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
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.

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:

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 */
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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
  
?>
chevron_right

Output:
1

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




Recommended Posts:


Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Article Tags :
Practice Tags :