Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Count elements whose sum with K is greater than max element

  • Last Updated : 21 May, 2021

Given an array arr[] and integer K, our task is to determine if the sum of each element in the array and K is greater than or equal to the maximum element that is present in the array that is arr[i] + k >= maxElement of array. Print the total count of all such elements. 
Examples:
 

Input : arr = [2, 3, 5, 1, 3], k = 3 
Output :
Explanations : 
In the given array the elements 2, 3, 5, 3 satisfy the condition because all of them on adding up with 3(=K) yields a value that is greater than the maximum element of the array which is 5.
Input : arr = [4, 2, 1, 1, 2], k = 1 
Output :
Explanations : 
In the given array the element 4 satisfy the condition because on adding 4 with 1(=K) we get a value that is greater than the maximum element of the array which is 4 itself. 
 

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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

 



Approach:
To solve the problem mentioned above we have to first store that maximum element that the array has. Then for every element check if the sum of the element and K gives a value greater than the maximum element then increment the count otherwise go to next element.
Below is the implementation of the above approach:
 

C++




// C++ implementation to Count of all the elements
// in the array whose summation with integer K returns
// a value that is greater than or equal to the
// maximum value present in the array
#include <bits/stdc++.h>
using namespace std;
 
// Function to count all the elements
int countNum(int arr[], int K, int n)
{
    int maxi = INT_MIN;
 
    // Store the maximum array element
    for (int i = 0; i < n; i++) {
        if (arr[i] > maxi)
            maxi = arr[i];
    }
 
    int cnt = 0;
 
    // Iterate in array
    for (int i = 0; i < n; i++) {
        // Check if current element and k gives
        // a greater sum than max element
        if (arr[i] + K >= maxi)
 
            // Increment the count
            cnt++;
        else
            continue;
    }
 
    // Return the final result
    return cnt;
}
 
// Driver code
int main()
{
    int arr[] = { 4, 2, 1, 1, 2 };
 
    int k = 1;
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << countNum(arr, k, n) << endl;
 
    return 0;
}

Java




// Java implementation to count of all the elements
// in the array whose summation with integer K returns
// a value that is greater than or equal to the
// maximum value present in the array
class GFG{
     
// Function to count all the elements
public static int countNum(int arr[], int K, int n)
{
    int maxi = 0;
 
    // Store the maximum array element
    for(int i = 0; i < n; i++)
    {
       if (arr[i] > maxi)
           maxi = arr[i];
    }
 
    int cnt = 0;
 
    // Iterate in array
    for(int i = 0; i < n; i++)
    {
         
       // Check if current element and k gives
       // a greater sum than max element
       if (arr[i] + K >= maxi)
        
           // Increment the count
           cnt++;
       else
           continue;
    }
 
    // Return the final result
    return cnt;
}
     
// Driver code   
public static void main(String[] args)
{
    int arr[] = { 4, 2, 1, 1, 2 };
    int k = 1;
    int n = arr.length;
 
    System.out.println(countNum(arr, k, n));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3




# Python3 implementation to Count of all the elements
# in the array whose summation with integer K returns
# a value that is greater than or equal to the
# maximum value present in the array
 
import sys
 
# Function to count all the elements
def countNum(arr, K, n):
     
    maxi = -sys.maxsize
     
    # Store the maximum array element
    for i in range(n) :
        if arr[i] > maxi:
            maxi = arr[i]
     
    cnt = 0
     
    # Iterate in array
    for i in range(n):
         
        # Check if current element and k gives
        # a greater sum than max element
        if (arr[i] + K) >= maxi:
             
            # Increment the count
            cnt += 1
        else :
            continue
     
    # Return the final result
    return cnt
     
# Driver code
if __name__=='__main__':
     
    arr = [ 4, 2, 1, 1, 2 ]
    k = 1
    n = len(arr)
     
    print(countNum(arr, k, n))
 
# This code is contributed by rutvik_56

C#




// C# implementation to count of all
// the elements in the array whose
// summation with integer K returns
// a value that is greater than or
// equal to the maximum value present
// in the array
using System;
 
class GFG{
     
// Function to count all the elements
public static int countNum(int[] arr, int K,
                                      int n)
{
    int maxi = 0;
 
    // Store the maximum array element
    for(int i = 0; i < n; i++)
    {
       if (arr[i] > maxi)
           maxi = arr[i];
    }
 
    int cnt = 0;
 
    // Iterate in array
    for(int i = 0; i < n; i++)
    {
        
       // Check if current element and k
       // gives a greater sum than max
       // element
       if (arr[i] + K >= maxi)
            
           // Increment the count
           cnt++;
       else
           continue;
    }
 
    // Return the final result
    return cnt;
}
     
// Driver code
public static void Main()
{
    int[] arr = { 4, 2, 1, 1, 2 };
    int k = 1;
    int n = arr.Length;
 
    Console.Write(countNum(arr, k, n));
}
}
 
// This code is contributed by chitranayal

Javascript




<script>
 
// Javascript implementation to Count of all the elements
// in the array whose summation with integer K returns
// a value that is greater than or equal to the
// maximum value present in the array
 
// Function to count all the elements
function countNum(arr, K, n)
{
    var maxi = -1000000000;
 
    // Store the maximum array element
    for (var i = 0; i < n; i++) {
        if (arr[i] > maxi)
            maxi = arr[i];
    }
 
    var cnt = 0;
 
    // Iterate in array
    for (var i = 0; i < n; i++) {
        // Check if current element and k gives
        // a greater sum than max element
        if (arr[i] + K >= maxi)
 
            // Increment the count
            cnt++;
        else
            continue;
    }
 
    // Return the final result
    return cnt;
}
 
// Driver code
var arr = [4, 2, 1, 1, 2];
var k = 1;
var n = arr.length;
document.write( countNum(arr, k, n));
 
</script>
Output: 
1

 

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




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!