Split the array into odd number of segments of odd lengths

Given an array of n length. The task is to check if the given array can be split into an odd number of sub-segment starting with odd integer and also the length of sub-segment must be odd. If possible the print ‘1’ else print ‘0’.

Examples:

Input: arr[] = {1, 3, 1}
Output: 1
1, 3, 1, can be split into 3 sub-segments of length odd i.e. 1
with starting and ending elements as odd.

Input: arr[] = {1, 3, 1, 1}
Output: 0

Approach:
Points to think before proceeding the actual solution:

• Order of sets made by adding an even number of the set with odd order is always even and vice-versa.
• Order of sets made by adding an odd number of the set with odd order is always odd and vice-versa.

Using the above lemma of number theory, the solution of the given problem can be easily calculated with below-proposed logic:
If the given array starts and ends with an odd integer and also the size of the array is odd then only the given array can be broken down into an odd number of sub-segments starting & ending with an odd number with odd size, else not.

Below is the implementation of the above approach:

C++

// CPP to check whether given
// array is breakable or not
#include <bits/stdc++.h>
using namespace std;

// Function to check
bool checkArray(int arr[], int n)
{
    // Check the result by processing
    // the first & last element and size
    return (arr[0] % 2) && (arr[n - 1] % 2) && (n % 2);
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << (int)checkArray(arr, n);

    return 0;
}

Java

// Java to check whether given
// array is breakable or not

class GFG
{

// Function to check
static int checkArray(int []arr, int n)
{
    // Check the result by processing
    // the first & last element and size
    return ((arr[0] % 2) > 0 &&
            (arr[n - 1] % 2) > 0 &&
            (n % 2) > 0) ? 1 : 0;
}

// Driver code
public static void main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5 };
    int n = arr.length;
    System.out.println(checkArray(arr, n));
}
}

// This code contributed by Rajput-Ji

Python3

# Python3 to check whether given
# array is breakable or not

# Function to check
def checkArray(arr, n):

    # Check the result by processing
    # the first & last element and size
    return ((arr[0] % 2) and
            (arr[n - 1] % 2) and (n % 2))

# Driver code
arr = [1, 2, 3, 4, 5 ]
n = len(arr);
if checkArray(arr, n):
    print(1)
else:
    print(0)

# This code is contributed
# by Mohit Kumar

C#

// C# to check whether given
// array is breakable or not
using System;

class GFG
{

// Function to check
static int checkArray(int []arr, int n)
{
    // Check the result by processing
    // the first & last element and size
    return ((arr[0] % 2) > 0 &&
            (arr[n - 1] % 2) > 0 &&
               (n % 2) > 0) ? 1 : 0;
}

// Driver code
static void Main()
{
    int []arr = { 1, 2, 3, 4, 5 };
    int n = arr.Length;
    Console.WriteLine(checkArray(arr, n));

}
}

// This code is contributed by mits

PHP

<?php
// PHP to check whether given
// array is breakable or not

// Function to check
function checkArray($arr, $n)
{
    // Check the result by processing
    // the first & last element and size
    return ($arr[0] % 2) &&
           ($arr[$n - 1] % 2) && ($n % 2);
}

// Driver code
$arr = array( 1, 2, 3, 4, 5 );
$n = sizeof($arr);
echo checkArray($arr, $n);

// This code is contributed by Ryuga
?>

Javascript

<script>

// javascript to check whether given
// array is breakable or not


// Function to check
function checkArray(arr,  n)
{
    // Check the result by processing
    // the first & last element and size
    return ((arr[0] % 2) > 0 &&
            (arr[n - 1] % 2) > 0 &&
               (n % 2) > 0) ? 1 : 0;
}

// Driver code

    var arr = [ 1, 2, 3, 4, 5 ];
    var n = arr.length;
    document.write(checkArray(arr, n));



</script>

Output:

1

Time Complexity : O(1), since there is only basic arithmetic that takes constant time.

Auxiliary Space: O(1), since no extra space has been taken.