Given two arrays A[] and B[] of size n. It is given that both array individually contains distinct elements. We need to find the sum of all elements that are not common.
Examples:
Input : A[] = {1, 5, 3, 8}
B[] = {5, 4, 6, 7}
Output : 29
1 + 3 + 4 + 6 + 7 + 8 = 29
Input : A[] = {1, 5, 3, 8}
B[] = {5, 1, 8, 3}
Output : 0
All elements are common.
Brute Force Method: One simple approach is that for each element in A[] check whether it is present in B[], if it is present in then add it to the result. Similarly, traverse B[] and for every element that is not present in B, add it to result.
Time Complexity: O(n2).
Auxiliary Space: O(1), As constant extra space is used.
Hashing concept: Create an empty hash and insert elements of both arrays into it. Now traverse hash table and add all those elements whose count is 1. (As per the question, both arrays individually have distinct elements)
Below is the implementation of the above approach:
C++
// CPP program to find Non-overlapping sum#include <bits/stdc++.h>using namespace std;// function for calculating// Non-overlapping sum of two arrayint findSum(int A[], int B[], int n){ // Insert elements of both arrays unordered_map<int, int> hash; for (int i = 0; i < n; i++) { hash[A[i]]++; hash[B[i]]++; } // calculate non-overlapped sum int sum = 0; for (auto x: hash) if (x.second == 1) sum += x.first; return sum;}// driver codeint main(){ int A[] = { 5, 4, 9, 2, 3 }; int B[] = { 2, 8, 7, 6, 3 }; // size of array int n = sizeof(A) / sizeof(A[0]); // function call cout << findSum(A, B, n); return 0;} |
Java
// Java program to find Non-overlapping sum import java.io.*;import java.util.*;class GFG { // function for calculating // Non-overlapping sum of two array static int findSum(int[] A, int[] B, int n) { // Insert elements of both arrays HashMap<Integer, Integer> hash = new HashMap<>(); for (int i = 0; i < n; i++) { if (hash.containsKey(A[i])) hash.put(A[i], 1 + hash.get(A[i])); else hash.put(A[i], 1); if (hash.containsKey(B[i])) hash.put(B[i], 1 + hash.get(B[i])); else hash.put(B[i], 1); } // calculate non-overlapped sum int sum = 0; for (Map.Entry entry : hash.entrySet()) { if (Integer.parseInt((entry.getValue()).toString()) == 1) sum += Integer.parseInt((entry.getKey()).toString()); } return sum; } // Driver code public static void main(String args[]) { int[] A = { 5, 4, 9, 2, 3 }; int[] B = { 2, 8, 7, 6, 3 }; // size of array int n = A.length; // function call System.out.println(findSum(A, B, n)); }}// This code is contributed by rachana soma |
Python3
# Python3 program to find Non-overlapping sum from collections import defaultdict# Function for calculating # Non-overlapping sum of two array def findSum(A, B, n): # Insert elements of both arrays Hash = defaultdict(lambda:0) for i in range(0, n): Hash[A[i]] += 1 Hash[B[i]] += 1 # calculate non-overlapped sum Sum = 0 for x in Hash: if Hash[x] == 1: Sum += x return Sum# Driver code if __name__ == "__main__": A = [5, 4, 9, 2, 3] B = [2, 8, 7, 6, 3] # size of array n = len(A) # Function call print(findSum(A, B, n)) # This code is contributed # by Rituraj Jain |
C#
// C# program to find Non-overlapping sumusing System;using System.Collections.Generic; class GFG { // function for calculating // Non-overlapping sum of two array static int findSum(int[] A, int[] B, int n) { // Insert elements of both arrays Dictionary<int, int> hash = new Dictionary<int, int>(); for (int i = 0; i < n; i++) { if (hash.ContainsKey(A[i])) { var v = hash[A[i]]; hash.Remove(A[i]); hash.Add(A[i], 1 + v); } else hash.Add(A[i], 1); if (hash.ContainsKey(B[i])) { var v = hash[B[i]]; hash.Remove(B[i]); hash.Add(B[i], 1 + v); } else hash.Add(B[i], 1); } // calculate non-overlapped sum int sum = 0; foreach(KeyValuePair<int, int> entry in hash) { if ((entry.Value) == 1) sum += entry.Key; } return sum; } // Driver code public static void Main(String []args) { int[] A = { 5, 4, 9, 2, 3 }; int[] B = { 2, 8, 7, 6, 3 }; // size of array int n = A.Length; // function call Console.WriteLine(findSum(A, B, n)); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript program to find Non-overlapping sum// function for calculating// Non-overlapping sum of two arrayfunction findSum(A, B, n) { // Insert elements of both arrays let hash = new Map(); for (let i = 0; i < n; i++) { if (hash.has(A[i])) hash.set(A[i], 1 + hash.get(A[i])); else hash.set(A[i], 1); if (hash.has(B[i])) hash.set(B[i], 1 + hash.get(B[i])); else hash.set(B[i], 1); } // calculate non-overlapped sum let sum = 0; for (let entry of hash) { if (parseInt((entry[1]).toString()) == 1) sum += parseInt((entry[0]).toString()); } return sum;}// Driver codelet A = [5, 4, 9, 2, 3];let B = [2, 8, 7, 6, 3];// size of arraylet n = A.length;// function calldocument.write(findSum(A, B, n));// This code is contributed by gfgking</script> |
Output
39
Time Complexity: O(n), since inserting in an unordered map is amortized constant.
