Find sum of nonrepeating (distinct) elements in an array (original) (raw)

Last Updated : 13 Sep, 2023

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Given an integer array with repeated elements, the task is to find the sum of all distinct elements in the array.
**Examples:

**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

**Naive Approach:

A **Simple Solution is to use two nested loops. The outer loop picks an element one by one starting from the leftmost element. The inner loop checks if the element is present on right side of it. If present, then ignores the element.

**Steps that were to follow the above approach:

• Make a variable sum and initialize it with 0. It is the variable that will contain the final answer
• Now traverse the input array
• While traversing the array pick an element and check all elements to its right by running an inner loop.
• If we get any element with the same value as that element then stop the inner loop
• If any element exists to the right of that element that has the same value then it is ok else add the value of that element to the sum.

**Code to implement the above approach:

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) { //Intialized a variable with 0 to contain final answer int sum = 0;

//Traverse the input array
for (int i=0; i<n; i++)
{
    int j=i+1;
    while(j<n){
        //if any element present on the right of arr[i] that has 
        //same value as arr[i] then break the loop
        if(arr[j]==arr[i]){break;}
        j++;
    }
    //If no such element exists then add this element's value into sum
    if(j==n){sum+=arr[i];}
}

//Finally return the answer 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;

public class Main {

// Find the sum of all non-repeated elements // in an array public static int findSum(int arr[], int n) {

// Intialize a variable with 0 to contain final answer
int sum = 0;

// Traverse the input array
for (int i = 0; i < n; i++) {
  int j = i + 1;
  while (j < n) 
  {

    // If any element present on the right of arr[i] that has 
    // same value as arr[i], then break the loop
    if (arr[j] == arr[i]) {
      break;
    }
    j++;
  }

  // If no such element exists then add this element's value into sum
  if (j == n) {
    sum += arr[i];
  }
}

// Finally return the answer
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

Find the sum of all non-repeated elements

in an array

def findSum(arr): # Initialize a variable with 0 to contain final answer sum = 0

# Traverse the input array
for i in range(len(arr)):
    j = i + 1
    while j < len(arr):
        # If any element present on the right of arr[i] that has
        # same value as arr[i] then break the loop
        if arr[j] == arr[i]:
            break
        j += 1
    # If no such element exists then add this element's value into sum
    if j == len(arr):
        sum += arr[i]

# Finally return the answer
return sum

Driver code

arr = [1, 2, 3, 1, 1, 4, 5, 6] print(findSum(arr))

C#

// C# code to find the sum of all non-repeated // elements in an array using System;

public class GFG {

// Find the sum of all non-repeated elements
// in an array
public static int FindSum(int[] arr, int n)
{
    // Initialized a variable with 0 to contain final
    // answer
    int sum = 0;
  
    // Traverse the input array
    for (int i = 0; i < n; i++) {
        int j = i + 1;
        while (j < n) {
            // if any element present on the right of
            // arr[i] that has same value as arr[i] then
            // break the loop
            if (arr[j] == arr[i]) {
                break;
            }
            j++;
        }
        // If no such element exists then add this
        // element's value into sum
        if (j == n) {
            sum += arr[i];
        }
    }

    // Finally return the answer
    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));
}

}

JavaScript

// JavaScript 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) { // Intialize a variable with 0 to contain final answer let sum = 0;

// Traverse the input array for (let i=0; i<arr.length; i++) { let j=i+1; while(j<arr.length){ // If any element present on the right of arr[i] that has // same value as arr[i] then break the loop if(arr[j]==arr[i]){break;} j++; } // If no such element exists then add this element's value into sum if(j==arr.length){sum+=arr[i];} }

// Finally return the answer return sum; }

// Driver code let arr = [1, 2, 3, 1, 1, 4, 5, 6]; console.log(findSum(arr));

`

**Output-

21

**Time Complexity : **O(n 2 ) ,because of two nested loop
**Auxiliary Space : O(1) , because no extra space has been used

A **Better Solution of this problem is that using sorting technique we firstly sort all elements of array in ascending order and find one by one distinct elements in array.

