Dada una array ordenada (con entradas únicas), tenemos que encontrar si existe un elemento (por ejemplo, X) que sea exactamente la mitad de la suma de todos los elementos de la array, incluido X.
Ejemplos:
Input : A = {1, 2, 3}
Output : YES
Sum of all the elements is 6 = 3*2;
Input : A = {2, 4}
Output : NO
Sum of all the elements is 6, and 3 is not present in the array.
1. Calcula la suma de todos los elementos del arreglo.
2. Puede haber dos casos
….a. La suma es impar, implica que no podemos encontrar tal X, ya que todas las entradas son enteras.
….b. La suma es par, si la mitad del valor de la suma existe en la array, la respuesta es SÍ, de lo contrario, NO.
3. Podemos usar la búsqueda binaria para encontrar si sum/2 existe en la array o no (ya que no tiene entradas duplicadas)
A continuación se muestra la implementación del enfoque anterior:
C++
// CPP program to check if array has an
// element whose value is half of array
// sum.
#include <bits/stdc++.h>
using namespace std;
// Function to check if answer exists
bool checkForElement(int array[], int n)
{
// Sum of all array elements
int sum = 0;
for (int i = 0; i < n; i++)
sum += array[i];
// If sum is odd
if (sum % 2)
return false;
sum /= 2; // If sum is Even
// Do binary search for the required element
int start = 0;
int end = n - 1;
while (start <= end)
{
int mid = start + (end - start) / 2;
if (array[mid] == sum)
return true;
else if (array[mid] > sum)
end = mid - 1;
else
start = mid + 1;
}
return false;
}
// Driver code
int main()
{
int array[] = { 1, 2, 3 };
int n = sizeof(array) / sizeof(array[0]);
if (checkForElement(array, n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java program to check if array has an
// element whose value is half of array
// sum.
import java.io.*;
class GFG {
// Function to check if answer exists
static boolean checkForElement(int array[], int n)
{
// Sum of all array elements
int sum = 0;
for (int i = 0; i < n; i++)
sum += array[i];
// If sum is odd
if (sum % 2>0)
return false;
sum /= 2; // If sum is Even
// Do binary search for the required element
int start = 0;
int end = n - 1;
while (start <= end)
{
int mid = start + (end - start) / 2;
if (array[mid] == sum)
return true;
else if (array[mid] > sum)
end = mid - 1;
else
start = mid + 1;
}
return false;
}
// Driver code
public static void main (String[] args) {
int array[] = { 1, 2, 3 };
int n = array.length;
if (checkForElement(array, n))
System.out.println( "Yes");
else
System.out.println( "No");
}
}
// This code is contributed by anuj_67..
Python3
# Python 3 program to check if array
# has an element whose value is half
# of array sum.
# Function to check if answer exists
def checkForElement(array, n):
# Sum of all array elements
sum = 0
for i in range(n):
sum += array[i]
# If sum is odd
if (sum % 2):
return False
sum //= 2 # If sum is Even
# Do binary search for the
# required element
start = 0
end = n - 1
while (start <= end) :
mid = start + (end - start) // 2
if (array[mid] == sum):
return True
elif (array[mid] > sum) :
end = mid - 1;
else:
start = mid + 1
return False
# Driver code
if __name__ == "__main__":
array = [ 1, 2, 3 ]
n = len(array)
if (checkForElement(array, n)):
print("Yes")
else:
print("No")
# This code is contributed
# by ChitraNayal
C#
// C# program to check if array has
// an element whose value is half
// of array sum.
using System;
class GFG
{
// Function to check if answer exists
static bool checkForElement(int[] array,
int n)
{
// Sum of all array elements
int sum = 0;
for (int i = 0; i < n; i++)
sum += array[i];
// If sum is odd
if (sum % 2 > 0)
return false;
sum /= 2; // If sum is Even
// Do binary search for the
// required element
int start = 0;
int end = n - 1;
while (start <= end)
{
int mid = start + (end - start) / 2;
if (array[mid] == sum)
return true;
else if (array[mid] > sum)
end = mid - 1;
else
start = mid + 1;
}
return false;
}
// Driver Code
static void Main()
{
int []array = { 1, 2, 3 };
int n = array.Length;
if (checkForElement(array, n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by ANKITRAI1
PHP
<?php
// PHP program to check if array has an
// element whose value is half of array
// sum.
// Function to check if answer exists
function checkForElement(&$array, $n)
{
// Sum of all array elements
$sum = 0;
for ($i = 0; $i < $n; $i++)
$sum += $array[$i];
// If sum is odd
if ($sum % 2)
return false;
$sum /= 2; // If sum is Even
// Do binary search for the
// required element
$start = 0;
$end = $n - 1;
while ($start <= $end)
{
$mid = $start + ($end - $start) / 2;
if ($array[$mid] == $sum)
return true;
else if ($array[$mid] > $sum)
$end = $mid - 1;
else
$start = $mid + 1;
}
return false;
}
// Driver code
$array = array(1, 2, 3 );
$n = sizeof($array);
if (checkForElement($array, $n))
echo "Yes";
else
echo "No";
// This code is contributed
// by Shivi_Aggarwal
?>
Javascript
<script>
// Javascript program to check if array has an
// element whose value is half of array
// sum.
