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

• Difficulty Level : Easy
• Last Updated : 09 Aug, 2021

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.

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 using namespace std; // Function to Check if the minimum element in the array is// greater than or equal to half of every other elementvoid 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 codeint main(){    int arr[] = { 2, 3, 4, 5 };     int len = sizeof(arr) / sizeof(arr);     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 elementsimport java.util.*;class GFG{ // Function to Check if the minimum// element in the array is greater// than or equal to half of every// other elementsstatic 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 codepublic 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 elementimport math # Function to Check if the minimum element# in the array is greater than or equal to# half of every other elementdef 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 codeif __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 elementsusing System; class GFG{ // Function to Check if the minimum// element in the array is greater// than or equal to half of every// other elementsstatic 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 codepublic 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


Output:
No

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

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