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:
- 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 。
- 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。
- 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