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++
// 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 |
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++
// 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 |
Output:
21
Time Complexity : O(n)
Auxiliary Space : O(n)
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