Check if minimum element in array is less than or equals half of every other element

Given an array arr[], the task is to check if the minimum element in the array is less than or equal to half of every other element. If it is then print “yes” otherwise print “no”.
Note: The minimum number in the given array is always unique.

Examples:

Input: arr = {2, 1, 4, 5} 
Output: Yes 
Explanation:
1 is the minimum element in the array arr[] and on dividing 2, 4, 5 by 2 we get 1, 2, 2.5 which is greater than or equal to the minimum number. Hence, print “yes”.

Input : arr = {2, 4, 5, 3} 
Output : No 
Explanation:
2 is the minimum element in the array arr[] and on dividing 4, 5, 3 by 2 we get 2, 2.5, 1.5 in which the integer 3 does not return a value which is greater than or equal to the minimum number ( 1.5 < 2). Hence, print “no”. 

Method 1:
To solve the problem mentioned above we have to find the smallest element with the help of loops and then scan through the entire array again and check if twice the smallest element is smaller than or equal to every other element. But this solution takes O(N) time using two loops and can be optimized further where only one iteration is involved.

Below is the implementation of the above approach:

C++

// C++ implementation to Check if the minimum element in the
// array is greater than or equal to half of every other
// elements
#include <bits/stdc++.h>
using namespace std;

// Function to Check if the minimum element in the array is
// greater than or equal to half of every other element
bool checkMin(int arr[], int len)
{

    // Initialise the variables to store the index of
    // smallest element
    int smallest = 0;

    for (int i = 0; i < len; i++) {

        // Check if current element is smaller than smallest
        if (arr[i] < arr[smallest]) {
            smallest = i;
        }
    }

    for (int i = 0; i < len; i++) {

        // Check if (current element / 2) is smaller than
        // smallest
        if (i != smallest && arr[i] / 2 < arr[smallest]) {
            return false;
        }
    }

    return true;
}

// Driver code
int main()
{
    int arr[] = { 2, 30, 4, 5 };

    int len = sizeof(arr) / sizeof(arr[0]);

    if (checkMin(arr, len)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
}

Java

/*package whatever //do not write package name here */

import java.io.*;

class GFG {
    public static boolean checkMin(int arr[], int len)
{

    // Initialise the variables to store the index of
    // smallest element
    int smallest = 0;

    for (int i = 0; i < len; i++) {

        // Check if current element is smaller than smallest
        if (arr[i] < arr[smallest]) {
            smallest = i;
        }
    }

    for (int i = 0; i < len; i++) {

        // Check if (current element / 2) is smaller than
        // smallest
        if (i != smallest && arr[i] / 2 < arr[smallest]) {
            return false;
          }
    }

    return true;
}

    public static void main (String[] args) {
    int arr[] = { 2, 30, 4, 5 };

    int len = arr.length;

    if (checkMin(arr, len)) {
        System.out.println("Yes");
    }
    else {
        System.out.println("No");
    }
    }
}

Python3

class GFG :
    @staticmethod
    def  checkMin( arr,  len) :

        # Initialise the variables to store the index of
        # smallest element
        smallest = 0
        i = 0
        while (i < len) :

            # Check if current element is smaller than smallest
            if (arr[i] < arr[smallest]) :
                smallest = i
            i += 1
        i = 0
        while (i < len) :

            # Check if (current element / 2) is smaller than
            # smallest
            if (i != smallest and int(arr[i] / 2) < arr[smallest]) :
                return False
            i += 1
        return True
    @staticmethod
    def main( args) :
        arr = [2, 30, 4, 5]
        len1 = len(arr)
        if (GFG.checkMin(arr, len1)) :
            print("Yes")
        else :
            print("No")

if __name__=="__main__":
    GFG.main([])

    # This code is contributed by aadityaburujwale.

C#

// C# implementation to Check if the minimum element in the
// array is greater than or equal to half of every other
// elements
using System;
class GFG {

