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 : 4
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 : 1
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.
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 |
1
Time Complexity: O(n)
Auxiliary Space: O(1)
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