Array Index with same count of even or odd numbers on both sides

0 0 11 minutes read

Given an array of N integers. We need to find an index such that Frequency of Even numbers on its left side is equal to the frequency of even numbers on its right sides Or frequency of odd numbers on its left side is equal to the frequency of Odd numbers on its right sides. If No such index exist in an array print -1 Else print required index. Note-(If more than one index exist then return index that comes first)

Examples:

Input : arr[] = {4, 3, 2, 1, 2, 4}  
Output : index = 2
Explanation: At index 2, there is one
odd number on its left and one odd on
its right.

Input :  arr[] = { 1, 2, 4, 5, 8, 3, 12}
Output : index = 3

Method 1 : (Simple Approach)

Run two loops. For every element, count evens and odds on its left and right sides.

Implementation:

C++

// CPP program to find an index which has
// same number of even elements on left and
// right, Or same number of odd elements on
// left and right.
#include <iostream>
using namespace std;
 
// Function to find index
int findIndex(int arr[], int n) {
 
  for (int i = 0; i < n; i++) {
    int odd_left = 0, even_left = 0;
    int odd_right = 0, even_right = 0;
 
    // To count Even and Odd numbers of left side
    for (int j = 0; j < i; j++) {
      if (arr[j] % 2 == 0)
        even_left++;
      else
        odd_left++;
    }
 
    // To count Even and Odd numbers of right side
    for (int k = n - 1; k > i; k--) {
      if (arr[k] % 2 == 0)
        even_right++;
      else
        odd_right++;
    }
 
    // To check Even Or Odd of Both sides are equal or not
    if (even_right == even_left || odd_right == odd_left)
      return i;
  }
 
  return -1;
}
 
// Driver's Function
int main() {
  int arr[] = {4, 3, 2, 1, 2};
  int n = sizeof(arr) / sizeof(arr[0]);
  int index = findIndex(arr, n);
 
  ((index == -1) ? cout << "-1" : 
          cout << "index = " << index);
  return 0;
}

Java

// Java program to find
// an index which has
// same number of even
// elements on left and
// right, Or same number
// of odd elements on
// left and right.
 
class GFG{
 
// Function to find index
static int findIndex(int arr[], int n) {
 
for (int i = 0; i < n; i++) {
 
    int odd_left = 0, even_left = 0;
    int odd_right = 0, even_right = 0;
 
    // To count Even and Odd
        // numbers of left side
    for (int j = 0; j < i; j++) {
    if (arr[j] % 2 == 0)
        even_left++;
    else
        odd_left++;
    }
 
    // To count Even and Odd
        // numbers of right side
    for (int k = n - 1; k > i; k--) {
    if (arr[k] % 2 == 0)
        even_right++;
    else
        odd_right++;
    }
 
    // To check Even Or Odd of Both
        // sides are equal or not
    if (even_right == even_left || odd_right == odd_left)
    return i;
}
 
return -1;
 
}
 
// Driver's Function
public static void main(String[] args) {
 
int arr[] = {4, 3, 2, 1, 2};
int n = arr.length;
int index = findIndex(arr, n);
 
if (index == -1) 
    System.out.println("-1"); 
else{
    System.out.print("index = ");
    System.out.print(index);
}
}
}
 
// This code is contributed by
// Smitha Dinesh Semwal

Python3

'''Python program to find
   an index which has
   same number of even
   elements on left and
   right, Or same number
   of odd elements on
   left and right.'''
  
 
  
# Function to find index
def findIndex(arr,n):
 
    for i in range(n):
  
        odd_left = 0
        even_left = 0
        odd_right = 0
        even_right = 0
  
        # To count Even and Odd
        # numbers of left side
        for j in range(i):
            if (arr[j] % 2 == 0):
                even_left=even_left+1
            else:
                odd_left=odd_left+1
     
  
        # To count Even and Odd
        # numbers of right side
        for k in range(n - 1, i, -1):
            if (arr[k] % 2 == 0):
                even_right=even_right+1
            else:
                odd_right=odd_right+1
     
  
        # To check Even Or Odd of Both
        # sides are equal or not
        if (even_right == even_left and odd_right == odd_left):
            return i
  
    return -1
  
 
  