  // Function to Check if the minimum element in the array
  // is greater than or equal to half of every other
  // element
  static bool checkMin(int[] arr, int len)
  {

    // Initialise the variables to store the index of
    // smallest element
    int smallest = 0;

    for (int i = 0; i < len; i++) {

      // Check if current element is smaller than
      // smallest
      if (arr[i] < arr[smallest]) {
        smallest = i;
      }
    }

    for (int i = 0; i < len; i++) {

      // Check if (current element / 2) is smaller
      // than smallest
      if (i != smallest
          && arr[i] / 2 < arr[smallest]) {
        return false;
      }
    }

    return true;
  }

  static void Main()
  {
    int[] arr = { 2, 30, 4, 5 };

    int len = arr.Length;

    if (checkMin(arr, len) == true) {
      Console.Write("Yes");
    }
    else {
      Console.Write("No");
    }
  }
}

// This code is contributed by garg28harsh.

Javascript

// Javascript implementation to Check if the minimum element
// in the array is greater than or equal to half of every
// other elements

    // Function to Check if the minimum element in the array
    // is greater than or equal to half of every other
    // element
    function checkMin(arr, len)
    {

    // Initialise the variables to store the index of
    // smallest element
    let smallest = 0;

    for (let i = 0; i < len; i++) {

        // Check if current element is smaller than smallest
        if (arr[i] < arr[smallest]) {
            smallest = i;
        }
    }

    for (let i = 0; i < len; i++) {

        // Check if (current element / 2) is smaller than
        // smallest
        if (i != smallest && arr[i] / 2 < arr[smallest]) {
            return false;
        }
    }

    return true;
}

// Driver code

let arr = [ 2, 30, 4, 5 ];

let len = arr.length;

if (checkMin(arr, len) == true) {
    console.log("Yes");
}
else {
    console.log("No");
}

// This code is contributed by garg28harsh.

Output

Yes

Time Complexity: O(N), where N is the length of the given array.
Auxiliary Space: O(1).

Method 2:
To optimize the above solution we can find the smallest as well as the second smallest element in a single iteration itself. Then simply check if twice of the smallest element is smaller than or equal to the second smallest element.

Below is the implementation of the above approach:  

C++

// C++ implementation to Check if the minimum element in the
// array is greater than or equal to half of every other elements
#include <bits/stdc++.h>
using namespace std;

// Function to Check if the minimum element in the array is
// greater than or equal to half of every other element
void checkMin(int arr[], int len)
{

    // Initialise the variables to store
    // smallest and second smallest
    int smallest = INT_MAX, secondSmallest = INT_MAX;

    for (int i = 0; i < len; i++) {

        // Check if current element is smaller than smallest,
        // the current smallest will become secondSmallest
        // and current element will be the new smallest
        if (arr[i] < smallest) {
            secondSmallest = smallest;
            smallest = arr[i];
        }

        // Check if current element is smaller than
        // secondSmallest simply update the latter
        else if (arr[i] < secondSmallest) {
            secondSmallest = arr[i];
        }
    }

    if (2 * smallest <= secondSmallest)
        cout << "Yes";
    else
        cout << "No";
}

// Driver code
int main()
{
    int arr[] = { 2, 3, 4, 5 };

    int len = sizeof(arr) / sizeof(arr[0]);

    checkMin(arr, len);
}

Java

// Java implementation to check
// if the minimum element in the
// array is greater than or equal
// to half of every other elements
import java.util.*;
class GFG{

// Function to Check if the minimum
// element in the array is greater
// than or equal to half of every
// other elements
static void checkMin(int arr[], int len)
{

    // Initialise the variables to store
    // smallest and second smallest
    int smallest = Integer.MAX_VALUE;
    int secondSmallest = Integer.MAX_VALUE;

    for(int i = 0; i < len; i++)
    {

       // Check if current element is smaller than
       // smallest, the current smallest will
       // become secondSmallest and current
       // element will be the new smallest
       if (arr[i] < smallest)
       {
           secondSmallest = smallest;
           smallest = arr[i];
       }

