Space-efficient algorithm for finding the largest balanced subarray?

2023-04-03 22:25 问答作者：

given an array of 0s and 1s, find maximum subarray such that number of zeros and 1s are equal. This needs to be done in O(n) time and O(1) space.

I have an algo which does it in O(n) time and O(n) space. It uses a prefix sum array and exploits the fact that if the number of 0s and 1s are same then sumOfSubarray = lengthOfSubarray/2

#include<iostream>
#define M 15

using namespace std;

void getSum(int arr[],int prefixsum[],int size) {
    int i;
    prefixsum[0]=arr[0]=0;
    prefixsum[1]=arr[1];
    for (i=2;i<=size;i++) {
        prefixsum[i]=prefixsum[i-1]+arr[i];
    }
}

void find(int a[],int &start,int &end) {
    while(start < end) {
        int mid = (start +end )/2;
        if((end-start+1) == 2 * (a[end] - a[start-1]))
                break;
        if((end-start+1) > 2 * (a[end] - a[start-1])) {
            if(a[start]==0 && a[end]==1)
                    start++; else
                    end--;
        } else {
            if(a[start]==1 && a[end]==0)
                    start++; else
                    end--;
        }
    }
}

int main() {
    int size,arr[M],ps[M],start=1,end,width;
    ;
    cin>>size;
    arr[0]=0;
    end=size;
    for (int i=1;i<=size;i++)
            cin>>arr[i];
    getSum(arr,ps,size);
 开发者_StackOverflow社区   find(ps,start,end);
    if(start!=end)
            cout<<(start-1)<<" "<<(end-1)<<endl; else cout<<"No soln\n";
    return 0;
}

Now my algorithm is O(n) time and O(Dn) space where Dn is the total imblance in the list.

This solution doesn't modify the list.

let D be the difference of 1s and 0s found in the list.

First, let's step linearily through the list and calculate D, just to see how it works:

I'm gonna use this list as an example : l=1100111100001110

Element   D
null      0
1         1
1         2   <-
0         1
0         0
1         1
1         2
1         3
1         4
0         3
0         2
0         1
0         0
1         1
1         2
1         3
0         2   <-

Finding the longest balanced subarray is equivalent to finding 2 equal elements in D that are the more far appart. (in this example the 2 2s marked with arrows.)

The longest balanced subarray is between first occurence of element +1 and last occurence of element. (first arrow +1 and last arrow : 00111100001110)

Remark:

The longest subarray will always be between 2 elements of D that are between [0,Dn] where Dn is the last element of D. (Dn = 2 in the previous example) Dn is the total imbalance between 1s and 0s in the list. (or [Dn,0] if Dn is negative)

In this example it means that I don't need to "look" at 3s or 4s

Proof:

Let Dn > 0 .

If there is a subarray delimited by P (P > Dn). Since 0 < Dn < P, before reaching the first element of D which is equal to P we reach one element equal to Dn. Thus, since the last element of the list is equal to Dn, there is a longest subarray delimited by Dns than the one delimited by Ps.And therefore we don't need to look at Ps

P cannot be less than 0 for the same reasons

the proof is the same for Dn <0

Now let's work on D, D isn't random, the difference between 2 consecutive element is always 1 or -1. Ans there is an easy bijection between D and the initial list. Therefore I have 2 solutions for this problem:

the first one is to keep track of first and last appearance of each element in D that are between 0 and Dn (cf remark).
second is to transform the list into D, and then work on D.

FIRST SOLUTION

For the time being I cannot find a better approach than the first one:

First calculate Dn (in O(n)) . Dn=2

Second instead of creating D, create a dictionnary where the keys are the value of D (between [0 and Dn]) and the value of each keys is a couple (a,b) where a is the first occurence of the key and b the last.

Element   D DICTIONNARY
null      0 {0:(0,0)}
1         1 {0:(0,0) 1:(1,1)}
1         2 {0:(0,0) 1:(1,1) 2:(2,2)}
0         1 {0:(0,0) 1:(1,3) 2:(2,2)}
0         0 {0:(0,4) 1:(1,3) 2:(2,2)}
1         1 {0:(0,4) 1:(1,5) 2:(2,2)}
1         2 {0:(0,4) 1:(1,5) 2:(2,6)}
1         3 { 0:(0,4) 1:(1,5) 2:(2,6)}
1         4 {0:(0,4) 1:(1,5) 2:(2,6)}  
0         3{0:(0,4) 1:(1,5) 2:(2,6) }
0         2 {0:(0,4) 1:(1,5) 2:(2,9) }
0         1 {0:(0,4) 1:(1,10) 2:(2,9) } 
0         0 {0:(0,11) 1:(1,10) 2:(2,9) } 
1         1 {0:(0,11) 1:(1,12) 2:(2,9) } 
1         2 {0:(0,11) 1:(1,12) 2:(2,13)}
1         3 {0:(0,11) 1:(1,12) 2:(2,13)} 
0         2 {0:(0,11) 1:(1,12) 2:(2,15)}

and you chose the element with the largest difference : 2:(2,15) and is l[3:15]=00111100001110 (with l=1100111100001110).

