XOR of Numbers Appearing Even Times in Given Range in C++

In this problem, we are given an array of n elements and there are some queries which are range from a start point to endpoint in the array. Our task is two finds the XOR of elements that appeared even a number of times in the range.

Let’s take an example to understand the problem,

Input −

array = {1, 2, 3, 1, 1, 2, 2, 3}
querries = 2
R = 4
L = 2, R = 5
L = 2, R = 7

Output −

0
1
0

The solution to this problem is quite easy, we will find the sum of XOR of all elements of the array within the given range by each query. For this, we will use prefix sum xor.

Example

Program to show the implementation of our solution,

 Live Demo

#include <bits/stdc++.h>
using namespace std;
struct que {
   int L, R, idx;
};
bool cmp(que a, que b){
   if (a.R != b.R)
      return a.R < b.R;
   else
      return a.L < b.L;
}
int findXORSum(int BIT[], int index){
   int xorSum = 0;
   index = index + 1;
   while (index > 0){
      xorSum ^= BIT[index];
      index -= index & (-index);
   }
   return xorSum;
}
void updateBIT(int BIT[], int N, int index, int val){
   index = index + 1;
   while (index <= N){
      BIT[index] ^= val;
      index += index & (-index);
   }
}
int* createBitTree(int arr[], int N){
   int* BIT = new int[N + 1];
   for (int i = 1; i <= N; i++)
      BIT[i] = 0;
   return BIT;
}
void findXORSolution(int arr[], int N, que queries[], int Q, int BIT[]){
   int* prefixXOR = new int[N + 1];
   map<int, int> XORval;
   for (int i = 0; i < N; i++) {
      if (!XORval[arr[i]])
         XORval[arr[i]] = i;
      if (i == 0)
         prefixXOR[i] = arr[i];
      else
         prefixXOR[i] = prefixXOR[i - 1] ^ arr[i];
   }
   int lastOcc[1000001];
   memset(lastOcc, -1, sizeof(lastOcc));
   sort(queries, queries + Q, cmp);
   int res[Q];
   int j = 0;
   for (int i = 0; i < Q; i++){
      while (j <= queries[i].R){
         if (lastOcc[XORval[arr[j]]] != -1)
            updateBIT(BIT, N, lastOcc[XORval[arr[j]]], arr[j]);
         updateBIT(BIT, N, j, arr[j]);
         lastOcc[XORval[arr[j]]] = j;
         j++;
      }
      int allXOR = prefixXOR[queries[i].R] ^ prefixXOR[queries[i].L - 1];
      int distinctXOR = findXORSum(BIT, queries[i].R) ^ findXORSum(BIT, queries[i].L - 1);
      res[queries[i].idx] = allXOR ^ distinctXOR;
   }
   for (int i = 0; i < Q; i++)
      cout << res[i] << endl;
}
int main() {
   int arr[] = {1, 2, 1, 1, 2, 2, 3, 1, 3};
   int N = sizeof(arr) / sizeof(arr[0]);
   int* BIT = createBitTree(arr, N);
   que queries[4];
   queries[0].L = 1;
   queries[0].R = 4; queries[0].idx = 0;
   queries[1].L = 2;
   queries[1].R = 7, queries[1].idx = 1;
   queries[2].L = 0;
   queries[2].R = 3, queries[2].idx = 2;
   queries[3].L = 3;
   queries[3].R = 6, queries[3].idx = 3;
   int Q = sizeof(queries) / sizeof(queries[0]);
   cout<<"Xor sum for all queries is \n";
   findXORSolution(arr, N, queries, Q, BIT);
   return 0;
}

Output

Xor sum for all queries is
3
2
0
2