Thanks for sharing valuable post Sabexch

This code defines a recursive function find(x, y) to determine whether the value of pow(A[x], find(x+1, y)) is odd or even. World 777

This is bullshit. Problem statement is vague and does not walk through any example. I wish I could give this negative star.

Python oneliner:

def solve(arr, queries):
    return ['Odd' if x>y or arr[x-1]%2 or x<y and arr[x]==0 else 'Even' for x,y in queries]

Some solutions published below omit the first condition (x>y) but it works only because there are no test cases with x>y and arr[x-1] even (you can make your own test case to see that it would fail in such event.

Explanation: x>y is odd, odd^anything is odd, anything^0 is odd, everything else is even.

Some things to notice: Since we only need to take note whether find(x,y) is odd or even, and since find(x,y) function is just repeated exponentiation with base arr[x]: 1. the result will be even if arr[x-1] (-1 because indices start from 1) is even and the next array element arr[x] is not zero (otherwise we end up getting 1). 2. the result will be odd if arr[x-1] is odd or if arr[x-1] is even and arr[x] is odd (as I explained in 1 above). Then the code is simple to be honest: In C++:

vector<string> solve(vector<int> arr, vector<vector<int>> queries) {
    vector<string> result;
    
    for (auto k : queries){
        int x = k[0], y = k[1];
        if (arr[x-1]%2 != 0) result.push_back("Odd");
        else if (x < y && arr[x] == 0) result.push_back("Odd");
        else result.push_back("Even");
    }
    return result;
}