Queries on probability of even or odd number in given ranges

Given an array A of size N, containing integers. We have to answer Q queries where each query is of the form:

  • K L R : If K = 0, then you have to find the probability of choosing an even number from the segment [L, R] (both inclusive) in the array A.
  • K L R : If K = 1, then you have to find the probability of choosing an odd number from the segment [L, R] (both inclusive) in the array A.

For each query print two integers p and q which represent the probability p/q. Both p and q are reduced to the minimal form.
If p is 0 print 0 or if p is equal to q print 1, otherwise print p and q alone.
Examples:

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
Explanation : 
First query is to find probability of even 
element in range [2, 2]. Since range contains 
a single element 5 which is odd, the answer 
is 0. Second query is to find probability of
odd element in range [2, 5]. There are 3
odd elements in range probability is 3/4.
Third query is for even elements in range
from 1 to 4. Since there are equal even
and odd elements, probability is 2/4
which is 1/2.

The idea is to maintain two arrays, say even[] and odd[], which maintain the number of even or odd element upto index i. Now, to answer each query, we can compute result denominator q by finding number of element in the given query range. To find result numerator, we remove number of elements upto l – 1 from elements upto r.
To output the answer in minimal form, we find the GCD of p and q and output p/gcd and q/gcd. For answer 0 and 1, we will explicitly specify the conditions.

Below is C++ implementation of this approach:

// CPP program to find probability of even
// or odd elements in a given range.
#include <bits/stdc++.h>
using namespace std;
// Number of tuples in a query
#define C 3
// Solve each query of K L R form
void solveQuery(int arr[], int n, int Q,
                           int query[][C])
{
    // To count number of odd and even
    // number upto i-th index.
    int even[n + 1];
    int odd[n + 1];
    even[0] = odd[0] = 0;
    // Counting number of odd and even
    // integer upto index i
    for (int i = 0; i < n; i++) {
        // If number is odd, increment the
        // count of odd frequency leave
        // even frequency same.
        if (arr[i] & 1) {
            odd[i + 1] = odd[i] + 1;
            even[i + 1] = even[i];
        }
        // If number is even, increment the
        // count of even frequency leave odd
        // frequency same.
        else {
            even[i + 1] = even[i] + 1;
            odd[i + 1] = odd[i];
        }
    }
    // To solve each query
    for (int i = 0; i < Q; i++) {
        int r = query[i][2];
        int l = query[i][1];
        int k = query[i][0];
        // Counting total number of element in
        // current query
        int q = r - l + 1;
        int p;
        // Counting number of odd or even element
        // in current query range
        if (k)
            p = odd[r] - odd[l - 1];
        else
            p = even[r] - even[l - 1];
        // If frequency is 0, output 0
        if (!p)
            cout << "0" << endl;
        // If frequency is equal to number of 
        // element in current range output 1.
        else if (p == q)
            cout << "1" << endl;
        // Else find the GCD of both. If yes,
        // output by dividing both number by gcd
        // to output the answer in reduced form.
        else {
            int g = __gcd(p, q);
            cout << p / g << " " << q / g << endl;
        }
    }
}
// Driven Program
int main()
{
    int arr[] = { 6, 5, 2, 1, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int Q = 2;
    int query[Q][C] = {
        { 0, 2, 2 },
        { 1, 2, 5 }
    };
    solveQuery(arr, n, Q, query);
    return 0;
}

Output:

0
3 4

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
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rakesh

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