Count ways to construct array with even product from given array such that absolute difference of same indexed elements is at most 1
Last Updated :
21 Apr, 2021
Given an array A[] of size N, the task is to count the numbers of ways to construct an array B[] of size N, such that the absolute difference at the same indexed elements must be less than or equal to 1, i.e. abs(A[i] - B[i]) ? 1, and the product of elements of the array B[] must be an even number.
Examples:
Input: A[] = { 2, 3 }
Output: 7
Explanation:
Possible values of the array B[] are { { 1, 2 }, { 1, 4 }, { 2, 2 }, { 2, 4 }, { 3, 2 }, { 3, 4 } }
Therefore, the required output is 7.
Input: A[] = { 90, 52, 56, 71, 44, 8, 13, 30, 57, 84 }
Output: 58921
Approach: The idea is to first count the number of ways to construct an array, B[] such that abs(A[i] - B[i]) <= 1 and then remove those arrays whose product of elements is not an even number. Follow the below steps to solve the problem:
- Possible values of B[i] such that abs(A[i] - B[i]) <= 1 are { A[i], A[i] + 1, A[i] - 1 }. Therefore, the total count of ways to construct an array, B[] such that abs(A[i] - B[i]) less than or equal to 1 is 3N.
- Traverse the array and store the count of even numbers in the array A[] say, X.
- If A[i] is an even number then (A[i] - 1) and (A[i] + 1) must be an odd number. Therefore, the total count of ways to the construct array, B[] whose product is not an even number is 2X.
- Finally, print the value of (3N - 2X).
Below is the implementation of the above approach:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find count the ways to construct
// an array, B[] such that abs(A[i] - B[i]) <=1
// and product of elements of B[] is even
void cntWaysConsArray(int A[], int N)
{
// Stores count of arrays B[] such
// that abs(A[i] - B[i]) <=1
int total = 1;
// Stores count of arrays B[] whose
// product of elements is not even
int oddArray = 1;
// Traverse the array
for (int i = 0; i < N; i++) {
// Update total
total = total * 3;
// If A[i] is an even number
if (A[i] % 2 == 0) {
// Update oddArray
oddArray *= 2;
}
}
// Print 3^N - 2^X
cout << total - oddArray << "\n";
}
// Driver Code
int main()
{
int A[] = { 2, 4 };
int N = sizeof(A) / sizeof(A[0]);
cntWaysConsArray(A, N);
return 0;
}
// Java Program to implement the
// above approach
import java.util.*;
class GFG
{
// Function to find count the ways to construct
// an array, B[] such that abs(A[i] - B[i]) <=1
// and product of elements of B[] is even
static void cntWaysConsArray(int A[], int N)
{
// Stores count of arrays B[] such
// that abs(A[i] - B[i]) <=1
int total = 1;
// Stores count of arrays B[] whose
// product of elements is not even
int oddArray = 1;
// Traverse the array
for (int i = 0; i < N; i++)
{
// Update total
total = total * 3;
// If A[i] is an even number
if (A[i] % 2 == 0)
{
// Update oddArray
oddArray *= 2;
}
}
// Print 3^N - 2^X
System.out.println( total - oddArray);
}
// Driver Code
public static void main(String[] args)
{
int A[] = { 2, 4 };
int N = A.length;
cntWaysConsArray(A, N);
}
}
// This code is contributed by code_hunt.
# Python3 program to implement
# the above approach
# Function to find count the ways to construct
# an array, B[] such that abs(A[i] - B[i]) <=1
# and product of elements of B[] is even
def cntWaysConsArray(A, N) :
# Stores count of arrays B[] such
# that abs(A[i] - B[i]) <=1
total = 1;
# Stores count of arrays B[] whose
# product of elements is not even
oddArray = 1;
# Traverse the array
for i in range(N) :
# Update total
total = total * 3;
# If A[i] is an even number
if (A[i] % 2 == 0) :
# Update oddArray
oddArray *= 2;
# Print 3^N - 2^X
print(total - oddArray);
# Driver Code
if __name__ == "__main__" :
A = [ 2, 4 ];
N = len(A);
cntWaysConsArray(A, N);
# This code is contributed by AnkThon
// C# program to implement the
// above approach
using System;
class GFG{
// Function to find count the ways to construct
// an array, []B such that abs(A[i] - B[i]) <=1
// and product of elements of []B is even
static void cntWaysConsArray(int []A, int N)
{
// Stores count of arrays []B such
// that abs(A[i] - B[i]) <=1
int total = 1;
// Stores count of arrays []B whose
// product of elements is not even
int oddArray = 1;
// Traverse the array
for(int i = 0; i < N; i++)
{
// Update total
total = total * 3;
// If A[i] is an even number
if (A[i] % 2 == 0)
{
// Update oddArray
oddArray *= 2;
}
}
// Print 3^N - 2^X
Console.WriteLine(total - oddArray);
}
// Driver Code
public static void Main(String[] args)
{
int []A = { 2, 4 };
int N = A.Length;
cntWaysConsArray(A, N);
}
}
// This code is contributed by shikhasingrajput
<script>
// Javascript Program to implement the
// above approach
// Function to find count the ways to construct
// an array, B such that abs(A[i] - B[i]) <=1
// and product of elements of B is even
function cntWaysConsArray(A, N)
{
// Stores count of arrays B such
// that abs(A[i] - B[i]) <=1
var total = 1;
// Stores count of arrays B whose
// product of elements is not even
var oddArray = 1;
// Traverse the array
for(i = 0; i < N; i++)
{
// Update total
total = total * 3;
// If A[i] is an even number
if (A[i] % 2 == 0)
{
// Update oddArray
oddArray *= 2;
}
}
// Print var 3^N - 2^X
document.write(total - oddArray);
}
// Driver Code
var A = [ 2, 4 ];
var N = A.length;
cntWaysConsArray(A, N);
// This code is contributed by umadevi9616
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
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