C++ Queries on Probability of Even or Odd Number in Given Ranges

To find the probability of numbers’ parity, i.e., is it even or odd, and for the given ranges. For each query, we need to print p and q representing the probability by p / q, for example.

Input : N = 5, arr[] = { 6, 5, 2, 1, 7 }
query 1: 0 2 2
query 2: 1 2 5
query 3: 0 1 4

Output : 0
3 4
1 2

In this problem, we will maintain two arrays containing the number of odd and even numbers present until that index. This simplifies our problems, and now we need to print their count and the number of elements present in that range.

Approach to Find the Solution

In this approach, we maintain two arrays. They contain the number of even and odd numbers found until ith index and solve this problem like prefix sum problems.

Example

#include <bits/stdc++.h>
using namespace std;
void solve(int arr[], int n, int Q,int query[][3]){
    int even[n + 1]; // our array for counting the number of evens find till ith index
    int odd[n + 1]; // our array for counting the number of odds find till ith index
    even[0] = 0; odd[0] = 0; // as we are doing 1 based indexing so we just set 0th index of both arrays to 0
    for (int i = 0; i < n; i++) {
        if (arr[i] & 1) { // if we found odd number we increment odd
            odd[i + 1] = odd[i] + 1;
            even[i + 1] = even[i];
        }
        else { // else we increment even
            even[i + 1] = even[i] + 1;
            odd[i + 1] = odd[i];
        }
    }
    for (int i = 0; i < Q; i++) { // traversing the queries
        int r = query[i][2]; // right range
        int l = query[i][1]; // left range
        int k = query[i][0]; // type of query
        int q = r - l + 1; // number of elements in the given range
        int p;
        if (k) // k is the type of query and we are finding the
            //number of elements with same parity in the given range
            p = odd[r] - odd[l - 1];
        else
            p = even[r] - even[l - 1];
        if (!p) // if p is zero we simply print 0
            cout << "0\n";
        else if (p == q) // if p == q we print 1
            cout << "1\n";
        else {
            int g = __gcd(p, q);
            cout << p / g << " " << q / g << "\n"; // as p and shouldn't have a common gcd so we divide the gcd
        }
    }
}
int main(){
    int arr[] = { 6, 5, 2, 1, 7 }; // given array
    int n = sizeof(arr) / sizeof(int); // size of our array
    int Q = 2; // number of our queries
    int query[Q][3] = {{ 0, 2, 2 },{ 1, 2, 5 }}; // given queries
    solve(arr, n, Q, query);
    return 0;
}

Output

0
3 4

Explanation of the Above Code

In the above approach, we count the number of even and odd numbers found to ith index by maintaining two arrays. Now we need to find the number of even or odd numbers present in the given range and print that number and print the total number of elements present.

Conclusion

In this tutorial, we solve the Queries on the probability of even or odd numbers in given ranges. We also learned the C++ program for this problem and the complete approach (Normal) by which we solved this problem. We can write the same program in other languages such as C, java, python, and other languages. We hope you find this tutorial helpful.