Given an integer array arr[], the task is to find the length of the longest subarray with an equal number of odd and even elements.
Examples:
Input: arr[] = {1, 2, 1, 2}
Output: 4
Explanation:
Subarrays in the given array are –
{{1}, {1, 2}, {1, 2, 1}, {1, 2, 1, 2}, {2}, {2, 1}, {2, 1, 2}, {1}, {1, 2}, {2}}
where the length of the longest subarray with an equal number of even and odd elements is 4 – {1, 2, 1, 2}Input: arr[] = {12, 4, 7, 8, 9, 2, 11, 0, 2, 13}
Output: 8
Naive Approach: Simple solution is to consider all subarrays one by one and check the count of even and odd elements in the subarray. Then find the maximum length of those subarray that contain an equal number of even elements and odd elements.
Steps to implement-
- Initialize a variable ans to store the final answer
- Run two loops to find all subarrays and simultaneously find their length
- After that count number of odd elements and the number of even elements in that
- If that subarray contains an equal number of odd and even elements then
- Update ans with maximum of ans and the length of that subarray
Code-
C++
// C++ program to find the length// of the longest sub-array with an// equal number of odd and even elements#include <bits/stdc++.h>using namespace std;// Function that returns the length of// the longest sub-array with an equal// number of odd and even elementsint maxSubarrayLength(int arr[], int N){ //To store answer int ans=0; //Find all subarray for(int i=0;i<N;i++){ //To store length of subarray int length=0; for(int j=i;j<N;j++){ //Increment the length length++; //To store count of even elements int even=0; //To store count of odd elements int odd=0; //Find the number of even elements and odd elements for(int k=i;k<=j;k++){ if(arr[k]%2==0){even++;} else if(arr[k]%2==1){odd++;} } //When subarray contains equal number of odd //and even elements if(odd==even){ ans=max(ans,length); } } } return ans;}// Driver Codeint main(){ int arr[] = { 12, 4, 7, 8, 9, 2, 11, 0, 2, 13 }; int n = sizeof(arr) / sizeof(arr[0]); cout << maxSubarrayLength(arr, n); return 0;} |
Java
import java.util.*;public class Main { // Function that returns the length of // the longest sub-array with an equal // number of odd and even elements static int maxSubarrayLength(int[] arr, int N) { // To store answer int ans = 0; // Find all subarrays for (int i = 0; i < N; i++) { // To store length of subarray int length = 0; for (int j = i; j < N; j++) { // Increment the length length++; // To store count of even elements int even = 0; // To store count of odd elements int odd = 0; // Find the number of even elements and odd // elements for (int k = i; k <= j; k++) { if (arr[k] % 2 == 0) { even++; } else if (arr[k] % 2 == 1) { odd++; } } // When subarray contains equal number of // odd and even elements if (odd == even) { ans = Math.max(ans, length); } } } return ans; } // Driver Code public static void main(String[] args) { int[] arr = { 12, 4, 7, 8, 9, 2, 11, 0, 2, 13 }; int n = arr.length; System.out.println(maxSubarrayLength(arr, n)); }} |
Python
# Function that returns the length of# the longest sub-array with an equal# number of odd and even elementsdef maxSubarrayLength(arr, N): # To store the answer ans = 0 # Find all subarrays for i in range(N): # To store the length of subarray length = 0 for j in range(i, N): # Increment the length length += 1 # To store the count of even elements even = 0 # To store the count of odd elements odd = 0 # Find the number of even elements and odd elements for k in range(i, j + 1): if arr[k] % 2 == 0: even += 1 elif arr[k] % 2 == 1: odd += 1 # When the subarray contains an equal number of odd # and even elements if odd == even: ans = max(ans, length) return ans# Driver Codeif __name__ == "__main__": arr = [12, 4, 7, 8, 9, 2, 11, 0, 2, 13] n = len(arr) print(maxSubarrayLength(arr, n))# This code is contributed by Taranpreet Singh. |
C#
