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
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
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 checkbool 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 codeint 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 checkstatic 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 codepublic 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 checkdef 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 codearr = [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 notusing System; class GFG{ // Function to checkstatic 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 codestatic 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 checkfunction 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.
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 0arr=[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)); }}} |
1
Time Complexity: O(n), where n is the length of the array.
Auxiliary Space: O(1)
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