// Function to check if answer exists
function checkForElement(array, n)
{
// Sum of all array elements
let sum = 0;
for (let i = 0; i < n; i++)
sum += array[i];
// If sum is odd
if (sum % 2)
return false;
sum = Math.floor(sum / 2); // If sum is Even
// Do binary search for the
// required element
let start = 0;
let end = n - 1;
while (start <= end)
{
let mid = Math.floor(start + (end - start) / 2);
if (array[mid] == sum)
return true;
else if (array[mid] > sum)
end = mid - 1;
else
start = mid + 1;
}
return false;
}
// Driver code
let array = new Array(1, 2, 3 );
let n = array.length;
if (checkForElement(array, n))
document.write("Yes");
else
document.write("No");
// This code is contributed by _saurabh_jaiswal
</script>
Producción:
Yes
Complejidad temporal: O(n)
Espacio auxiliar: O(1)
Otra solución eficiente que también funciona para arreglos no clasificados.
La idea es usar hashing.
C++
// CPP program to check if array has an
// element whose value is half of array
// sum.
#include <bits/stdc++.h>
using namespace std;
// Function to check if answer exists
bool checkForElement(int array[], int n)
{
// Sum of all array elements
// and storing in a hash table
unordered_set<int> s;
int sum = 0;
for (int i = 0; i < n; i++) {
sum += array[i];
s.insert(array[i]);
}
// If sum/2 is present in hash table
if (sum % 2 == 0 && s.find(sum/2) != s.end())
return true;
else
return false;
}
// Driver code
int main()
{
int array[] = { 1, 2, 3 };
int n = sizeof(array) / sizeof(array[0]);
if (checkForElement(array, n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java program to check if array has an
// element whose value is half of array
// sum.
import java.util.*;
class GFG {
// Function to check if answer exists
static boolean checkForElement(int array[], int n) {
// Sum of all array elements
// and storing in a hash table
Set<Integer> s = new LinkedHashSet<>();
int sum = 0;
for (int i = 0; i < n; i++) {
sum += array[i];
s.add(array[i]);
}
// If sum/2 is present in hash table
if (sum % 2 == 0 && s.contains(sum / 2)
&& (sum / 2 )== s.stream().skip(s.size() - 1).findFirst().get()) {
return true;
} else {
return false;
}
}
// Driver code
public static void main(String[] args) {
int array[] = {1, 2, 3};
int n = array.length;
System.out.println(checkForElement(array, n) ? "Yes" : "No");
}
}
// This code is contributed by 29AjayKumar
Python3
# Python 3 program to check if array has an
# element whose value is half of array
# sum.
# Function to check if answer exists
def checkForElement(array, n):
# Sum of all array elements
# and storing in a hash table
s = set()
sum = 0
for i in range(n):
sum += array[i]
s.add(array[i])
# If sum/2 is present in hash table
f = int(sum / 2)
if (sum % 2 == 0 and f in s):
return True
else:
return False
# Driver code
if __name__ == '__main__':
array = [1, 2, 3]
n = len(array)
if (checkForElement(array, n)):
print("Yes")
else:
print("No")
# This code is contributed by
# Surendra_Gangwar
C#
// C# program to check if array has an
// element whose value is half of array
// sum.
using System;
using System.Collections.Generic;
class GFG
{
// Function to check if answer exists
static Boolean checkForElement(int []array, int n)
{
// Sum of all array elements
// and storing in a hash table
HashSet<int> s = new HashSet<int>();
int sum = 0;
for (int i = 0; i < n; i++)
{
sum += array[i];
s.Add(array[i]);
}
// If sum/2 is present in hash table
if (sum % 2 == 0 && s.Contains(sum / 2))
{
return true;
}
else
{
return false;
}
}
// Driver code
public static void Main(String[] args)
{
int []array = {1, 2, 3};
int n = array.Length;
Console.WriteLine(checkForElement(array, n) ? "Yes" : "No");
}
}
// This code is contributed by Princi Singh
Javascript
<script>
// Javascript program to check if array has an
// element whose value is half of array
// sum.
// Function to check if answer exists
function checkForElement(array, n)
{
// Sum of all array elements
// and storing in a hash table
let s = new Set();
let sum = 0;
for(let i = 0; i < n; i++)
{
sum += array[i];
s.add(array[i]);
}
// If sum/2 is present in hash table
if (sum % 2 == 0 && s.has(sum / 2))
{
return true;
}
else
{
return false;
}
}
// Driver code
let array = [ 1, 2, 3 ];
let n = array.length;
document.write(
checkForElement(array, n) ? "Yes" : "No");
// This code is contributed by rag2127
</script>
Producción:
Yes
Tiempo Complejidad : O(n)
Espacio Auxiliar : O(n)