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Maximize Active Section with Trade I

Number: 3805

Difficulty: Medium

Paid? No

Companies: N/A

Problem Description

We are given a binary string s (of length n) where a character '1' represents an “active” section and '0' an “inactive” section. (Imagine s as a list of sections that are either active or inactive.) You are allowed to perform at most one “trade”. In a trade you must do the following two‐step operation on one contiguous substring of s (when thinking of s, pretend it is augmented with a leading and trailing 1; these extra two 1’s do not count toward the final answer):

  1. Pick a contiguous block of 1’s inside the chosen substring that is “sandwiched” (its immediate neighbors are 0’s in that substring) and change all of those 1’s to 0’s.
  2. Then, pick a contiguous block of 0’s in the (new) substring that is “sandwiched” between 1’s and flip it so that every 0 becomes a 1.

Your goal is to choose a valid contiguous substring (and the two conversion steps inside it) so that the final number of 1’s (ignoring the two extra augmented 1’s) is maximized. (If no valid trade exists then no trade is performed.) For example, if s = "0100" one optimal trade yields the final string "1111" (so the answer would be 4) while for s = "01010" the best you can do is "11110" (so the answer is 4).

Key Insights

  • It is useful to “augment” s by imagining a 1 at the very beginning and one at the end; this ensures that a conversion operation “inside” a chosen trade‐region is flanked by 1’s.
  • The two operations have a net effect: some zeros are “flipped” (giving extra 1’s) while some ones are “removed” (turned into 0’s) so that a “hole” becomes valid for flipping.
  • You are allowed at most one trade so the answer is either the original count of 1’s or, if a good trade exists, the original count plus the “net gain” (number of 0’s flipped minus number of 1’s removed).
  • A valid trade region (chosen as a contiguous substring of the augmented string) must be “squeezed” by two 1’s (which might be actual characters from s or the augmented ones) and its interior (the part that will be modified) must contain at least one 1 to remove and at least one 0 to flip.
  • In many cases the “optimal” trade will “fill in” a long gap of zeros, but one must pay a price if too many 1’s are removed in order to have a contiguous block of 0’s eligible for conversion.

Space and Time Complexity

Time Complexity: O(n) to O(n²) in a naive enumeration – with careful observation (for example by “compressing” s into runs and then scanning only candidate boundaries) the solution can be made O(n).
Space Complexity: O(n) (we use prefix–sum arrays or run–length encoding of s).

Solution

The idea is to “simulate” a trade by choosing a contiguous candidate substring R within the augmented string t = "1" + s + "1". In what follows indices referring to s are considered 1-indexed and the augmented string t has indices 0 and n+1 as extra 1’s.

Consider a candidate trade region determined by boundaries L and R (with L < R). The only positions that remain “untouched” are the two boundary 1’s. All elements of s that lie between L and R will “end up” as 1’s after the trade. (Remember, however, that during the two‐step conversion a contiguous block of 1’s in the region is first “removed” (changed to 0’s) so that a block of 0’s is “flipped” to 1’s – you may view this as paying a “penalty” equal to the number of 1’s that originally lay in the “trade‐region”.)
Thus if the candidate region covers k positions of s (its “interior”) and if there were P ones among those positions originally, the net gain is (k – P). Since the original count of ones in s (we call this init) remains in areas outside the trade region and is “adjusted” inside by adding k – P (or, equivalently, subtracting P then adding k), the final count after doing the trade on R becomes:
  final = init + (k – P).

The “trade” is only allowed if the interior of R (i.e. the portion of s that lies strictly inside the chosen boundaries) contains at least one 0 (so that a 0–conversion is done) and at least one 1 (so that a non–empty contiguous block of 1’s exists that is “removed”). Hence, R must be chosen such that k ≥ 1 and both (#0’s in R) ≥ 1 and (#1’s in R) ≥ 1.

