Odd-Even Subarrays | Editorial by aksingh007

Modify the input array as follows:
1. If A[i] is odd, then replace it by -1
2. If A[i] is even, then repace it by 1
Now, if a subarray [L,R] contains equal odd and even numbers, then the the net Sum of the subarray will always be 0.In order to find all such subarray one should 
calculate prefix sum of the new updated array and calculate the  total number of ways to select any two elements from the prefix sum array such that the selected
elements are equal.

```
#include<stdio.h>
#include<math.h>
#include<iostream>
using namespace std;
#define s(n) scanf("%lld",&n)
#define ll long
ll sum[500004];
    ll pos[500000],neg[500000];
    ll arr[500002];
 ll get(int n){
   ll ans=0;
    pos[0]=1;
    for(int i=0;i<n;i++){
        sum[i+1]=sum[i]+arr[i];
        if(sum[i+1]>=0)
            pos[(int)sum[i+1]]++; 
        else
            neg[abs(sum[i+1])]++;
    }
    for(int i=0;i<300001;i++){
        if(pos[i]>=2)
            ans+=pos[i]*(pos[i]-1)/2;
        if(neg[i]>=2 )
            ans+=neg[i]*(neg[i]-1)/2;
    }
   return ans;
}
    
    
int  main() {
      int n;
      cin>>n;
       for(int i=0;i<n;i++){
           cin>>arr[i];
           if(arr[i]%2==0) arr[i]=-1;
           else arr[i]=1;
       }
       printf("%lld\n",get(n));
       return 0;

}
```

Odd-Even Subarrays | Editorial by aksingh007

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