Get the kth smallest number using the digits of the given number

Check if an array contains all elements of a given range

Last Updated : 14 Sep, 2023

Try it on GfG Practice

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

Method 1 : (Intuitive)

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

Algorithm

1. Check if A > B. If yes, return false as it is an invalid range.
2. Loop through each integer i in the range [A, B] (inclusive):
    a. Initialize a boolean variable found to false.
    b. Loop through each element j in the array arr:
        i. If j is equal to i, set found to true and break out of the inner loop.
    c. If found is still false after looping through all elements of arr, return false as i is not in the array.
3. If the loop completes without returning false, return true as all elements in the range [A, B] are found in arr.

Implementation of above approach

C++

#include <iostream>
#include <algorithm> // for std::sort
 
using namespace std;
  
 
bool check_elements_in_range(int arr[], int n, int A, int B) {
    if (A > B) {
        return false; // invalid range
    }
 
    for (int i = A; i <= B; i++) {
        bool found = false;
        for (int j = 0; j < n; j++) {
            if (arr[j] == i) {
                found = true;
                break;
            }
        }
        if (!found) {
            return false; // element not found in array
        }
    }
    return true; // all elements in range found in array
}
 
int main() {
    int arr[] = {1, 4, 5, 2, 7, 8, 3};
    int n = sizeof(arr) / sizeof(arr[0]);
    int A = 2, B = 5;
    if (check_elements_in_range(arr, n, A, B)) {
        cout << "Yes" << endl;
    } else {
        cout << "No" << endl;
    }
    return 0;
}

Java

public class Main {
    public static boolean checkElementsInRange(int[] arr, int A, int B) {
        if (A > B) {
            return false; // invalid range
        }
 
        for (int i = A; i <= B; i++) {
            boolean found = false;
            for (int j = 0; j < arr.length; j++) {
                if (arr[j] == i) {
                    found = true;
                    break;
                }
            }
            if (!found) {
                return false; // element not found in array
            }
        }
        return true; // all elements in range found in array
    }
 
    public static void main(String[] args) {
        int[] arr = {1, 4, 5, 2, 7, 8, 3};
        int A = 2, B = 5;
        if (checkElementsInRange(arr, A, B)) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
}

Python

def check_elements_in_range(arr, n, A, B):
    if A > B:
        return False  # invalid range
     
    for i in range(A, B+1):
        found = False
        for j in range(n):
            if arr[j] == i:
                found = True
                break
        if not found:
            return False  # element not found in array
     
    return True  # all elements in range found in array
 
 
arr = [1, 4, 5, 2, 7, 8, 3]
n = len(arr)
A, B = 2, 5
if check_elements_in_range(arr, n, A, B):
    print("Yes")
else:
    print("No")

C#

using System;
 
namespace RangeCheckApp
{
    class Program
    {
        static bool CheckElementsInRange(int[] arr, int A, int B)
        {
            if (A > B)
            {
                return false; // invalid range
            }
 
            for (int i = A; i <= B; i++)
            {
                bool found = false;
                foreach (int num in arr)
                {
                    if (num == i)
                    {
                        found = true;
                        break;
                    }
                }
                if (!found)
                {
                    return false; // element not found in array
                }
            }
            return true; // all elements in range found in array
        }
 
        static void Main(string[] args)
        {
            int[] arr = { 1, 4, 5, 2, 7, 8, 3 };
            int A = 2, B = 5;
            if (CheckElementsInRange(arr, A, B))
            {
                Console.WriteLine("Yes");
            }
            else
            {
                Console.WriteLine("No");
            }
        }
    }
}

Javascript

function checkElementsInRange(arr, A, B) {
    if (A > B) {
        return false; // invalid range
    }
 
    for (let i = A; i <= B; i++) {
        let found = false;
        for (let j = 0; j < arr.length; j++) {
            if (arr[j] === i) {
                found = true;
                break;
            }
        }
        if (!found) {
            return false; // element not found in array
        }
    }
    return true; // all elements in range found in array
}
 
const arr = [1, 4, 5, 2, 7, 8, 3];
const A = 2, B = 5;
if (checkElementsInRange(arr, A, B)) {
    console.log("Yes");
} else {
    console.log("No");
}

Output

Yes

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

Method 2 : (Hashing)

We can maintain a count array or a hash table that 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.

Algorithm:

Initialize an empty unordered set.
Insert all the elements of the array into the set.
Traverse all the elements between A and B, inclusive, and check if each element is present in the set or not.
If any element is not present in the set, return false.
If all the elements are present in the set, return true.