Time complexity :

2 passes, the first one to caclulate Dn, the second one to build the dictionnary. find the max in the dictionnary.

Total is O(n)

Space complexity:

the current element in D : O(1) the dictionnary O(Dn)

I don't take 3 and 4 in the dictionnary because of the remark

The complexity is O(n) time and O(Dn) space (in average case Dn << n).

I guess there is may be a better way than a dictionnary for this approach.

Any suggestion is welcome.

Hope it helps

SECOND SOLUTION (JUST AN IDEA NOT THE REAL SOLUTION)

The second way to proceed would be to transform your list into D. (since it's easy to go back from D to the list it's ok). (O(n) time and O(1) space, since I transform the list in place, even though it might not be a "valid" O(1) )

Then from D you need to find the 2 equal element that are the more far appart.

it looks like finding the longest cycle in a linked list, A modification of Richard Brent algorithm might return the longest cycle but I don't know how to do it, and it would take O(n) time and O(1) space.

Once you find the longest cycle, go back to the first list and print it.

This algorithm would take O(n) time and O(1) space complexity.

Different approach but still O(n) time and memory. Start with Neil's suggestion, treat 0 as -1.

Notation: A[0, …, N-1] - your array of size N, f(0)=0, f(x)=A[x-1]+f(x-1) - a function

If you'd plot f, you'll see, that what you look for are points for which f(m)=f(n), m=n-2k where k-positive natural. More precisely, only for x such that A[x]!=A[x+1] (and the last element in an array) you must check whether f(x) already occurred. Unfortunately, now I see no improvement over having array B[-N+1…N-1] where such information would be stored.

To complete my thought: B[x]=-1 initially, B[x]=p when p = min k: f(k)=x . And the algorithm is (double-check it, as I'm very tired):

fx = 0
B = new array[-N+1, …, N-1]
maxlen = 0
B[0]=0
for i=1…N-1 :
    fx = fx + A[i-1]
    if B[fx]==-1 :
        B[fx]=i
    else if ((i==N-1) or (A[i-1]!=A[i])) and (maxlen < i-B[fx]):
        We found that A[B[fx], …, i] is best than what we found so far
        maxlen = i-B[fx]

Edit: Two bed-thoughts (= figured out while laying in bed :P ):

1) You could binary search the result by the length of subarray, which would give O(n log n) time and O(1) memory algorithm. Let's use function g(x)=x - x mod 2 (because subarrays which sum to 0 are always of even length). Start by checking, if the whole array sums to 0. If yes -- we're done, otherwise continue. We now assume 0 as starting point (we know there's subarray of such length and "summing-to-zero property") and g(N-1) as ending point (we know there's no such subarray). Let's do

    a = 0
    b = g(N-1)
    while a<b : 
        c = g((a+b)/2)
        check if there is such subarray in O(n) time
        if yes:
            a = c
        if no:
            b = c
    return the result: a (length of maximum subarray)

Checking for subarray with "summing-to-zero property" of some given length L is simple:

    a = 0
    b = L
    fa = fb = 0
    for i=0…L-1:
        fb = fb + A[i]
    while (fa != fb) and (b<N) :
        fa = fa + A[a]
        fb = fb + A[b]
        a = a + 1
        b = b + 1
    if b==N:
        not found
    found, starts at a and stops at b

2) …can you modify input array? If yes and if O(1) memory means exactly, that you use no additional space (except for constant number of elements), then just store your prefix table values in your input array. No more space used (except for some variables) :D

And again, double check my algorithms as I'm veeery tired and could've done off-by-one errors.

Like Neil, I find it useful to consider the alphabet {±1} instead of {0, 1}. Assume without loss of generality that there are at least as many +1s as -1s. The following algorithm, which uses O(sqrt(n log n)) bits and runs in time O(n), is due to "A.F."

Note: this solution does not cheat by assuming the input is modifiable and/or has wasted bits. As of this edit, this solution is the only one posted that is both O(n) time and o(n) space.

