检查数组中的最小元素是否小于或等于其他所有元素的一半

给定数组arr [] ，任务是检查数组中的最小元素是否小于或等于其他元素的一半。如果是，则打印“是”，否则打印“否”。
注意：给定数组中的最小数目始终是唯一的。

例子：

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”. 
 

方法1：
为了解决上述问题，我们必须借助循环找到最小的元素，然后再次扫描整个数组，并检查最小元素是否小于或等于其他所有元素的两倍。但是，该解决方案使用两个循环需要O(N)时间，并且在仅涉及一次迭代的情况下可以进一步优化。

方法2：
为了优化上述解决方案，我们可以在一次迭代中找到最小和第二小的元素。然后，只需检查最小元素的两倍是否小于或等于第二个最小元素。

下面是上述方法的实现：

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 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

 
// 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
 

输出：

No

时间复杂度： O(n)

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