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

Last Updated : 03 Mar, 2022

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:

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

Below is the implementation of the above approach:

C++

// 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;
}

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# 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

<?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

?>

JavaScript

<script>

// Javascript program to implement above approach

// Function that will find out
// the valid position
function validPosition(arr, N, K)
{
    var count = 0, sum = 0;

    // find sum of all the elements
    for (var 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 (var i = 0; i < N; i++) {
        if ((arr[i] + K) > (sum - arr[i]))
            count++;
    }

    return count;
}

// Driver code
var arr = [ 2, 1, 6, 7 ], K = 4;
var N = arr.length;
document.write( validPosition(arr, N, K));


</script>

Output:

Time Complexity : O(N)
Auxiliary Space : O(1)

Find all possible values of K such that the sum of first N numbers starting from K is G

Naman_Garg

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

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