# Queries to check whether all the elements in the given range occurs even number of times

Given an array arr[] containing N integers and there are Q queries where each query consists of a range [L, R]. The task is to find whether all the elements from the given index range have even frequency or not.

Examples:

Input: arr[] = {1, 1, 2, 2, 1}, Q[][] = {{1, 5}, {1, 4}, {3, 4}}
Output:
No
Yes
Yes

Input: arr[] = {100, 100, 200, 100}, Q[][] = {{1, 4}, {1, 2}}
Output:
No
Yes

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive approach: A simple approach will be to iterate from the index range [L, R] for each of the query and count the frequency of each element in that range and check that every element is occurring even number of times or not. The worst case time complexity of this approach will be O(Q * N) where Q is the number of queries and N is the number of elements in the array.

Efficient approach: Since XOR of two equal number is 0 i.e. if all the elements of the array appears even number of times then the XOR of the complete array will be 0. The same can be said about the elements in the given range [L, R]. Now, to check if the XOR of all the elements in the given range is 0 or not, a prefix XOR array can be created where preXor[i] will store the XOR of the elements arr[0…i]. And the XOR of the elements arr[i…j] can be easily calculated as preXor[j] ^ preXor[i – 1].

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to perform the given queries ` `void` `performQueries(vector<``int``>& A, ` `                    ``vector >& q) ` `{ ` ` `  `    ``int` `n = (``int``)A.size(); ` ` `  `    ``// Making array 1-indexed ` `    ``A.insert(A.begin(), 0); ` ` `  `    ``// To store the cumulative xor ` `    ``vector<``int``> pref_xor(n + 1, 0); ` ` `  `    ``// Taking cumulative Xor ` `    ``for` `(``int` `i = 1; i <= n; ++i) { ` `        ``pref_xor[i] ` `            ``= pref_xor[i - 1] ^ A[i]; ` `    ``} ` ` `  `    ``// Iterating over the queries ` `    ``for` `(``auto` `i : q) { ` `        ``int` `L = i.first, R = i.second; ` `        ``if` `(L > R) ` `            ``swap(L, R); ` ` `  `        ``// If both indices are different and xor ` `        ``// in the range [L, R] is 0 ` `        ``if` `(L != R and pref_xor[R] == pref_xor[L - 1]) ` `            ``cout << ``"Yes\n"``; ` `        ``else` `            ``cout << ``"No\n"``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``vector<``int``> Arr = { 1, 1, 2, 2, 1 }; ` ` `  `    ``vector > q = { ` `        ``{ 1, 5 }, ` `        ``{ 1, 4 }, ` `        ``{ 3, 4 } ` ` `  `    ``}; ` ` `  `    ``performQueries(Arr, q); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the approach ` `import` `java.util.*; ` `     `  `class` `GFG ` `{ ` `static` `class` `pair ` `{  ` `    ``int` `first, second;  ` `    ``public` `pair(``int` `first, ``int` `second)  ` `    ``{  ` `        ``this``.first = first;  ` `        ``this``.second = second;  ` `    ``}  ` `}  ` ` `  `// Function to perform the given queries ` `static` `void` `performQueries(``int` `[]A, ` `                           ``pair[] q) ` `{ ` `    ``int` `n = A.length; ` ` `  `    ``// To store the cumulative xor ` `    ``int` `[]pref_xor = ``new` `int``[n + ``1``]; ` ` `  `    ``// Taking cumulative Xor ` `    ``for` `(``int` `i = ``1``; i <=n; ++i)  ` `    ``{ ` `        ``pref_xor[i] = pref_xor[i - ``1``] ^  ` `                             ``A[i - ``1``]; ` `    ``} ` ` `  `    ``// Iterating over the queries ` `    ``for` `(pair i : q) ` `    ``{ ` `        ``int` `L = i.first, R = i.second; ` `        ``if` `(L > R) ` `        ``{ ` `            ``int` `temp = L; ` `            ``L = R; ` `            ``R = temp; ` `        ``} ` `         `  `        ``// If both indices are different and xor ` `        ``// in the range [L, R] is 0 ` `        ``if` `(L != R && pref_xor[R] == pref_xor[L - ``1``]) ` `            ``System.out.println(``"Yes"``); ` `    ``else` `        ``System.out.println(``"No"``); ` `    ``} ` `} ` ` `  `// Driver code ` `static` `public` `void` `main (String []arg) ` `{ ` `    ``int` `[]Arr = { ``1``, ``1``, ``2``, ``2``, ``1` `}; ` `    ``pair[] q = { ``new` `pair(``1``, ``5` `), ` `                 ``new` `pair(``1``, ``4` `), ` `                 ``new` `pair(``3``, ``4` `) }; ` ` `  `    ``performQueries(Arr, q); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

