ALTARAY - Editorial

PROBLEM LINK:

Contest
Practice

Authors: Jaub Safin
Testers: Vasya Antoniuk
Translators: Vasya Antoniuk (Russian), Team VNOI (Vietnamese) and Hu Zecong (Mandarin)
Editorialist: Praveen Dhinwa

DIFFICULTY:

Simple

PREREQUISITES:

Simple observations, dp

PROBLEM:

An array is called alternating if any two consecutive numbers in it have opposite signs (i.e. one of them should be negative, whereas the other should be positive).

You are given an array A of size N consisting of non-zero elements. For each index i from 1 to N, you have to find the length of longest subarray starting at i.

QUICK EXPLANATION:

We can observe that if an array is alternating, then all of its subarrays are also alternating.

So, we can divide the array into maximal alternating subarrays.
For doing that, we will iterate over array from left to right, if the sign of current number is different than previous number, then this number can be used to extend previous alternating subarray, Otherwise we will have to start constructing a new maximal alternating subarray.

In this way, we can partition the array into various maximal alternating subarrays.

After this, finding length of longest subarray starting at index i is quite easy, as it can be done easily by finding the position of i in the corresponding maximal alternating subarray. If p be position and L be the length of the maximal subarray, then L - p + 1 will be the length of longest subarray starting at index i.

EXPLANATION:

Observation 1:
Values of the number don’t matter, only their signs do.

So, we can change the array A such that it consists of -1 and 1’s only.

Observation 2:
If an array A is alternating, then all of it’s subarrays are alternating.

Let us take an example to understand how we can apply the above observation to solve the problem.
Let A = [1, 1, -1, 1, 1, -1, -1]
So, we start from A_1 and note that A_1 is equal A_2.
So the maximal subarray starting from 1 will be [1] itself.

Now, we go to A_2, we can see that A_2, A_3, A_4 have different signs, and A_4 has same sign as A_5.
So the maximal subarray starting from index 2 will be [1, -1, 1].

So, we break the array into several parts such that each part is maximal alternating subarray. In our example, the parts will be
[1] [1, -1, 1] [1, -1], [-1]

We can formalize the above idea to write a pseudo code to break an array into various parts of maximal alternating subarrays.

vector<vector<int> > parts;
vector<int> curPart;
subpart.push_back(a[0]);
for (int i = 1; i < n; i++) {
	// If signs of current and previous number are different, 
	// then it means that we can extend the current part.
	if (a[i] * a[i - 1] == -1) {
		curPart.push_back(a[i]);
	} else {
		// We add the curPart into parts.
		parts.push_back(curPart);
	}
} 
// Check at the end whether the curPart has non-zero length or not.
// If it has, then add curPart into parts.
if (curPart.size() > 0) {
	parts.push_back(subpart);
}

Now, let us find the length of longest alternating subarray ending at each index i for our example subarray A.
We get [1] [3, 2, 1], [2, 1], [1]

So, this means that for an maximal alternating subarray of length L, the answers (length of longest alternating subarray start from that index) will be L, L-1, \dots, 1.

We can use this idea to solve our problem.

// i denotes the current index of array at which we currently are.
int i = 1;
for (vector<int> curPart : parts) {
	int L = curPart.size();
	while (L > 0) {
		answer[i] = L;
		// increment i
		i++;
		// decrement L
		L--;
	}
}
// Note the fact that we didn't use the explicit values of curPart, only its size matter.

Dynamic programming based Solution

You can also solve the problem by a very simple dp.

Let len[i] denote the maximum length of alternating subarray starting at position i.
We can observer that if a[i] and a[i + 1] has opposite signs, then len[i] will be 1 more than len[i + 1].
Otherwise in the case when they have same sign, then len[i] will be just 1.

len[N] = 1;
for (int i = N - 1; i >= 1; i--) {
	// a[i] and a[i + 1] have different signs. 
	// Note that the a[i] can go upto 10^9, 
	// So if a is stored in int data type, then the a[i] * a[i + 1] might not fit in int.
	// So, we cast it to long long
	if (a[i] * (long long) a[i + 1]< 0) {
		len[i] = len[i + 1] + 1;
	} else {
		len[i] = 1;
	}
}

Time Complexity:

As we have seen in both the solutions we have to iterate over the array A only once or constant number of times. So, time complexity of the algorithm will be \mathcal{O}(N) which will easily fit in time with N = 10^5 and T = 10.