# Driver's Function
 
arr = [4, 3, 2, 1, 2]
n = len(arr)
index = findIndex(arr, n)
  
if (index == -1):
    print("-1") 
else:
    print("index = ", index)
 
 
 
# This code is contributed
# by Anant Agarwal.

C#

// C# program to find an index which has
// same number of even elements on left 
// and right, Or same number of odd 
// elements on left and right.
using System;
 
class GFG{
 
    // Function to find index
    static int findIndex(int []arr, int n) {
     
        for (int i = 0; i < n; i++) {
         
            int odd_left = 0, even_left = 0;
            int odd_right = 0, even_right = 0;
         
            // To count Even and Odd
            // numbers of left side
            for (int j = 0; j < i; j++) {
                if (arr[j] % 2 == 0)
                    even_left++;
                else
                    odd_left++;
            }
         
            // To count Even and Odd
            // numbers of right side
            for (int k = n - 1; k > i; k--) {
                if (arr[k] % 2 == 0)
                    even_right++;
                else
                    odd_right++;
            }
         
            // To check Even Or Odd of Both
            // sides are equal or not
            if (even_right == even_left ||
                        odd_right == odd_left)
                return i;
        }
         
        return -1;
    }
     
    // Driver's Function
    public static void Main() {
     
        int []arr = {4, 3, 2, 1, 2};
        int n = arr.Length;
         
        int index = findIndex(arr, n);
         
        if (index == -1) 
            Console.Write("-1"); 
        else{
            Console.Write("index = ");
            Console.Write(index);
        }
    }
}
 
// This code is contributed by vt_m.

PHP

<?php
// PHP program to find an index which has
// same number of even elements on left and
// right, Or same number of odd elements on
// left and right.
 
// Function to find index
function findIndex($arr, $n) 
{
 
    for ($i = 0; $i < $n; $i++)
    {
        $odd_left = 0; 
        $even_left = 0;
        $odd_right = 0; 
        $even_right = 0;
 
        // To count Even and Odd 
        // numbers of left side
        for ($j = 0; $j < $i; $j++)
        {
            if ($arr[$j] % 2 == 0)
                $even_left++;
            else
                $odd_left++;
        }
 
            // To count Even and Odd 
            // numbers of right side
            for ($k = $n - 1; $k > $i; $k--)
            {
                if ($arr[$k] % 2 == 0)
                    $even_right++;
                else
                    $odd_right++;
            }
 
        // To check Even Or Odd of
        // Both sides are equal or not
        if ($even_right == $even_left || 
                $odd_right == $odd_left)
        return $i;
    }
 
    return -1;
}
 
// Drivers Code
{
    $arr = array(4, 3, 2, 1, 2);
    $n = sizeof($arr) / sizeof($arr[0]);
    $index = findIndex($arr, $n);
     
    if($index == -1) 
    echo "-1";
    else
    echo ("index = $index" );
    return 0;
}
 
// This code is contributed by nitin mittal.
?>

Javascript

<script>
 
// Javascript program to find an index which has
// same number of even elements on left and
// right, Or same number of odd elements on
// left and right.
 
 
// Function to find index
function findIndex(arr, n) {
 
for (let i = 0; i < n; i++) {
    let odd_left = 0, even_left = 0;
    let odd_right = 0, even_right = 0;
 
    // To count Even and Odd numbers of left side
    for (let j = 0; j < i; j++) {
    if (arr[j] % 2 == 0)
        even_left++;
    else
        odd_left++;
    }
 
    // To count Even and Odd numbers of right side
    for (let k = n - 1; k > i; k--) {
    if (arr[k] % 2 == 0)
        even_right++;
    else
        odd_right++;
    }
 
    // To check Even Or Odd of Both sides are equal or not
    if (even_right == even_left || odd_right == odd_left)
    return i;
}
 
return -1;
}
 
// Driver's Function
 
let arr = [4, 3, 2, 1, 2];
let n = arr.length;
let index = findIndex(arr, n);
 
if (index == -1) 
            document.write("-1"); 
        else{
            document.write("index = ");
            document.write(index);
        }
 
 
//This code is contributed by Mayank Tyagi
</script>

Output:

index = 2

Time Complexity : O(n*n)
Auxiliary Space : O(1)

Method 2:(Efficient solution)