 `# Python3 implementation of the approach ` ` `  `# Function to perform the given queries ` `def` `performQueries(A, q): ` ` `  `    ``n ``=` `len``(A) ` `     `  `    ``# To store the cumulative xor ` `    ``pref_xor ``=` `[``0` `for` `i ``in` `range``(n ``+` `1``)] ` ` `  `    ``# Taking cumulative Xor ` `    ``for` `i ``in` `range``(``1``, n ``+` `1``): ` `        ``pref_xor[i] ``=` `pref_xor[i ``-` `1``] ^ A[i ``-` `1``] ` ` `  `    ``# Iterating over the queries ` `    ``for` `i ``in` `q: ` `        ``L ``=` `i[``0``] ` `        ``R ``=` `i[``1``] ` `        ``if` `(L > R): ` `            ``L, R ``=` `R, L ` ` `  `        ``# If both indices are different and  ` `        ``# xor in the range [L, R] is 0 ` `        ``if` `(L !``=` `R ``and`  `            ``pref_xor[R] ``=``=` `pref_xor[L ``-` `1``]): ` `            ``print``(``"Yes"``) ` `        ``else``: ` `            ``print``(``"No"``) ` ` `  `# Driver code ` `Arr ``=` `[``1``, ``1``, ``2``, ``2``, ``1``] ` ` `  `q ``=` `[[ ``1``, ``5` `], ` `     ``[ ``1``, ``4` `], ` `     ``[ ``3``, ``4` `]] ` ` `  `performQueries(Arr, q); ` ` `  `# This code is contributed by Mohit Kumar `

 `// C# implementation of the approach ` `using` `System; ` `     `  `class` `GFG ` `{ ` `public` `class` `pair ` `{  ` `    ``public` `int` `first, second;  ` `    ``public` `pair(``int` `first, ``int` `second)  ` `    ``{  ` `        ``this``.first = first;  ` `        ``this``.second = second;  ` `    ``}  ` `}  ` ` `  `// Function to perform the given queries ` `static` `void` `performQueries(``int` `[]A, ` `                           ``pair[] q) ` `{ ` `    ``int` `n = A.Length; ` ` `  `    ``// To store the cumulative xor ` `    ``int` `[]pref_xor = ``new` `int``[n + 1]; ` ` `  `    ``// Taking cumulative Xor ` `    ``for` `(``int` `i = 1; i <= n; ++i)  ` `    ``{ ` `        ``pref_xor[i] = pref_xor[i - 1] ^  ` `                             ``A[i - 1]; ` `    ``} ` ` `  `    ``// Iterating over the queries ` `    ``foreach` `(pair k ``in` `q) ` `    ``{ ` `        ``int` `L = k.first, R = k.second; ` `        ``if` `(L > R) ` `        ``{ ` `            ``int` `temp = L; ` `            ``L = R; ` `            ``R = temp; ` `        ``} ` `         `  `        ``// If both indices are different and xor ` `        ``// in the range [L, R] is 0 ` `        ``if` `(L != R && pref_xor[R] == pref_xor[L - 1]) ` `            ``Console.WriteLine(``"Yes"``); ` `    ``else` `        ``Console.WriteLine(``"No"``); ` `    ``} ` `} ` ` `  `// Driver code ` `static` `public` `void` `Main (String []arg) ` `{ ` `    ``int` `[]Arr = { 1, 1, 2, 2, 1 }; ` `    ``pair[] q = {``new` `pair(1, 5 ), ` `                ``new` `pair(1, 4 ), ` `                ``new` `pair(3, 4 )}; ` ` `  `    ``performQueries(Arr, q); ` `} ` `} ` `     `  `// This code is contributed by Rajput-Ji `

Output:
```No
Yes
Yes
```

Time Complexity: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :