Given a array such that all its terms is either 0 or 1.You need to tell the number represented by a subarray a[l..r] is odd or even

Examples:

Input : arr = {1, 1, 0, 1} l = 1, r = 3 Output : odd number represented by arr[l...r] is 101 which 5 in decimal form which is odd Input : arr = {1, 1, 1, 1} l = 0, r = 3 Output : odd

The important point to note here is all the odd numbers in binary form have 1 as their rightmost bit and all even numbers have 0 as their rightmost bit.

The reason is simple all other bits other than rightmost bit have even values and sum of even numbers is always even .Now the rightmost bit can have value either 1 or 0 as we know even + odd = odd so when right most bit is 1 the number is odd and when it is 0 the number is even.

So to solve this problem we have to just check if a[r] is 0 or 1 and accordingly print odd or even

// C++ program to find if a subarray // is even or odd. #include<bits/stdc++.h> using namespace std; // prints if subarray is even or odd void checkEVENodd (int arr[], int n, int l, int r) { // if arr[r] = 1 print odd if (arr[r] == 1) cout << "odd" << endl; // if arr[r] = 0 print even else cout << "even" << endl; } // driver code int main() { int arr[] = {1, 1, 0, 1}; int n = sizeof(arr)/sizeof(arr[0]); checkEVENodd (arr, n, 1, 3); return 0; }

Output:

odd

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