Auxiliary Space: O(n).
Another method: Using set data structure
- Insert elements of Array A in the set data structure and add into sum
- Check if B’s elements are there in set if exist then remove current element from set, otherwise add current element to sum
- Finally, return sum
Below is the implementation of the above approach:
C++
// CPP program to find Non-overlapping sum#include <bits/stdc++.h>using namespace std;// function for calculating// Non-overlapping sum of two arrayint findSum(int A[], int B[], int n){ int sum = 0; // Insert elements of Array A in set // and add into sum set<int> st; for (int i = 0; i < n; i++) { st.insert(A[i]); sum += A[i]; } // Check if B's element are there in set // if exist then remove current element from // set, otherwise add current element into sum for (int i = 0; i < n; i++) { if (st.find(B[i]) == st.end()) { sum += B[i]; } else { sum -= B[i]; } } // Finally, return sum return sum;}// Driver codeint main(){ int A[] = { 5, 4, 9, 2, 3 }; int B[] = { 2, 8, 7, 6, 3 }; // size of array int n = sizeof(A) / sizeof(A[0]); // function call cout << findSum(A, B, n); return 0;}// This code is contributed by hkdass001 |
Java
// Java program to find Non-overlapping sumimport java.io.*;import java.util.*;class GFG { // function for calculating // Non-overlapping sum of two array public static int findSum(int[] A, int[] B, int n) { int sum = 0; // Insert elements of Array A in set // and add into sum Set<Integer> st = new HashSet<>(); for (int i = 0; i < n; i++) { st.add(A[i]); sum += A[i]; } // Check if B's element are there in set // if exist then remove current element from // set, otherwise add current element into sum for (int i = 0; i < n; i++) { if (!st.contains(B[i])) { sum += B[i]; } else { sum -= B[i]; } } // Finally, return sum return sum; } public static void main (String[] args) { int[] A = { 5, 4, 9, 2, 3 }; int[] B = { 2, 8, 7, 6, 3 }; // size of array int n = A.length; // function call System.out.println(findSum(A, B, n)); }}// This code is contributed by lokesh. |
Python3
# python program to find Non-overlapping sum# function for calculating# Non-overlapping sum of two arraydef findSum(A, B, n): sum = 0; # Insert elements of Array A in set # and add into sum st = set(); for i in range(0,n): st.add(A[i]); sum += A[i]; # Check if B's element are there in set # if exist then remove current element from # set, otherwise add current element into sum for i in range (0, n): if (B[i] in st): sum -= B[i]; else : sum += B[i]; # Finally, return sum return sum;# Driver codeA = [ 5, 4, 9, 2, 3 ];B = [ 2, 8, 7, 6, 3 ];# size of arrayn = len(A);# function callprint(findSum(A, B, n)); |
C#
// C# code for the above approachusing System;using System.Collections.Generic;public class GFG { // function for calculating // Non-overlapping sum of two array public static int FindSum(int[] A, int[] B, int n) { int sum = 0; // Insert elements of Array A in set // and add into sum HashSet<int> st = new HashSet<int>(); for (int i = 0; i < n; i++) { st.Add(A[i]); sum += A[i]; } // Check if B's element are there in set // if exist then remove current element from // set, otherwise add current element into sum for (int i = 0; i < n; i++) { if (!st.Contains(B[i])) { sum += B[i]; } else { sum -= B[i]; } } // Finally, return sum return sum; } static public void Main() { // Code int[] A = { 5, 4, 9, 2, 3 }; int[] B = { 2, 8, 7, 6, 3 }; // size of array int n = A.Length; // function call Console.WriteLine(FindSum(A, B, n)); }}// This code is contributed by lokeshmvs21. |
Javascript
// Javascript program to find Non-overlapping sum// function for calculating// Non-overlapping sum of two arrayfunction findSum(A, B, n){ let sum = 0; // Insert elements of Array A in set // and add into sum let st = new Set(); for (let i = 0; i < n; i++) { st.add(A[i]); sum += A[i]; } // Check if B's element are there in set // if exist then remove current element from // set, otherwise add current element into sum for (let i = 0; i < n; i++) { if (!st.has(B[i])) { sum += B[i]; } else { sum -= B[i]; } } // Finally, return sum return sum;}// Driver code let A = [ 5, 4, 9, 2, 3 ]; let B = [ 2, 8, 7, 6, 3 ]; // size of array let n = A.length; // function call document.write(findSum(A, B, n)); |
Output
39
Time Complexity: O(n*log n)
Auxiliary Space: O(n)
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