Dada una array arr[] y el entero K , nuestra tarea es determinar si la suma de cada elemento de la array y K es mayor o igual que el elemento máximo que está presente en la array que es arr[i] + k > = maxElement de la array . Imprime el recuento total de todos esos elementos.
Ejemplos:
Entrada : arr = [2, 3, 5, 1, 3], k = 3
Salida : 4
Explicaciones :
En el arreglo dado los elementos 2, 3, 5, 3 cumplen la condición porque todos ellos al sumar 3( =K) produce un valor mayor que el elemento máximo de la array, que es 5.
Entrada: arr = [4, 2, 1, 1, 2], k = 1
Salida: 1
Explicaciones:
en la array dada, el elemento 4 cumplen la condición porque al sumar 4 con 1(=K) obtenemos un valor mayor que el elemento máximo del arreglo que es el mismo 4.
Enfoque:
para resolver el problema mencionado anteriormente, primero debemos almacenar ese elemento máximo que tiene la array. Luego, para cada elemento, verifique si la suma del elemento y K da un valor mayor que el elemento máximo, luego incremente el conteo; de lo contrario, vaya al siguiente elemento.
A continuación se muestra la implementación del enfoque anterior:
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>
1
Complejidad temporal: O(n)
Espacio auxiliar: O(1)