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.

Implementation:

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

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>

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
?>

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.

Approach#2: Using Greedy Approach

We can use a greedy approach to split the array into odd number of segments of odd lengths. We can start with the first element of the array and keep adding odd-length segments until we reach the end of the array.

Algorithm

1. Initialize a variable ‘count’ to 0.
2. Initialize a variable ‘i’ to 0.
3. Iterate over the array from index 0 to n-1, where n is the length of the array.
4. If the current element is odd and the sum of its index and value is odd, increment ‘count’.
5. If ‘count’ is odd and we have reached the end of the array, return 1, else return 0.

C++

#include <iostream>
#include <vector>

using namespace std;

int splitArrayIntoOddSegments(vector<int>& arr) {
    int count = 0;
    for (int i = 0; i < arr.size(); i++) {
        if (arr[i] % 2 != 0 && (i + arr[i]) % 2 != 0) {
            count += 1;
        }
    }
    if (count % 2 != 0 && arr.size() % 2 != 0) {
        return 1;
    } else {
        return 0;
    }
}

int main() {
    vector<int> arr = {1, 2, 3, 4, 5};
    cout << splitArrayIntoOddSegments(arr) << endl;
    return 0;
}

Java

public class Main {
    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 4, 5};
        System.out.println(splitArrayIntoOddSegments(arr));
    }

    public static int splitArrayIntoOddSegments(int[] arr) {
        int count = 0;
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] % 2 != 0 && (i + arr[i]) % 2 != 0) {
                count++;
            }
        }
        if (count % 2 != 0 && arr.length % 2 != 0) {
            return 1;
        } else {
            return 0;
        }
    }
}

Python3

def split_array_into_odd_segments(arr):
    count = 0
    for i in range(len(arr)):
        if arr[i] % 2 != 0 and (i + arr[i]) % 2 != 0:
            count += 1
    if count % 2 != 0 and len(arr) % 2 != 0:
        return 1
    else:
        return 0
arr=[1, 2, 3, 4, 5 ]
print( split_array_into_odd_segments(arr))

C#

using System;
using System.Collections.Generic;

namespace SplitArrayIntoOddSegments {
class Program {
    static int SplitArrayIntoOddSegments(List<int> arr)
    {
        int count = 0;
        for (int i = 0; i < arr.Count; i++) {
            if (arr[i] % 2 != 0 && (i + arr[i]) % 2 != 0) {
                count += 1;
            }
        }
        if (count % 2 != 0 && arr.Count % 2 != 0) {
            return 1;
        }
        else {
            return 0;
        }
    }
    static void Main(string[] args)
    {
        List<int> arr = new List<int>{ 1, 2, 3, 4, 5 };
        Console.WriteLine(SplitArrayIntoOddSegments(arr));
    }
}
}

Javascript

// Function to split an array into odd segments
function splitArrayIntoOddSegments(arr) {
    let count = 0;  // Initialize a count to track valid segments

    for (let i = 0; i < arr.length; i++) {
        // Check if the current element is odd and its position is such that
        // it will create an odd-sized segment
        if (arr[i] % 2 !== 0 && (i + arr[i]) % 2 !== 0) {
            count += 1;
        }
    }

    // Check if the total count of valid segments and the array length are both odd
    if (count % 2 !== 0 && arr.length % 2 !== 0) {
        return 1;
    } else {
        return 0;
    }
}

const arr = [1, 2, 3, 4, 5];  // Example array
console.log(splitArrayIntoOddSegments(arr));

Output

1

Time Complexity: O(n), where n is the length of the array.
Auxiliary Space: O(1)