Dada una array de enteros con elementos repetidos, la tarea es encontrar la suma de todos los elementos distintos en la array.
Ejemplos:
Input : arr[] = {12, 10, 9, 45, 2, 10, 10, 45,10};
Output : 78
Here we take 12, 10, 9, 45, 2 for sum
because it's distinct elements
Input : arr[] = {1, 10, 9, 4, 2, 10, 10, 45 , 4};
Output : 71
Una solución simple es usar dos bucles anidados. El bucle exterior elige un elemento uno por uno comenzando desde el elemento más a la izquierda. El bucle interno verifica si el elemento está presente en el lado izquierdo. Si está presente, ignora el elemento.
Complejidad de Tiempo : O(n 2 )
Espacio Auxiliar : O(1)
Una mejor solución a este problema es que, al usar la técnica de clasificación, primero clasificamos todos los elementos de la array en orden ascendente y encontramos elementos distintos uno por uno en la array.
Implementación:
C++
// C++ Find the sum of all non-repeated
// elements in an array
#include<bits/stdc++.h>
using namespace std;
// Find the sum of all non-repeated elements
// in an array
int findSum(int arr[], int n)
{
// sort all elements of array
sort(arr, arr + n);
int sum = 0;
for (int i=0; i<n; i++)
{
if (arr[i] != arr[i+1])
sum = sum + arr[i];
}
return sum;
}
// Driver code
int main()
{
int arr[] = {1, 2, 3, 1, 1, 4, 5, 6};
int n = sizeof(arr)/sizeof(int);
cout << findSum(arr, n);
return 0;
}
Java
import java.util.Arrays;
// Java Find the sum of all non-repeated
// elements in an array
public class GFG {
// Find the sum of all non-repeated elements
// in an array
static int findSum(int arr[], int n) {
// sort all elements of array
Arrays.sort(arr);
int sum = arr[0];
for (int i = 0; i < n-1; i++) {
if (arr[i] != arr[i + 1]) {
sum = sum + arr[i+1];
}
}
return sum;
}
// Driver code
public static void main(String[] args) {
int arr[] = {1, 2, 3, 1, 1, 4, 5, 6};
int n = arr.length;
System.out.println(findSum(arr, n));
}
}
Python3
# Python3 Find the sum of all non-repeated # elements in an array # Find the sum of all non-repeated elements # in an array def findSum(arr, n): # sort all elements of array arr.sort() sum = arr[0] for i in range(0,n-1): if (arr[i] != arr[i+1]): sum = sum + arr[i+1] return sum # Driver code def main(): arr= [1, 2, 3, 1, 1, 4, 5, 6] n = len(arr) print(findSum(arr, n)) if __name__ == '__main__': main() # This code is contributed by 29AjayKumar
C#
// C# Find the sum of all non-repeated
// elements in an array
using System;
class GFG
{
// Find the sum of all non-repeated elements
// in an array
static int findSum(int []arr, int n)
{
// sort all elements of array
Array.Sort(arr);
int sum = arr[0];
for (int i = 0; i < n - 1; i++)
{
if (arr[i] != arr[i + 1])
{
sum = sum + arr[i + 1];
}
}
return sum;
}
// Driver code
public static void Main()
{
int []arr = {1, 2, 3, 1, 1, 4, 5, 6};
int n = arr.Length;
Console.WriteLine(findSum(arr, n));
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// JavaScript Program to find the sum of all non-repeated
// elements in an array
// Find the sum of all non-repeated elements
// in an array
function findSum(arr, n)
{
// sort all elements of array
arr.sort();
let sum = 0;
for (let i=0; i<n; i++)
{
if (arr[i] != arr[i+1])
sum = sum + arr[i];
}
return sum;
}
// Driver code
let arr = [1, 2, 3, 1, 1, 4, 5, 6];
let n = arr.length;
document.write(findSum(arr, n));
// This code is contributed by Surbhi Tyagi
</script>
21
Complejidad de tiempo: O(n log n)
Espacio auxiliar: O(1)
Una solución eficiente de este problema es que usando unordered_set ejecutamos un solo bucle for y qué valor viene la primera vez que se agrega en la variable de suma y se almacena en la tabla hash para que la próxima vez no usemos este valor.
