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

2.5

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.

CPP

// 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
Output:

index = 2

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

Method 2:(Efficient solution)
1- Create two vectors of pair types i.e v_left and v_right
2- v_left[i] stores the frequency of odd and even numbers of its left sides
3- v_right[i] stores the frequency of odd and even numbers of its right sides
4- Now check (v_left[i].first == v_right[i].first || v_left[i].second == v_right[i].second)
if True return i
5- At last if no such index exist return -1

// 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;
}
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.


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