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Number of positions such that adding K to the element is greater than sum of all other elements
  • Last Updated : 15 Apr, 2021

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: 
 

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#




// 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: 
1

 

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