作者：大帅哥明日夜 | 来源：互联网 | 2023-10-11 21:02
检查数组中的最小元素是否小于或等于每隔一个元素的一半
原文:https://www . geesforgeks . org/check-如果数组中的最小元素小于或等于其他元素的一半/
给定一个数组 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:给定数组中的最小数总是唯一的。
示例:
输入: arr = {2，1，4，5}
输出:是
解释:
1 是 arr[]数组中的最小元素，2，4，5 除以 2 得到 1，2，2.5，大于或等于最小数。因此，打印“是”。
输入: arr = {2，4，5，3}
输出:否
解释:
2 是 arr[]数组中的最小元素，4，5，3 除以 2 得到 2，2.5，1.5，其中整数 3 不返回大于或等于最小数的值(1.5 <2)。因此，打印“否”。
方法 1:
要解决上面提到的问题，我们必须借助循环找到最小元素，然后再次扫描整个数组，检查两次最小元素是否小于或等于每隔一个元素。但是这个解决方案使用两个循环花费了 O(N)个时间，并且可以在只涉及一次迭代的情况下进一步优化。
方法 2:
为了优化上述解，我们可以在单次迭代中找到最小的以及第二小的元素本身。然后简单地检查最小元素的两倍是否小于或等于第二小元素。
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 <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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<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:
时间复杂度: O(N)，其中 N 为给定数组的长度。
辅助空间: O(N)。
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