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(n^2).
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 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 array int 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 code int 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 |
Output:
39
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