**Implementation:

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

`

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

An **Efficient solution to this problem is that using unordered_set we run a single for loop and in which the value comes the first time it's an add-in sum variable and stored in a hash table that for the next time we do not use this value.

**Implementation:

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

`

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

**Method #3:Using Built-in python and javascript functions:

**Approach for python:

• Calculate the frequencies using **Counter() function
• Convert the frequency keys to the list.
• Calculate the sum of the list.

**Approach for Javascript:

• The Counter function from the collections module in Python has been replaced with an empty object.
• The **keys() method is used to extract the keys of the object as an array.
• The **reduce() method is used to calculate the sum of the array.

**Below is the implementation of the **above approach.

C++ `

// c++ program for the above approach #include #include #include

using namespace std;

// Function to return the sum of distinct elements int sumOfElements(vector arr, int n) {

// Creating an unordered_map to store the frequency of each element unordered_map<int, int> freq;

for(int i=0; i<n; i++) { freq[arr[i]]++; }

// Creating a vector to store the unique elements vector lis;

for(auto it=freq.begin(); it!=freq.end(); it++) { lis.push_back(it->first); }

// Calculating the sum of unique elements int sum = 0; for(int i=0; i<lis.size(); i++) { sum += lis[i]; }

return sum; }

// Driver code int main() {

vector arr = {1, 2, 3, 1, 1, 4, 5, 6}; int n = arr.size();

cout << sumOfElements(arr, n);

return 0; }

// This code is contributed by Prince Kumar

Java

// Java program for the above approach

import java.util.*;

public class Main {

// Function to return the sum of distinct elements public static int sumOfElements(List arr, int n) {

// Creating a HashMap to store the frequency of each element
HashMap<Integer, Integer> freq = new HashMap<>();

for(int i=0; i<n; i++) {
  freq.put(arr.get(i), freq.getOrDefault(arr.get(i), 0) + 1);
}

// Creating a list to store the unique elements
List<Integer> lis = new ArrayList<>();

for(Map.Entry<Integer, Integer> entry : freq.entrySet()) {
  lis.add(entry.getKey());
}

// Calculating the sum of unique elements
int sum = 0;
for(int i=0; i<lis.size(); i++) {
  sum += lis.get(i);
}

return sum;

}

// Driver code public static void main(String[] args) {

List<Integer> arr = Arrays.asList(1, 2, 3, 1, 1, 4, 5, 6);
int n = arr.size();

System.out.println(sumOfElements(arr, n));

} }

// This code is contributed by adityashatmfh

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

C#

using System; using System.Collections.Generic; using System.Linq;

public class Program { // Function to return the sum of distinct elements public static int SumOfElements(List arr, int n) { // Creating a Dictionary to store the frequency of each element Dictionary<int, int> freq = new Dictionary<int, int>();

    for (int i = 0; i < n; i++)
    {
        if (freq.ContainsKey(arr[i]))
            freq[arr[i]]++;
        else
            freq[arr[i]] = 1;
    }

    // Creating a list to store the unique elements
    List<int> lis = new List<int>();

    foreach (KeyValuePair<int, int> entry in freq)
    {
        lis.Add(entry.Key);
    }

    // Calculating the sum of unique elements
    int sum = 0;
    for (int i = 0; i < lis.Count; i++)
    {
        sum += lis[i];
    }

    return sum;
}

// Driver code
public static void Main(string[] args)
{
    List<int> arr = new List<int> { 1, 2, 3, 1, 1, 4, 5, 6 };
    int n = arr.Count;

    Console.WriteLine(SumOfElements(arr, n));
}

}

JavaScript

// JavaScript program for the above approach function sumOfElements(arr, n) {

// Creating an empty object
let freq = {};

// Loop to create frequency object
for(let i = 0; i < n; i++) {
    freq[arr[i]] = (freq[arr[i]] || 0) + 1;
}

// Converting keys of freq object to array
let lis = Object.keys(freq).map(Number);

// Return sum of array
return lis.reduce((a, b) => a + b, 0);

}

// Driver code let arr = [1, 2, 3, 1, 1, 4, 5, 6]; let n = arr.length;

console.log(sumOfElements(arr, n));

`

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