Create two vectors of pair types i.e v_left and v_right
v_left[i] stores the frequency of odd and even numbers of its left sides
v_right[i] stores the frequency of odd and even numbers of its right sides
Now check (v_left[i].first == v_right[i].first || v_left[i].second == v_right[i].second)
if True return i
At last if no such index exist return -1

Implementation:

C++

// CPP program to find an index which has
// same number of even elements on left and
// right, Or same number of odd elements on
// left and right.
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
 
// Function to Find index
int Find_Index(int n, int arr[]) {
 
  int odd = 0, even = 0;
 
  // Create two vectors of pair type
  vector<pair<int, int>> v_left, v_right;
 
  v_left.push_back(make_pair(odd, even));
  for (int i = 0; i < n - 1; i++) {
    if (arr[i] % 2 == 0)
      even++;
    else
      odd++;
 
    v_left.push_back(make_pair(odd, even));
  }
 
  odd = 0, even = 0;
  v_right.push_back(make_pair(odd, even));
  for (int i = n - 1; i > 0; i--) {
    if (arr[i] % 2 == 0)
      even++;
    else
      odd++;
 
    v_right.push_back(make_pair(odd, even));
  }
 
  reverse(v_right.begin(), v_right.end());
 
  for (int i = 0; i < v_left.size(); i++) {
 
    // To check even or odd of Both sides are 
    // equal or not
    if (v_left[i].first == v_right[i].first ||
        v_left[i].second == v_right[i].second)
      return i;
  }
  return -1;
}
 
// Driver's Function
int main() {
  int arr[] = {4, 3, 2, 1, 2};
  int n = sizeof(arr) / sizeof(arr[0]);
  int index = Find_Index(n, arr);
  ((index == -1) ? cout << "-1" : cout << "index = " << index);
  return 0;
}

Java

// Java program to find an index which has
// same number of even elements on left and
// right, Or same number of odd elements on
// left and right.
import java.util.*;
 
class GFG
{
 
    public static class pair
    {
 
        int first, second;
 
        pair(int f, int s) 
        {
            first = f;
            second = s;
        }
    };
 
    // Function to Find index
    static int Find_Index(int n, int arr[]) 
    {
 
        int odd = 0, even = 0;
 
        // Create two vectors of pair type
        Vector<pair> v_left = new Vector<>();;
        Vector<pair> v_right = new Vector<>();
 
        v_left.add(new pair(odd, even));
        for (int i = 0; i < n - 1; i++) 
        {
            if (arr[i] % 2 == 0) 
            {
                even++;
            } 
            else
            {
                odd++;
            }
 
            v_left.add(new pair(odd, even));
        }
 
        odd = 0;
        even = 0;
        v_right.add(new pair(odd, even));
        for (int i = n - 1; i > 0; i--)
        {
            if (arr[i] % 2 == 0)
            {
                even++;
            } 
            else
            {
                odd++;
            }
 
            v_right.add(new pair(odd, even));
        }
        Collections.reverse(v_right);
 
        for (int i = 0; i < v_left.size(); i++)
        {
 
            // To check even or odd of Both sides are 
            // equal or not
            if (v_left.get(i).first == v_right.get(i).first
                    || v_left.get(i).second == v_right.get(i).second)
            {
                return i;
            }
        }
        return -1;
    }
 
    // Driver code
    public static void main(String[] args) 
    {
        int arr[] = {4, 3, 2, 1, 2};
        int n = arr.length;
        int index = Find_Index(n, arr);
        if (index == -1)
        {
            System.out.println("-1");
        }
        else
        {
            System.out.println("index = " + index);
        }
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3

# Python program to find an index which has
# same number of even elements on left and
# right, Or same number of odd elements on
# left and right.
class pair:
    def __init__(self, f, s):
        self.first = f
        self.second = s
 
# Function to Find index
def Find_Index(n, arr):
 
    odd, even = 0, 0
 
    # Create two vectors of pair type
    v_left = []
    v_right = []
 
    v_left.append(pair(odd, even))
    for i in range(n-1):
 
        if (arr[i] % 2 == 0):
            even += 1
             
        else:
            odd += 1
 
        v_left.append(pair(odd, even))
 
    odd = 0
    even = 0
    v_right.append(pair(odd, even))
    for i in range(n-1,0,-1):
        if (arr[i] % 2 == 0):
            even += 1
        else:
            odd += 1
 
        v_right.append(pair(odd, even))
         
    v_right = v_right[::-1]
 
    for i in range(len(v_left)):
 
