Maximum distance between two even integers in a given array

Given an array arr[] having N integers, the task is to find the maximum distance between any two occurrences of even integers.

Examples:

Input: arr[] = {1, 2, 3, 4, 5, 6, 7}
Output: 4
Explanation: The distance between arr[1] = 2 and arr[5] = 6 is 4 which is the maximum distance between two even integers present in the given array.

Input: arr[] = {3, 5, 6, 9, 11}
Output: 0
Explanation: The given array contains less than 2 even integers. Hence, the maximum distance is 0.

Naive Approach: The given problem can be solved by checking the distance between all pairs of even integers occurring in the array and maintaining the maximum in them which will be the required answer.

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach: This problem can be solved using Two Pointer approach. 

• Handle the case with less than 2 even integers separately.
• The index of maximum distance even integers will be the index of the first and the last occurrence of even integers.
• This can be done simply by traversing the array using two pointers – one from start and one from end.
• As soon an even integer is reached from both pointers, return the distance between them.

Below is the implementation of the above approach:

C++

// C++ program to of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to calculate the maximum
// distance between any two occurrence
// of even integers in the given array
int maximizeDistance(int arr[], int n)
{
    // Stores the index of
    // 1st even integer
    int i = 0;

    // Traverse array arr[]
    while (i < n) {
        if (arr[i] % 2 == 0) {
            break;
        }
        i++;
    }
    // Stores the index of
    // last even integer
    int j = n - 1;

    // Traverse array arr[]
    // in reverse direction
    while (j >= 0) {
        if (arr[j] % 2 == 0) {
            break;
        }
        j--;
    }

    // Case where arr[] has less
    // that two even integers
    if (i >= j) {
        return 0;
    }

    // Return Answer
    return j - i;
}

// Driver Code
int main()
{
    int arr[] = { 3, 4, 5, 6, 7, 8 };
    int N = sizeof(arr) / sizeof(int);

    cout << maximizeDistance(arr, N);

    return 0;
}

Java

// Java program to of the above approach
import java.util.*;
public class GFG {

  // Function to calculate the maximum
  // distance between any two occurrence
  // of even integers in the given array
  static int maximizeDistance(int arr[], int n)
  {
    // Stores the index of
    // 1st even integer
    int i = 0;

    // Traverse array arr[]
    while (i < n) {
      if (arr[i] % 2 == 0) {
        break;
      }
      i++;
    }
    // Stores the index of
    // last even integer
    int j = n - 1;

    // Traverse array arr[]
    // in reverse direction
    while (j >= 0) {
      if (arr[j] % 2 == 0) {
        break;
      }
      j--;
    }

    // Case where arr[] has less
    // that two even integers
    if (i >= j) {
      return 0;
    }

    // Return Answer
    return j - i;
  }

  // Driver Code
  public static void main(String args[])
  {
    int arr[] = { 3, 4, 5, 6, 7, 8 };
    int N = arr.length;

    System.out.print(maximizeDistance(arr, N));
  }
}

// This code is contributed by Samim Hossain Mondal.

Python3

# python3 program to of the above approach

# Function to calculate the maximum
# distance between any two occurrence
# of even integers in the given array
def maximizeDistance(arr, n):

    # Stores the index of
    # 1st even integer
    i = 0

    # Traverse array arr[]
    while (i < n):
        if (arr[i] % 2 == 0):
            break

        i += 1

    # Stores the index of
    # last even integer
    j = n - 1

    # Traverse array arr[]
    # in reverse direction
    while (j >= 0):
        if (arr[j] % 2 == 0):
            break

        j -= 1

    # Case where arr[] has less
    # that two even integers
    if (i >= j):
        return 0

    # Return Answer
    return j - i

# Driver Code
if __name__ == "__main__":

    arr = [3, 4, 5, 6, 7, 8]
    N = len(arr)

    print(maximizeDistance(arr, N))

# This code is contributed by rakeshsahni

C#

// C# program to of the above approach
using System;
class GFG {

  // Function to calculate the maximum
  // distance between any two occurrence
  // of even integers in the given array
  static int maximizeDistance(int[] arr, int n)
  {

    // Stores the index of
    // 1st even integer
    int i = 0;

    // Traverse array arr[]
    while (i < n) {
      if (arr[i] % 2 == 0) {
        break;
      }
      i++;
    }
    // Stores the index of
    // last even integer
    int j = n - 1;

    // Traverse array arr[]
    // in reverse direction
    while (j >= 0) {
      if (arr[j] % 2 == 0) {
        break;
      }
      j--;
    }

    // Case where arr[] has less
    // that two even integers
    if (i >= j) {
      return 0;
    }

    // Return Answer
    return j - i;
  }

  // Driver Code
  public static void Main()
  {
    int[] arr = { 3, 4, 5, 6, 7, 8 };
    int N = arr.Length;

    Console.Write(maximizeDistance(arr, N));
  }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
        // JavaScript code for the above approach

        // Function to calculate the maximum
        // distance between any two occurrence
        // of even integers in the given array
        function maximizeDistance(arr, n)
        {

            // Stores the index of
            // 1st even integer
            let i = 0;

            // Traverse array arr[]
            while (i < n) {
                if (arr[i] % 2 == 0) {
                    break;
                }
                i++;
            }
            // Stores the index of
            // last even integer
            let j = n - 1;

            // Traverse array arr[]
            // in reverse direction
            while (j >= 0) {
                if (arr[j] % 2 == 0) {
                    break;
                }
                j--;
            }

            // Case where arr[] has less
            // that two even integers
            if (i >= j) {
                return 0;
            }

            // Return Answer
            return j - i;
        }

        // Driver Code

        let arr = [3, 4, 5, 6, 7, 8];
        let N = arr.length;

        document.write(maximizeDistance(arr, N));

     // This code is contributed by Potta Lokesh
    </script>

Output

4

Time Complexity: O(N)
Auxiliary Space: O(1)