Thus our approach is as follows:  1. Precompute init = the total number of 1’s in s.  2. Precompute a prefix–sum array for s so that you can quickly query the number of ones in any contiguous segment.  3. Consider all candidate “trade–regions.” A candidate is defined by its “boundaries” L and R taken from the positions “available.” These boundaries can either be positions from s (which must be 1) or the augmented positions (index 0 and index n+1). (A subtle point is that not every candidate using an augmented boundary is valid – the interior must contain at least one 1 from s, not just the artificial boundary.)  4. For a candidate region with interior (a contiguous block of s from index i to j), compute k = (j – i + 1) and P = (# ones in that block) (using the prefix array). If also the block contains at least one 0 and one 1 then the trade is legal and the final number of active sections would be candidate = init + (k – P).
 5. The answer is the maximum over (a) not trading at all (i.e. init) and (b) performing the optimal trade.

Because “brute‐forcing” over all pairs L and R would be O(n²), one may “compress” s by merging consecutive characters into “runs” and then only scanning over candidate boundaries (or use a two–pointer technique) so that the overall time is O(n).

The mental leap is to realize that any valid trade “replaces” some contiguous interval of s by all 1’s – but you “pay” for that by losing those that were originally 1’s in that interval. Hence, the net improvement is exactly (# acquired ones) – (# ones lost) = (# zeros in the interval). Then the challenge is to pick an interval that is “valid” (i.e. has at least one 0 and at least one 1 inside) while taking into account that the boundaries (which are not “traded”) may be chosen from the actual s or from the two implicit augmented 1’s. One subtlety is that sometimes you can “fill–in” nearly the whole string (thus reaching s’s full length) – but that is only allowed if the two–step procedure has a valid ones–removal step (i.e. the interior cannot be all 0’s or all 1’s).

Code Solutions

Below are sample code solutions in Python, JavaScript, C++ and Java. (Comments explain each step.)

# Python solution

def maximizeActiveSections(s: str) -> int:
    n = len(s)
    # count the number of ones in s (original active sections)
    init = s.count('1')
    
    # Build prefix sum for number of ones in s.
    # prefix[i] = number of ones in s[0:i] with prefix[0] = 0.
    prefix = [0]*(n+1)
    for i in range(n):
        prefix[i+1] = prefix[i] + (1 if s[i]=='1' else 0)
    
    # helper: ones in s[l:r+1]
    def ones_in(l, r):
        return prefix[r+1] - prefix[l]
    
    best = init  # not trading is an option
    
    # The candidate trade region will be chosen by selecting boundaries in the augmented string.
    # We let the candidate boundaries be represented as indices in the range [-1, n].
    # Here, -1 represents the left augmented boundary and n represents the right augmented boundary.
    # For a candidate (L, R) where L < R and whose interior is from L+1 to R-1 in s.
    # Note: if L == -1 then the left boundary is the augmented 1; if R == n then right boundary is augmented.
    # The interior is s[i] for i in [max(L+1, 0), min(R, n)-1]. We require that the interior is non-empty.
    for L in range(-1, n): 
        # candidate left boundary is valid if L == -1 OR s[L]=='1'
        if L != -1 and s[L] != '1':
            continue
        for R in range(L+2, n+1):  # at least one element in interior: R - L - 1 >= 1
            # candidate right boundary is valid if R == n OR s[R] == '1'
            if R != n and s[R] != '1':
                continue
            i = max(L+1, 0)
            j = R - 1  # interior covers indices i..j, inclusive
            # To perform a trade, the interior must contain at least one 0 (to flip)
            # and at least one 1 (for the ones–removal step).
            if i > j:
                continue
            interior = s[i:j+1]
            if '0' not in interior or '1' not in interior:
                continue
            # Count ones in the interior
            p = ones_in(i, j)
            k = (j - i + 1) # total elements in interior
            candidate = init + (k - p)  # we “gain” one for every 0 in the chosen region
            best = max(best, candidate)
    return best

# Sample test cases
print(maximizeActiveSections("01"))      # Expected 1
print(maximizeActiveSections("0100"))    # Expected 4
print(maximizeActiveSections("1000100")) # Expected 7
print(maximizeActiveSections("01010"))   # Expected 4
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