AUTHOR’S AND TESTER’S SOLUTIONS:

setter
tester

3 Likes

CAn u provide us with ur test cases and corresponding answers pls…
since I am getting wrong answer .

#include
using namespace std;

int main() 
{
int t,i,j,x,k;
int a[100][100];
int n[10];
x=0;
cin>>t;
for(i=0;i<t;i++)
{
cin>>n[i];
for(j=0;j<n[i];j++)
{
cin>>a[i][j];
}
}
for(k=0;k<t;k++)
{
for(i=0;i<n[k]-1;i++)
{
x=0;
for(j=i;j<n[k];j++)
{
if(((a[k][j]<0)&&(a[k][j+1]>0))||((a[k][j]>0)&&(a[k][j+1]<0)))
{
x=x+1;
}
else
{
break;
}
}
if(x==0)
{
cout<<"1";
}
else
{
cout<<x+1;
}
}
cout<<"1";
cout<<"\n";
}
	return 0;
}

3
4
1 2 3 4
4
-1 2 -3 4
6
-2 -2 -3 5 -2 -1

My Approach:

Everytime you find a subarray whose longest alternating count is greater than one, all you have to do is print it, and decrement and print it till it becomes one. Then skip the next count-1 elements.

My Solution

1 Like

can tester tell me ur test cases and corresponding answers pls

In the dp solution why do you write a[i]*1ll*[i+1] instead of simple a[i]*a[i+1],
I wrote the latter during contest and got wrong answer and when I changed it to former it was accepted later. I thought maybe it is to prevent int overflow but I tried making array as array of long long ints but still I got wrong answer.

There is something wrong in your claims. Can you give links of both the solutions?

@dpraveen

https://www.codechef.com/viewsolution/9713110 <-- this is solution where a is long long int(WA)
https://www.codechef.com/viewsolution/9713154 <-- this is the other one(AC)

The Above solutions are similar to solution in editorial

https://www.codechef.com/viewsolution/9711723 <-- This is the one I submitted during contest.(WA)

https://www.codechef.com/viewsolution/9713715 <-- This is with a is long long int (WA)

https://www.codechef.com/viewsolution/9713162 <-- This is the same submission with only one change as mentioned above in previous comment(i.e. including 1ll). (AC)

If your solution is wrong, then try generating subarrays of length <= 10 (with +1, -1) only and run against a correct solution.

Also if you have some overflow bug due to not using long long at desired places, then for the same test cases, change +1 to 10^9 and -1 to -10^9

https://www.codechef.com/viewsolution/9712232
This is link of my solution… getting wrong answer continuously…Still i cant found any bugg… Pls help me out

Tried inputting many cases and getting every tym correct answer.Still i am getting WA in ur compiler

https://www.codechef.com/viewsolution/9712232
This is solution… And not able to get any bugg

Getting right answer for every sample input

Did exactly the same way which is mentioned in Quick Explanation.

My Solution : https://www.codechef.com/viewsolution/9708007

@dpraveen did you see the solutions above? What do you think?

is my program correct?
someone pl tell
https://www.codechef.com/viewsolution/9714935

someone plzzzz help!!

You should not print anything other than output, So remove printf("enter no of cases ");

#include<stdio.h>
int main ()
{
int T=0,i=0,n=0,flag=0,c=0;
long long int j=0,count=0,A[100001];
scanf("%d",&T);
for(i=0;i<T;i++)
{
scanf("%d",&n);
for(j=1;j<=n;j++)
{
scanf("%lld",&A[j]);
}
for(j=1;j<=n;j++)
{
count=1;
flag=0;
c=j;
if(j==n)
{
printf("%d",1);
}
if(j!=n)
{
if(A[j]*A[j+1]>0&&j!=n)
{
printf("%d",1);
printf(" “);
}
else
{
while(A[j]*A[j+1]<0)
{
count++;
j++;
flag=1;
}
if(flag==1)
{
printf(”%lld",count);
printf(" ");
}
}
j=c;

        }
        }
        printf("\n");
    }
return(0);

}

can you please check where i got wrong?

//