(arr, n, A, B)) : 
    print("Yes")  
      
# False denotes any element  
# was not present  
else: 
    print("No")  
  
# This code is contributed  
# by Yatin Gupta 

C#

// C# Code for Check if an array contains 
// all elements of a given range 
using System; 
  
class GFG { 
      
    // Function to check the array for  
    // elements in given range 
    public static bool check_elements(int []arr, int n, 
                                      int A, int B) 
    { 
        // Range is the no. of elements  
        // that are to be checked 
        int range = B - A; 
      
        // Traversing the array 
        for (int i = 0; i < n; i++) 
        { 
      
            // If an element is in range 
            if (Math.Abs(arr[i]) >= A &&  
                Math.Abs(arr[i]) <= B)  
                { 
                    int z = Math.Abs(arr[i]) - A; 
                    if (arr[z] > 0) 
                    { 
                      arr[z] = arr[z] * - 1; 
                    } 
                } 
        } 
      
        // Checking whether elements in  
        // range 0-range are negative 
        int count=0; 
  
        for (int i = 0; i <= range  
             && i < n; i++)  
        { 
      
            // Element from range is  
            // missing from array 
            if (arr[i] > 0) 
                return false; 
            else
                count++; 
        } 
  
        if(count != (range + 1)) 
            return false; 
  
        // All range elements are present 
        return true; 
    } 
      
    // Driver Code 
    public static void Main(String []args)  
    { 
        // Defining Array and size 
        int []arr = {1, 4, 5, 2, 7, 8, 3}; 
        int n = arr.Length; 
      
        // A is lower limit and B is  
        // the upper limit of range 
        int A = 2, B = 5; 
      
        // True denotes all elements were present 
        if (check_elements(arr, n, A, B)) 
         Console.WriteLine("Yes"); 
      
        // False denotes any element was not present 
        else
            Console.WriteLine("No"); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// Function to check the  
// array for elements in  
// given range 
function check_elements($arr, $n,  
                        $A, $B) 
{ 
    // Range is the no. of  
    // elements that are to 
    // be checked 
    $range = $B - $A; 
  
    // Traversing the array 
    for ($i = 0; $i < $n; $i++)  
    { 
  
        // If an element is in range 
        if (abs($arr[$i]) >= $A &&  
            abs($arr[$i]) <= $B)  
        { 
  
            // Negating at index 
            // ‘element – A’ 
            $z = abs($arr[$i]) - $A; 
            if ($arr[$z] > 0)  
            { 
                $arr[$z] = $arr[$z] * -1; 
            } 
        } 
    } 
  
    // Checking whether elements  
    // in range 0-range are negative 
    $count = 0; 
    for ($i = 0; $i <= $range &&  
                 $i< $n; $i++)  
    { 
  
        // Element from range is 
        // missing from array 
        if ($arr[$i] > 0) 
            return -1; 
        else
            $count++; 
    } 
    if($count!= ($range + 1)) 
        return -1; 
    // All range elements 
    // are present 
    return true; 
} 
  
// Driver code 
  
// Defining Array and size 
$arr = array(1, 4, 5, 2, 
             7, 8, 3); 
$n = sizeof($arr); 
  
// A is lower limit and 
// B is the upper limit 
// of range 
$A = 2; $B = 5; 
  
// True denotes all 
// elements were present 
if ((check_elements($arr, $n, 
                    $A, $B)) == true) 
  
    echo "Yes"; 
  
// False denotes any  
// element was not present 
else
    echo "No"; 
  
// This code is contributed by aj_36 
?> 

Output :

Yes

Time complexity : O(n)
Auxiliary space : O(1)

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My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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