A easier version, which uses O(n) bits, streams the array of prefix sums and marks the first occurrence of each value. It then scans backward, considering for each height between 0 and sum(arr) the maximal subarray at that height. Some thought reveals that the optimum is among these (remember the assumption). In Python:

sum = 0
min_so_far = 0
max_so_far = 0
is_first = [True] * (1 + len(arr))
for i, x in enumerate(arr):
    sum += x
    if sum < min_so_far:
        min_so_far = sum
    elif sum > max_so_far:
        max_so_far = sum
    else:
        is_first[1 + i] = False

sum_i = 0
i = 0
while sum_i != sum:
    sum_i += arr[i]
    i += 1
sum_j = sum
j = len(arr)
longest = j - i
for h in xrange(sum - 1, -1, -1):
    while sum_i != h or not is_first[i]:
        i -= 1
        sum_i -= arr[i]
    while sum_j != h:
        j -= 1
        sum_j -= arr[j]
    longest = max(longest, j - i)

The trick to get the space down comes from noticing that we're scanning is_first sequentially, albeit in reverse order relative to its construction. Since the loop variables fit in O(log n) bits, we'll compute, instead of is_first, a checkpoint of the loop variables after each O(√(n log n)) steps. This is O(n/√(n log n)) = O(√(n/log n)) checkpoints, for a total of O(√(n log n)) bits. By restarting the loop from a checkpoint, we compute on demand each O(√(n log n))-bit section of is_first.

(P.S.: it may or may not be my fault that the problem statement asks for O(1) space. I sincerely apologize if it was I who pulled a Fermat and suggested that I had a solution to a problem much harder than I thought it was.)

If indeed your algorithm is valid in all cases (see my comment to your question noting some corrections to it), notice that the prefix array is the only obstruction to your constant memory goal.

Examining the find function reveals that this array can be replaced with two integers, thereby eliminating the dependence on the length of the input and solving your problem. Consider the following:

You only depend on two values in the prefix array in the find function. These are a[start - 1] and a[end]. Yes, start and end change, but does this merit the array?
Look at the progression of your loop. At the end, start is incremented or end is decremented only by one.
Considering the previous statement, if you were to replace the value of a[start - 1] by an integer, how would you update its value? Put another way, for each transition in the loop that changes the value of start, what could you do to update the integer accordingly to reflect the new value of a[start - 1]?
Can this process can be repeated with a[end]?
If, in fact, the values of a[start - 1] and a[end] can be reflected with two integers, doesn't the whole prefix array no longer serve a purpose? Can't it therefore be removed?

With no need for the prefix array and all storage dependencies on the length of the input removed, your algorithm will use a constant amount of memory to achieve its goal, thereby making it O(n) time and O(1) space.

I would prefer you solve this yourself based on the insights above, as this is homework. Nevertheless, I have included a solution below for reference:

#include <iostream>
using namespace std;

void find( int *data, int &start, int &end )
{
    // reflects the prefix sum until start - 1
    int sumStart = 0;

    // reflects the prefix sum until end
    int sumEnd = 0;
    for( int i = start; i <= end; i++ )
        sumEnd += data[i];

    while( start < end )
    {
        int length = end - start + 1;
        int sum = 2 * ( sumEnd - sumStart );

        if( sum == length )
            break;
        else if( sum < length )
        {
            // sum needs to increase; get rid of the lower endpoint
            if( data[ start ] == 0 && data[ end ] == 1 )
            {
                // sumStart must be updated to reflect the new prefix sum
                sumStart += data[ start ];
                start++;
            }
            else
            {
                // sumEnd must be updated to reflect the new prefix sum
                sumEnd -= data[ end ];
                end--;
            }
        }
        else
        {
            // sum needs to decrease; get rid of the higher endpoint
            if( data[ start ] == 1 && data[ end ] == 0 )
            {
                // sumStart must be updated to reflect the new prefix sum
                sumStart += data[ start ];
                start++;
            }
            else
            {
                // sumEnd must be updated to reflect the new prefix sum
                sumEnd -= data[ end ];
                end--;
            }
        }
    }
}

int main() {
    int length;
    cin >> length;

    // get the data
    int data[length];
    for( int i = 0; i < length; i++ )
        cin >> data[i];

    // solve and print the solution
    int start = 0, end = length - 1;
    find( data, start, end );

    if( start == end )
        puts( "No soln" );
    else
        printf( "%d %d\n", start, end );

    return 0;
}

This algorithm is O(n) time and O(1) space. It may modify the source array, but it restores all the information back. So it is not working with const arrays. If this puzzle has several solutions, this algorithm picks the solution nearest to the array beginning. Or it might be modified to provide all solutions.