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codeforces #493_div2B.cpp

// Your file with changes

#include<bits/stdc++.h>
using namespace std;
//no need to pass the whole array every time as :
//	you are not changing the array
// 	rather make the array global .

int i,j,k,c=0,n,en=0,on=0,b;
int f(int b,int i,int c,int a[]){
	if(i==n)//if reached the end of array
		return c;
	if(a[i]%2)
		on++;
	else 
		en++;
	if(en==on && (i+1)!=n && abs(a[i+1]-a[i])<=b){
		//if even count =odd count and the index is not last (we are not allowed 
		//to cut at the end and the cost of cut is affordable)
		en=0;on=0;//start counting again

		// c=max(c,max(f(b-abs(a[i+1]-a[i]),i+1,c+1,a),f(b,i+1,c,a)));

		// better way to take max of more numbers
		c=max({c, f(b-abs(a[i+1]-a[i]),i+1,c+1,a), f(b,i+1,c,a) });
/*The point to note here is that function 'f' doesn't return a negative value. 
It either adds something to the value of c it received, or doesn't do anything. 
so, instead of c=max({c,....});	 you can write c+=max(f(), f());	 and make function 
return just the excess value of c.
*/		
		//update max number of cuts as maximum of (current value of cuts,
		//value of current cut is applied, current cut is not applied)
	}
	else
		c=f(b,i+1,c,a);
	//if even count!=odd count or cost is not affordable (note we already checked for
	//reaching end of array)
	
	return c;
}

int main(){

cin>>n>>b;
int a[n];
for(i=0;i<n;i++)
cin>>a[i];

cout<<f(b,0,0,a)<<"\n";
return 0;
}
/*
6 100
1 2 3 4 5 6
*/

codeforces #493_div2B_accepted.cpp

// Accepted solution

#include<bits/stdc++.h>
using namespace std;

int a[1000004];
int dp[102][102];
int i,j,k,c=0,n,en=0,on=0,b;
//see how we reduced the number of reqired parameters,these are what we call states
//in dp, and these decide the dimention of dp array
// https://qr.ae/TWGZ9O read this answer for more details
// 

int f(int b,int i){
	// cout<<" b and i "<<b<<" "<<i<<"\n";
	int c=0;
	if(i==n || b<=0)//if reached the end of array
		return c;
	else if(dp[b][i]!=-1)//if we have calculated the same thing earlier
		return dp[b][i];//don't recompute, return it else compute it,store and then return

	if(a[i]%2)
		on++;
	else 
		en++;
	if(en==on && (i+1)!=n && abs(a[i+1]-a[i])<=b){
		//if even count =odd count and the index is not last (we are not allowed 
		//to cut at the end and the cost of cut is affordable)
		en=0;on=0;//start counting again

		// c=max(c,max(f(b-abs(a[i+1]-a[i]),i+1,c+1,a),f(b,i+1,c,a)));
		// better way to take max of more numbers
		c=max( 1+f(b-abs(a[i+1]-a[i]),i+1), f(b,i+1) );
	// max of (if cut is made, cut isn't made)
	}
	else
		c=f(b,i+1);
	//if even count!=odd count or cost is not affordable (note we already checked for
	//reaching end of array)
	
	return dp[b][i]=c;
}

int main(){
memset(dp,-1,sizeof(dp));
cin>>n>>b;

for(i=0;i<n;i++)
cin>>a[i];

cout<<f(b,0)<<"\n";
return 0;
}
/*
6 100
1 2 3 4 5 6
*/

instruction.txt

Please learn to write commments as much as possible, this will fasten the helping process a lot.
Also go through the comments added in both codes carefully as this will help you to understand better.

link to accepted solution