        # To check even or odd of Both sides are
        # equal or not
        if (v_left[i].first == v_right[i].first or v_left[i].second == v_right[i].second):
            return i
         
    return -1
 
# Driver code
arr = [4, 3, 2, 1, 2]
n = len(arr)
index = Find_Index(n, arr)
if (index == -1):
    print("-1")
else:
    print("index = " + str(index))
 
# This code is contributed by shinjanpatra

C#

// C# program to find an index which has
// same number of even elements on left and
// right, Or same number of odd elements on
// left and right.
using System;
using System.Collections.Generic;
 
class GFG
{
    public class pair
    {
        public int first, second;
 
        public pair(int f, int s) 
        {
            first = f;
            second = s;
        }
    };
 
    // Function to Find index
    static int Find_Index(int n, int []arr) 
    {
        int odd = 0, even = 0;
 
        // Create two vectors of pair type
        List<pair> v_left = new List<pair>();;
        List<pair> v_right = new List<pair>();
 
        v_left.Add(new pair(odd, even));
        for (int i = 0; i < n - 1; i++) 
        {
            if (arr[i] % 2 == 0) 
            {
                even++;
            } 
            else
            {
                odd++;
            }
 
            v_left.Add(new pair(odd, even));
        }
 
        odd = 0;
        even = 0;
        v_right.Add(new pair(odd, even));
        for (int i = n - 1; i > 0; i--)
        {
            if (arr[i] % 2 == 0)
            {
                even++;
            } 
            else
            {
                odd++;
            }
 
            v_right.Add(new pair(odd, even));
        }
        v_right.Reverse();
 
        for (int i = 0; i < v_left.Count; i++)
        {
 
            // To check even or odd of Both sides are 
            // equal or not
            if (v_left[i].first == v_right[i].first || 
                v_left[i].second == v_right[i].second)
            {
                return i;
            }
        }
        return -1;
    }
 
    // Driver code
    public static void Main(String[] args) 
    {
        int []arr = {4, 3, 2, 1, 2};
        int n = arr.Length;
        int index = Find_Index(n, arr);
        if (index == -1)
        {
            Console.WriteLine("-1");
        }
        else
        {
            Console.WriteLine("index = " + index);
        }
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
// Javascript program to find an index which has
// same number of even elements on left and
// right, Or same number of odd elements on
// left and right.
class pair
{
    constructor(f,s)
    {
        this.first = f;
        this.second = s;
    }
}
 
// Function to Find index
function Find_Index(n,arr)
{
    let odd = 0, even = 0;
  
        // Create two vectors of pair type
        let v_left = [];
        let v_right = [];
  
        v_left.push(new pair(odd, even));
        for (let i = 0; i < n - 1; i++)
        {
            if (arr[i] % 2 == 0)
            {
                even++;
            }
            else
            {
                odd++;
            }
  
            v_left.push(new pair(odd, even));
        }
  
        odd = 0;
        even = 0;
        v_right.push(new pair(odd, even));
        for (let i = n - 1; i > 0; i--)
        {
            if (arr[i] % 2 == 0)
            {
                even++;
            }
            else
            {
                odd++;
            }
  
            v_right.push(new pair(odd, even));
        }
        v_right.reverse();
  
        for (let i = 0; i < v_left.length; i++)
        {
  
            // To check even or odd of Both sides are
            // equal or not
            if (v_left[i].first == v_right[i].first
                    || v_left[i].second == v_right[i].second)
            {
                return i;
            }
        }
        return -1;
}
 
// Driver code
let arr = [4, 3, 2, 1, 2];
let n = arr.length;
let index = Find_Index(n, arr);
if (index == -1)
{
    document.write("-1");
}
else
{
    document.write("index = " + index);
}
 
// This code is contributed by rag2127
</script>

Output:

index = 2

Time Complexity : O(n)
Auxiliary Space : O(n)

Further optimization : We can optimize the space used in the program. Instead of making vectors of pairs, we can make vectors of integers. We can use the fact that number of odd elements is equal total elements minus total number of even elements. Similarly, number of even elements is equal total elements minus total number of odd elements.

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!