Number of positions such that adding K to the element is greater than sum of all other elements

Given an array arr[] and a number K. The task is to find out the number of valid positions i such that (arr[i] + K) is greater than sum of all elements of array excluding arr[i].

Examples:

Input: arr[] = {2, 1, 6, 7} K = 4
Output: 1
Explaination: 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
Explaination: There is no valid position.


Approach:

  1. First of all find the sum of all the elements of the array and store it in a variable say sum.
  2. Now, traverse the array and for every position i check if the condition (arr[i] + K) > (sum – arr[i]) holds.
  3. If YES then increase the counter and finally print the value of counter.

Below is the implementation of the above approach:

C++

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// C++ program to implement above approach
  
#include <bits/stdc++.h>
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;
}

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Java

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// 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 */

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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#

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// 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

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PHP

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<?php
// PHP program to implement above approach 
  
// Function that will find out 
// the valid position 
function validPosition($arr, $N, $K
    $count = 0; $sum = 0; 
  
    // find sum of all the elements 
    for ($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 ($i = 0; $i < $N; $i++)
    
        if (($arr[$i] + $K) > ($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
  
?>

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Output:

1


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