C++

// C++ code for the following approach
 
#include <bits/stdc++.h>
using namespace std;
 
// function that check all elements between A and B including
// them are present in set or not
bool check_elements(int arr[] , int n , int A, int B){
  unordered_set<int>st;
   
  // put all the elements of array into set
  for(int i=0 ;i<n ;i++){
      st.insert(arr[i]);
  }
   
  // now check every between between A to B including them also, that they are 
  // present in set or not
  for(int i=A ;i<=B ; i++){
      // element not present in set so return false
      // and no need to traverse further
      if(st.find(i) == st.end()){
          return false;
      }
  }
   
  // all elements between A and B including them are 
  // present in set so return true
  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;
}
 
// this code is contributed by bhardwajji

Java

// Java code for the following approach
 
import java.util.*;
 
class Main {
 
    // function that check all elements between A and B
    // including them are present in set or not
    public static boolean checkElements(int arr[], int n,
                                        int A, int B)
    {
        Set<Integer> st = new HashSet<Integer>();
 
        // put all the elements of array into set
        for (int i = 0; i < n; i++) {
            st.add(arr[i]);
        }
 
        // now check every between between A to B including
        // them also, that they are present in set or not
        for (int i = A; i <= B; i++) {
            // element not present in set so return false
            // and no need to traverse further
            if (!st.contains(i)) {
                return false;
            }
        }
 
        // all elements between A and B including them are
        // present in set so return true
        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 (checkElements(arr, n, A, B))
            System.out.println("Yes");
        // False denotes any element was not present
        else
            System.out.println("No");
    }
}

Python3

# Python code for the following approach
import collections
 
# function that check all elements between A and B
# including them are present in set or not
def checkElements(arr, n, A, B):
    st = set()
 
    # put all the elements of array into set
    for i in range(n):
        st.add(arr[i])
 
    # now check every between between A to B including
    # them also, that they are present in set or not
    for i in range(A, B+1):
        # element not present in set so return false
        # and no need to traverse further
        if i not in st:
            return False
 
    # all elements between A and B including them are
    # present in set so return true
    return True
 
# Driver code
if __name__ == "__main__":
   
    # 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 checkElements(arr, n, A, B):
        print("Yes")
         
    # False denotes any element was not present
    else:
        print("No")

C#

using System;
using System.Collections.Generic;
 
class MainClass {
    // function that check all elements between A and B
    // including them are present in set or not
    static bool CheckElements(int[] arr, int n, int A,
                              int B)
    {
        HashSet<int> set = new HashSet<int>();
 
        // put all the elements of array into set
        for (int i = 0; i < n; i++) {
            set.Add(arr[i]);
        }
 
        // now check every between between A to B including
        // them also, that they are present in set or not
        for (int i = A; i <= B; i++) {
            // element not present in set so return false
            // and no need to traverse further
            if (!set.Contains(i)) {
                return false;
            }
        }
 
        // all elements between A and B including them are
        // present in set so return true
        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 (CheckElements(arr, n, A, B))
            Console.WriteLine("Yes");
        // False denotes any element was not present
        else
            Console.WriteLine("No");
    }
}

Javascript

// JavaScript code for the following approach
 
// function that check all elements between A and B
// including them are present in set or not
function checkElements(arr, n, A, B) {
    let st = new Set();
 
    // put all the elements of array into set
    for (let i = 0; i < n; i++) {
        st.add(arr[i]);
    }
 
    // now check every between between A to B including
    // them also, that they are present in set or not
    for (let i = A; i <= B; i++) {
        // element not present in set so return false
        // and no need to traverse further
        if (!st.has(i)) {
            return false;
        }
    }
 
    // all elements between A and B including them are
    // present in set so return true
    return true;
}
 
// Defining Array and size
let arr = [1, 4, 5, 2, 7, 8, 3];
let n = arr.length;
 
// A is lower limit and B is the upper limit of range
let A = 2;
let B = 5;
 
// True denotes all elements were present
if (checkElements(arr,n,A,B)) {
    console.log("Yes");
}
// False denotes any element was not present 
else {
    console.log("No");
}

Output

Yes

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

Method 3 : (Best):

Every element(x) in the range (A to B) has a corresponding unique index (x-A)

Do a linear traversal of the array. If an element is found such that in the given range, i.e., |arr[i]| >= A and |arr[i]| <=B
Negate the element at index (arr[i]-A) corresponding to this element arr[i].(do this only the element at that index is positive)
Now, count the number of number of elements which are negative .This count must be equal to B-A+1.
As, an element at an index is negative indicates that element corresponding to that index is present in array.

Implementation:

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)
{
    //Array should contain atleast B-A+1 elements
    if(n<B-A+1) return false;
    // 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;
}
 
// This code is contributed by Sania Kumari Gupta

C

#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
 
// 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))
        printf("Yes");
    // False denotes any element was not present
    else
        printf("No");
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta

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.

Javascript

<script>
 
    // Javascript Code for Check
    // if an array contains
    // all elements of a given range
     
    // 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
        let range = B - A;
       
        // Traversing the array
        for (let i = 0; i < n; i++)
        {
       
            // If an element is in range
            if (Math.abs(arr[i]) >= A && 
                Math.abs(arr[i]) <= B) 
                {
                    let z = Math.abs(arr[i]) - A;
                    if (arr[z] > 0)
                    {
                      arr[z] = arr[z] * - 1;
                    }
                }
        }
       
        // Checking whether elements in 
        // range 0-range are negative
        let count=0;
   
        for (let 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;
    }
     
    // Defining Array and size
    let arr = [1, 4, 5, 2, 7, 8, 3];
    let n = arr.length;
 
    // A is lower limit and B is 
    // the upper limit of range
    let A = 2, B = 5;
 
    // True denotes all elements were present
    if (check_elements(arr, n, A, B))
      document.write("Yes");
 
    // False denotes any element was not present
    else
      document.write("No");
     
</script>

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)

Get the kth smallest number using the digits of the given number

GeeksforGeeks

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