计算与 K 之和大于最大元素的元素

给定一个数组 arr[]和整数 K ，我们的任务是确定数组中每个元素与 K 的总和是否大于或等于数组中存在的最大元素arr[i] + k > = 数组的最大元素。打印所有这些元素的总数。
例子：

Input : arr = [2, 3, 5, 1, 3], k = 3 
Output : 4 
Explanations : 
In the given array the elements 2, 3, 5, 3 satisfy the condition because all of them on adding up with 3(=K) yields a value that is greater than the maximum element of the array which is 5.
Input : arr = [4, 2, 1, 1, 2], k = 1 
Output : 1 
Explanations : 
In the given array the element 4 satisfy the condition because on adding 4 with 1(=K) we get a value that is greater than the maximum element of the array which is 4 itself. 
 

方法：
为了解决上面提到的问题，我们必须首先存储数组具有的最大元素。然后对于每个元素检查元素和 K 的总和是否给出大于最大元素的值，然后增加计数，否则转到下一个元素。
下面是上述方法的实现：

C++

// C++ implementation to Count of all the elements
// in the array whose summation with integer K returns
// a value that is greater than or equal to the
// maximum value present in the array
#include 
using namespace std;
 
// Function to count all the elements
int countNum(int arr[], int K, int n)
{
    int maxi = INT_MIN;
 
    // Store the maximum array element
    for (int i = 0; i < n; i++) {
        if (arr[i] > maxi)
            maxi = arr[i];
    }
 
    int cnt = 0;
 
    // Iterate in array
    for (int i = 0; i < n; i++) {
        // Check if current element and k gives
        // a greater sum than max element
        if (arr[i] + K >= maxi)
 
            // Increment the count
            cnt++;
        else
            continue;
    }
 
    // Return the final result
    return cnt;
}
 
// Driver code
int main()
{
    int arr[] = { 4, 2, 1, 1, 2 };
 
    int k = 1;
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << countNum(arr, k, n) << endl;
 
    return 0;
}

Java

// Java implementation to count of all the elements
// in the array whose summation with integer K returns
// a value that is greater than or equal to the
// maximum value present in the array
class GFG{
     
// Function to count all the elements
public static int countNum(int arr[], int K, int n)
{
    int maxi = 0;
 
    // Store the maximum array element
    for(int i = 0; i < n; i++)
    {
       if (arr[i] > maxi)
           maxi = arr[i];
    }
 
    int cnt = 0;
 
    // Iterate in array
    for(int i = 0; i < n; i++)
    {
         
       // Check if current element and k gives
       // a greater sum than max element
       if (arr[i] + K >= maxi)
        
           // Increment the count
           cnt++;
       else
           continue;
    }
 
    // Return the final result
    return cnt;
}
     
// Driver code   
public static void main(String[] args)
{
    int arr[] = { 4, 2, 1, 1, 2 };
    int k = 1;
    int n = arr.length;
 
    System.out.println(countNum(arr, k, n));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 implementation to Count of all the elements
# in the array whose summation with integer K returns
# a value that is greater than or equal to the
# maximum value present in the array
 
import sys
 
# Function to count all the elements
def countNum(arr, K, n):
     
    maxi = -sys.maxsize
     
    # Store the maximum array element
    for i in range(n) :
        if arr[i] > maxi:
            maxi = arr[i]
     
    cnt = 0
     
    # Iterate in array
    for i in range(n):
         
        # Check if current element and k gives
        # a greater sum than max element
        if (arr[i] + K) >= maxi:
             
            # Increment the count
            cnt += 1
        else :
            continue
     
    # Return the final result
    return cnt
     
# Driver code
if __name__=='__main__':
     
    arr = [ 4, 2, 1, 1, 2 ]
    k = 1
    n = len(arr)
     
    print(countNum(arr, k, n))
 
# This code is contributed by rutvik_56

C#

// C# implementation to count of all
// the elements in the array whose
// summation with integer K returns
// a value that is greater than or
// equal to the maximum value present
// in the array
using System;
 
class GFG{
     
// Function to count all the elements
public static int countNum(int[] arr, int K,
                                      int n)
{
    int maxi = 0;
 
    // Store the maximum array element
    for(int i = 0; i < n; i++)
    {
       if (arr[i] > maxi)
           maxi = arr[i];
    }
 
    int cnt = 0;
 
    // Iterate in array
    for(int i = 0; i < n; i++)
    {
        
       // Check if current element and k
       // gives a greater sum than max
       // element
       if (arr[i] + K >= maxi)
            
           // Increment the count
           cnt++;
       else
           continue;
    }
 
    // Return the final result
    return cnt;
}
     
// Driver code
public static void Main()
{
    int[] arr = { 4, 2, 1, 1, 2 };
    int k = 1;
    int n = arr.Length;
 
    Console.Write(countNum(arr, k, n));
}
}
 
// This code is contributed by chitranayal

Javascript

 
// Javascript implementation to Count of all the elements
// in the array whose summation with integer K returns
// a value that is greater than or equal to the
// maximum value present in the array
 
// Function to count all the elements
function countNum(arr, K, n)
{
    var maxi = -1000000000;
 
    // Store the maximum array element
    for (var i = 0; i < n; i++) {
        if (arr[i] > maxi)
            maxi = arr[i];
    }
 
    var cnt = 0;
 
    // Iterate in array
    for (var i = 0; i < n; i++) {
        // Check if current element and k gives
        // a greater sum than max element
        if (arr[i] + K >= maxi)
 
            // Increment the count
            cnt++;
        else
            continue;
    }
 
    // Return the final result
    return cnt;
}
 
// Driver code
var arr = [4, 2, 1, 1, 2];
var k = 1;
var n = arr.length;
document.write( countNum(arr, k, n));
 

输出：

1

时间复杂度： O(n)
辅助空间： O(1)

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