using System;class GFG{ // Function that returns the length of // the longest sub-array with an equal // number of odd and even elements static int MaxSubarrayLength(int[] arr, int N) { // To store the answer int ans = 0; // Find all subarrays for (int i = 0; i < N; i++) { // To store the length of the subarray int length = 0; for (int j = i; j < N; j++) { // Increment the length length++; // To store count of even elements int even = 0; // To store count of odd elements int odd = 0; // Find the number of even elements and odd elements for (int k = i; k <= j; k++) { if (arr[k] % 2 == 0) { even++; } else if (arr[k] % 2 == 1) { odd++; } } // When subarray contains an equal number of odd // and even elements if (odd == even) { ans = Math.Max(ans, length); } } } return ans; } // Driver Code static void Main() { int[] arr = { 12, 4, 7, 8, 9, 2, 11, 0, 2, 13 }; int n = arr.Length; Console.WriteLine(MaxSubarrayLength(arr, n)); }} |
Javascript
// JavaScript program to find the length// of the longest sub-array with an// equal number of odd and even elements // Function that returns the length of// the longest sub-array with an equal// number of odd and even elements function maxSubarrayLength(arr, N) { //To store answer let ans=0; //Find all subarray for(let i=0;i<N;i++){ //To store length of subarray let length=0; for(let j=i;j<N;j++){ //Increment the length length++; //To store count of even elements let even=0; //To store count of odd elements let odd=0; //Find the number of even elements and odd elements for(let k=i;k<=j;k++){ if(arr[k]%2==0){even++;} else if(arr[k]%2==1){odd++;} } //When subarray contains equal number of odd //and even elements if(odd==even){ ans=Math.max(ans,length); } } } return ans; }// Driver Codevar arr = [12, 4, 7, 8, 9, 2, 11, 0, 2, 13 ];var n = arr.length;console.log(maxSubarrayLength(arr, n)); |
Output
8
Time Complexity: O(N3), because of two nested loops to find all subarrays and a third loop to find the number of odd and even elements in a subarray
Auxiliary Space: O(1), because no extra space has been used
Efficient Approach: The idea is to consider the odd elements as 1 and even elements as -1 and return the length of the longest sub-array with the sum equal to 0. The subarray with a given sum can be found using this method.
Time Complexity: O(N)
Below is the implementation of the above approach:
C++
// C++ program to find the length// of the longest sub-array with an// equal number of odd and even elements#include <bits/stdc++.h>using namespace std;// Function that returns the length of// the longest sub-array with an equal// number of odd and even elementsint maxSubarrayLength(int* A, int N){ // Initialize variable to store result int maxLen = 0; // Initialize variable to store sum int curr_sum = 0; // Create an empty map to store // index of the sum unordered_map<int, int> hash; // Loop through the array for (int i = 0; i < N; i++) { if (A[i] % 2 == 0) curr_sum -= 1; else curr_sum += 1; // Check if number of even and // odd elements are equal if (curr_sum == 0) maxLen = max(maxLen, i + 1); // If curr_sum already exists in map // we have a subarray with 0 sum, i.e, // equal number of odd and even number if (hash.find(curr_sum) != hash.end()) maxLen = max(maxLen, i - hash[curr_sum]); // Store the index of the sum else hash[curr_sum] = i; } return maxLen;}// Driver Codeint main(){ int arr[] = { 12, 4, 7, 8, 9, 2, 11, 0, 2, 13 }; int n = sizeof(arr) / sizeof(arr[0]); cout << maxSubarrayLength(arr, n); return 0;} |
Java
// Java program to find the length// of the longest sub-array with an// equal number of odd and even elementsimport java.util.