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📜  元素加K大于所有其他元素之和的位置数

📅  最后修改于: 2021-04-26 19:36:11             🧑  作者: Mango

给定一个数组arr []和一个数字K。任务是找出有效位置i的数量，以使(arr [i] + K)大于数组中除arr [i]之外的所有元素的总和。
例子： 

Input: arr[] = {2, 1, 6, 7} K = 4
Output: 1
Explanation: There is only 1 valid position i.e 4th. 
After adding 4 to the element at 4th position 
it is greater than the sum of all other 
elements of the array.

Input: arr[] = {2, 1, 5, 4} K = 2
Output: 0
Explanation: There is no valid position.

方法：

  1. 首先，找到数组所有元素的总和，并将其存储在变量中，例如sum 。
  2. 现在，遍历阵列和用于每个位置i检查是否满足条件(ARR [I] + K)>(总和- ARR [I])是否成立。
  3. 如果是，则增加计数器，最后打印计数器的值。

下面是上述方法的实现：

C++
// C++ program to implement above approach
 
#include 
using namespace std;
 
// Function that will find out
// the valid position
int validPosition(int arr[], int N, int K)
{
    int count = 0, sum = 0;
 
    // find sum of all the elements
    for (int i = 0; i < N; i++) {
        sum += arr[i];
    }
 
    // adding K to the element and check
    // whether it is greater than sum of
    // all other elements
    for (int i = 0; i < N; i++) {
        if ((arr[i] + K) > (sum - arr[i]))
            count++;
    }
 
    return count;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 1, 6, 7 }, K = 4;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << validPosition(arr, N, K);
 
    return 0;
}

Java
// Java implementation of the approach
class GFG
{
 
// Function that will find out
// the valid position
static int validPosition(int arr[], int N, int K)
{
    int count = 0, sum = 0;
 
    // find sum of all the elements
    for (int i = 0; i < N; i++)
    {
        sum += arr[i];
    }
 
    // adding K to the element and check
    // whether it is greater than sum of
    // all other elements
    for (int i = 0; i < N; i++)
    {
        if ((arr[i] + K) > (sum - arr[i]))
            count++;
    }
 
    return count;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 1, 6, 7 }, K = 4;
    int N = arr.length;
    System.out.println(validPosition(arr, N, K));
}
}
 
/* This code contributed by PrinciRaj1992 */

Python3
# Python3 program to implement
# above approach
 
# Function that will find out
# the valid position
def validPosition(arr, N, K):
    count = 0; sum = 0;
 
    # find sum of all the elements
    for i in range(N):
        sum += arr[i];
 
    # adding K to the element and check
    # whether it is greater than sum of
    # all other elements
    for i in range(N):
        if ((arr[i] + K) > (sum - arr[i])):
            count += 1;
 
    return count;
 
# Driver code
arr = [2, 1, 6, 7 ];
K = 4;
N = len(arr);
 
print(validPosition(arr, N, K));
 
# This code is contributed by 29AjayKumar

C#
// C# implementation of the approach
using System;
     
class GFG
{
  
// Function that will find out
// the valid position
static int validPosition(int []arr, int N, int K)
{
    int count = 0, sum = 0;
  
    // find sum of all the elements
    for (int i = 0; i < N; i++)
    {
        sum += arr[i];
    }
  
    // adding K to the element and check
    // whether it is greater than sum of
    // all other elements
    for (int i = 0; i < N; i++)
    {
        if ((arr[i] + K) > (sum - arr[i]))
            count++;
    }
  
    return count;
}
  
// Driver code
public static void Main(String[] args)
{
    int []arr = { 2, 1, 6, 7 };int K = 4;
    int N = arr.Length;
    Console.WriteLine(validPosition(arr, N, K));
}
}
 
// This code has been contributed by 29AjayKumar

PHP
 ($sum - $arr[$i]))
            $count++;
    }
 
    return $count;
}
 
    // Driver code
    $arr = array( 2, 1, 6, 7 );
    $K = 4;
    $N = count($arr) ;
 
    echo validPosition($arr, $N, $K);
     
    // This code is contributed by AnkitRai01
 
?>

Javascript

输出： 
1