Find sum of non-repeating (distinct) elements in an array

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

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 left side of it. If present, then ignores the element.

Time Complexity : O(n2)
Auxiliary Space : O(1)

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

C++

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// 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;
}

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Java

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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));
  
    }
}

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Python3

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# 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

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C#

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// C# 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 
    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*/

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Output:

21

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

An Efficient solution of this problem is that using unordered_set we run a single for loop and which value comes first time its add in sum variable and store in hash table that for next time we not use this value.

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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Output:

21

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

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