*;class GFG{// Function that returns the length of// the longest sub-array with an equal// number of odd and even elementsstatic int maxSubarrayLength(int []A, int N){ // Initialize variable to store result int maxLen = 0; // Initialize variable to store sum int curr_sum = 0; // Create an empty map to store // index of the sum HashMap<Integer,Integer> hash = new HashMap<Integer,Integer>(); // Loop through the array for (int i = 0; i < N; i++) { if (A[i] % 2 == 0) curr_sum -= 1; else curr_sum += 1; // Check if number of even and // odd elements are equal if (curr_sum == 0) maxLen = Math.max(maxLen, i + 1); // If curr_sum already exists in map // we have a subarray with 0 sum, i.e, // equal number of odd and even number if (hash.containsKey(curr_sum)) maxLen = Math.max(maxLen, i - hash.get(curr_sum)); // Store the index of the sum else { hash.put(curr_sum, i); } } return maxLen;}// Driver Codepublic static void main(String[] args){ int arr[] = { 12, 4, 7, 8, 9, 2, 11, 0, 2, 13 }; int n = arr.length; System.out.print(maxSubarrayLength(arr, n));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to find the length # of the longest sub-array with an # equal number of odd and even elements # Function that returns the length of # the longest sub-array with an equal # number of odd and even elements def maxSubarrayLength(A, N) : # Initialize variable to store result maxLen = 0; # Initialize variable to store sum curr_sum = 0; # Create an empty map to store # index of the sum hash = {}; # Loop through the array for i in range(N) : if (A[i] % 2 == 0) : curr_sum -= 1; else : curr_sum += 1; # Check if number of even and # odd elements are equal if (curr_sum == 0) : maxLen = max(maxLen, i + 1); # If curr_sum already exists in map # we have a subarray with 0 sum, i.e, # equal number of odd and even number if curr_sum in hash : maxLen = max(maxLen, i - hash[curr_sum]); # Store the index of the sum else : hash[curr_sum] = i; return maxLen; # Driver Code if __name__ == "__main__" : arr = [ 12, 4, 7, 8, 9, 2, 11, 0, 2, 13 ]; n = len(arr); print(maxSubarrayLength(arr, n)); # This code is contributed by AnkitRai01 |
C#
// C# program to find the length// of the longest sub-array with an// equal number of odd and even elementsusing System;using System.Collections.Generic;class GFG{// Function that returns the length of// the longest sub-array with an equal// number of odd and even elementsstatic int maxSubarrayLength(int []A, int N){ // Initialize variable to store result int maxLen = 0; // Initialize variable to store sum int curr_sum = 0; // Create an empty map to store // index of the sum Dictionary<int, int> hash = new Dictionary<int, int>(); // Loop through the array for (int i = 0; i < N; i++) { if (A[i] % 2 == 0) curr_sum -= 1; else curr_sum += 1; // Check if number of even and // odd elements are equal if (curr_sum == 0) maxLen = Math.Max(maxLen, i + 1); // If curr_sum already exists in map // we have a subarray with 0 sum, i.e, // equal number of odd and even number if (hash.ContainsKey(curr_sum)) maxLen = Math.Max(maxLen, i - hash[curr_sum]); // Store the index of the sum else { hash.Add(curr_sum, i); } } return maxLen;}// Driver Codepublic static void Main(String[] args){ int []arr = { 12, 4, 7, 8, 9, 2, 11, 0, 2, 13 }; int n = arr.Length; Console.Write(maxSubarrayLength(arr, n));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program to find the length// of the longest sub-array with an// equal number of odd and even elements// Function that returns the length of// the longest sub-array with an equal// number of odd and even elementsfunction maxSubarrayLength(A, N){ // Initialize variable to store result var maxLen = 0; // Initialize variable to store sum var curr_sum = 0; // Create an empty map to store // index of the sum var hash = new Map(); // Loop through the array for (var i = 0; i < N; i++) { if (A[i] % 2 == 0) curr_sum -= 1; else curr_sum += 1; // Check if number of even and // odd elements are equal if (curr_sum == 0) maxLen = Math.max(maxLen, i + 1); // If curr_sum already exists in map // we have a subarray with 0 sum, i.e, // equal number of odd and even number if (hash.has(curr_sum)) maxLen = Math.max(maxLen, i - hash.get(curr_sum)); // Store the index of the sum else hash.set(curr_sum, i); } return maxLen;}// Driver Codevar arr = [12, 4, 7, 8, 9, 2, 11, 0, 2, 13 ];var n = arr.length;document.write( maxSubarrayLength(arr, n));</script> |
Output
8
Time Complexity: O(n)
Auxiliary Space: O(n), where n is the size of the given array.
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