       // Check if current element is smaller than
       // secondSmallest simply update the latter
       else if (arr[i] < secondSmallest)
       {
           secondSmallest = arr[i];
       }
    }
    if (2 * smallest <= secondSmallest)
        System.out.print("Yes");
    else
        System.out.print("No");
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 3, 4, 5 };
    int len = arr.length;

    checkMin(arr, len);
}
}

// This code is contributed by amal kumar choubey

Python3

# Python3 implementation to Check if
# the minimum element in the array
# is greater than or equal to half
# of every other element
import math

# Function to Check if the minimum element
# in the array is greater than or equal to
# half of every other element
def checkMin(arr, n):

    # Initialise the variables to store
    # smallest and second smallest
    smallest = math.inf
    secondSmallest = math.inf

    for i in range(n):

        # Check if current element is
        # smaller than smallest,
        # the current smallest will become
        # secondSmallest and current element
        # will be the new smallest
        if(arr[i] < smallest):
            secondSmallest = smallest
            smallest = arr[i]

        # Check if current element is smaller than
        # secondSmallest simply update the latter
        elif(arr[i] < secondSmallest):
            secondSmallest = arr[i]

    if(2 * smallest <= secondSmallest):
        print("Yes")
    else:
        print("No")

# Driver code
if __name__ == '__main__':
    arr = [ 2, 3, 4, 5 ]

    n = len(arr)

    checkMin(arr, n)

# This code is contributed by Shivam Singh.

C#

// C# implementation to check
// if the minimum element in the
// array is greater than or equal
// to half of every other elements
using System;

class GFG{

// Function to Check if the minimum
// element in the array is greater
// than or equal to half of every
// other elements
static void checkMin(int []arr, int len)
{

    // Initialise the variables to store
    // smallest and second smallest
    int smallest = int.MaxValue;
    int secondSmallest = int.MaxValue;

    for(int i = 0; i < len; i++)
    {

       // Check if current element is smaller than
       // smallest, the current smallest will
       // become secondSmallest and current
       // element will be the new smallest
       if (arr[i] < smallest)
       {
           secondSmallest = smallest;
           smallest = arr[i];
       }

       // Check if current element is smaller than
       // secondSmallest simply update the latter
       else if (arr[i] < secondSmallest)
       {
           secondSmallest = arr[i];
       }
    }
    if (2 * smallest <= secondSmallest)
        Console.Write("Yes");
    else
        Console.Write("No");
}

// Driver code
public static void Main(String[] args)
{
    int []arr = { 2, 3, 4, 5 };
    int len = arr.Length;

    checkMin(arr, len);
}
}

// This code is contributed by amal kumar choubey

Javascript

<script>

// Javascript implementation to check
// if the minimum element in the
// array is greater than or equal
// to half of every other elements

// Function to Check if the minimum
// element in the array is greater
// than or equal to half of every
// other element
function checkMin(arr, len)
{

    // Initialise the variables to store
    // smallest and second smallest
    var smallest = Number.INFINITY,
        secondSmallest = Number.INFINITY;

    for(var i = 0; i < len; i++)
    {

        // Check if current element is
        // smaller than smallest, the
        // current smallest will become
        // secondSmallest and current
        // element will be the new smallest
        if (arr[i] < smallest)
        {
            secondSmallest = smallest;
            smallest = arr[i];
        }

        // Check if current element is smaller than
        // secondSmallest simply update the latter
        else if (arr[i] < secondSmallest)
        {
            secondSmallest = arr[i];
        }
    }

    if (2 * smallest <= secondSmallest)
        document.write("Yes");
    else
        document.write("No");
}

// Driver code
var arr = [ 2, 3, 4, 5 ];
var len = 4;

checkMin(arr, len);

// This code is contributed by akshitsaxenaa09

</script>

Output:

No

Time Complexity: O(N), where N is the length of the given array.
Auxiliary Space: O(1).