Implementación:
C++
// C++ Find the sum of all non- repeated
// elements in an array
#include<bits/stdc++.h>
using namespace std;
// Find the sum of all non-repeated elements
// in an array
int findSum(int arr[],int n)
{
int sum = 0;
// Hash to store all element of array
unordered_set< int > s;
for (int i=0; i<n; i++)
{
if (s.find(arr[i]) == s.end())
{
sum += arr[i];
s.insert(arr[i]);
}
}
return sum;
}
// Driver code
int main()
{
int arr[] = {1, 2, 3, 1, 1, 4, 5, 6};
int n = sizeof(arr)/sizeof(int);
cout << findSum(arr, n);
return 0;
}
Java
// Java Find the sum of all non- repeated
// elements in an array
import java.util.*;
class GFG
{
// Find the sum of all non-repeated elements
// in an array
static int findSum(int arr[], int n)
{
int sum = 0;
// Hash to store all element of array
HashSet<Integer> s = new HashSet<Integer>();
for (int i = 0; i < n; i++)
{
if (!s.contains(arr[i]))
{
sum += arr[i];
s.add(arr[i]);
}
}
return sum;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {1, 2, 3, 1, 1, 4, 5, 6};
int n = arr.length;
System.out.println(findSum(arr, n));
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 Find the sum of all # non- repeated elements in an array # Find the sum of all non-repeated # elements in an array def findSum(arr, n): s = set() sum = 0 # Hash to store all element # of array for i in range(n): if arr[i] not in s: s.add(arr[i]) for i in s: sum = sum + i return sum # Driver code arr = [1, 2, 3, 1, 1, 4, 5, 6] n = len(arr) print(findSum(arr, n)) # This code is contributed by Shrikant13
C#
// C# Find the sum of all non- repeated
// elements in an array
using System;
using System.Collections.Generic;
class GFG
{
// Find the sum of all non-repeated elements
// in an array
static int findSum(int []arr, int n)
{
int sum = 0;
// Hash to store all element of array
HashSet<int> s = new HashSet<int>();
for (int i = 0; i < n; i++)
{
if (!s.Contains(arr[i]))
{
sum += arr[i];
s.Add(arr[i]);
}
}
return sum;
}
// Driver code
public static void Main(String[] args)
{
int []arr = {1, 2, 3, 1, 1, 4, 5, 6};
int n = arr.Length;
Console.WriteLine(findSum(arr, n));
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// Javascript program Find the sum of all non- repeated
// elements in an array
// Find the sum of all non-repeated elements
// in an array
function findSum(arr, n)
{
let sum = 0;
// Hash to store all element of array
let s = new Set();
for (let i = 0; i < n; i++)
{
if (!s.has(arr[i]))
{
sum += arr[i];
s.add(arr[i]);
}
}
return sum;
}
// Driver code
let arr = [1, 2, 3, 1, 1, 4, 5, 6];
let n = arr.length;
document.write(findSum(arr, n));
</script>
21
Complejidad temporal: O(n)
Espacio auxiliar: O(n)
Método n.º 3: uso de las funciones integradas de python:
Acercarse:
- Calcule las frecuencias usando la función Counter()
- Convierte las claves de frecuencia a la lista.
- Calcular la suma de la lista.
A continuación se muestra la implementación del enfoque anterior.
Python3
# Python program for the above approach from collections import Counter # Function to return the sum of distinct elements def sumOfElements(arr, n): # Counter function is used to # calculate frequency of elements of array freq = Counter(arr) # Converting keys of freq dictionary to list lis = list(freq.keys()) # Return sum of list return sum(lis) # Driver code if __name__ == "__main__": arr = [1, 2, 3, 1, 1, 4, 5, 6] n = len(arr) print(sumOfElements(arr, n)) # This code is contributed by vikkycirus
21
Complejidad temporal: O(n)
Espacio auxiliar: O(n)
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA