Maximum Number of Non-Overlapping Substrings

Maximum Number of Non-Overlapping Substrings

What is this problem about?

The Maximum Number of Non-Overlapping Substrings coding problem provides a string s and asks you to find the maximum number of non-overlapping substrings such that each substring contains all occurrences of its starting character, and all characters within the substring appear an equal or greater number of times than their occurrences outside the substring. The substrings must be chosen to maximize their count.

Why is this asked in interviews?

Amazon uses this problem to test a combination of string processing, greedy algorithms, and careful interval management. The constraint about containing all occurrences of a starting character, and the non-overlapping requirement, makes it tricky. It evaluates a candidate's ability to identify valid substring candidates and then apply a greedy selection strategy.

Algorithmic pattern used

Greedy with interval management: First, precompute the first and last occurrence index for every character in the string. Then, iterate through the string to identify "valid candidate" substrings. A valid candidate substring s[i...j] (starting at i and ending at j) must satisfy:

1. It contains all occurrences of s[i]. This means first_occurrence[s[i]] must be i.
2. For every character c within s[i...j], first_occurrence[c] must be ≥ i and last_occurrence[c] must be ≤ j. This effectively "expands" the substring boundaries if needed. After finding all such candidate substrings, sort them by their end indices. Then, greedily select non-overlapping substrings: pick the first one, then pick the next one whose start index is greater than the previously picked substring's end index. This greedy approach maximizes the count.

Example explanation

s = "abacaba", no explicit k.

• 'a': first=0, last=6
• 'b': first=1, last=5
• 'c': first=3, last=3

Candidates:

• Start 'a' at 0: must include all 'a's, expands to [0,6]. s[0..6]="abacaba".
  • 'b' is in [0,6], first['b']=1 > 0.
  • 'c' is in [0,6], first['c']=3 > 0. Valid candidate: [0,6]
• Start 'b' at 1: must include all 'b's. Extends to [1,5]. s[1..5]="bacab".
  • 'a' is in [1,5]. first['a']=0 < 1. Invalid. This example demonstrates the complexity of candidate expansion. Let's simplify and assume the problem meant for the chosen substrings only to contain all instances of their starting character, and all other characters within the substring must also have all their instances within that substring.

A simpler interpretation for "all occurrences of its starting character" AND "all characters within the substring appear an equal or greater number of times than their occurrences outside the substring" means any chosen substring s[i...j] must satisfy that for any character c present in s[i...j], its first occurrence must be >= i and its last occurrence must be <= j.

Let's refine. For each i:

1. Find first[s[i]] and last[s[i]]. The valid start for a substring containing s[i] would be first[s[i]].
2. Determine right_bound = last[s[i]].
3. Iterate ptr from first[s[i]] to right_bound: right_bound = max(right_bound, last[s[ptr]]). If first[s[ptr]] < first[s[i]], then this segment is invalid as s[ptr] starts before s[i] but is included.
4. If valid, [first[s[i]], right_bound] is a candidate.

For s = "ababa", k=1.

• first = {'a':0, 'b':1}, last = {'a':4, 'b':3}
• i=0, s[0]='a':
  • start = first['a'] = 0, end = last['a'] = 4
  • ptr=0, s[0]='a': end=max(4, last['a']=4) = 4.
  • ptr=1, s[1]='b': first['b']=1. first['b'] >= start. end=max(4, last['b']=3) = 4.
  • ptr=2, s[2]='a': first['a']=0. first['a'] >= start. end=max(4, last['a']=4) = 4. Candidate: [0,4] ("ababa")

This problem is very subtle in its definition of "valid substring". Assuming the example's common usage for "all occurrences of char in the substring are within the substring": s = "abacaba"

• Candidate "abacaba" from s[0] is valid (all 'a's, 'b's, 'c's within it)
• Candidate "bacab" from s[1] would be invalid because s[0]='a' is outside its left bound.
• Candidate "c" from s[3] is valid (s[3..3])
• Candidate "b" from s[1] is valid if we only consider this specific 'b' instance.

Given the example output for similar problems, the typical approach is:

1. For each character c, find its min_idx[c] and max_idx[c].
2. Iterate i from 0 to n-1. If i == min_idx[s[i]] (meaning s[i] is the first occurrence of this character), then it's a potential start of a valid substring.
3. For such an i, find the potential end_idx. This end_idx is initially max_idx[s[i]]. Then, iterate k from i to end_idx. For each character s[k], we must ensure min_idx[s[k]] >= i. If min_idx[s[k]] < i, then this s[i] cannot be the start of a valid substring that encompasses all s[k] occurrences starting from i. Also, update end_idx = max(end_idx, max_idx[s[k]]). If end_idx exceeds n-1 at any point, this i is not a valid start.
4. Collect all such valid [start_idx, end_idx] intervals.
5. Sort these intervals by their end_idx.
6. Greedily select non-overlapping intervals to maximize count.

Common mistakes candidates make

• Incorrectly identifying valid substrings: The definition of "valid" is crucial and often misconstrued. Failing to expand boundaries correctly or mishandling character occurrences can lead to incorrect candidates.
• Non-optimal greedy selection: Sorting intervals by start time and picking the shortest, or by end time and picking the one that ends earliest, is critical for interval scheduling problems.
• High complexity for precomputation: Repeated scans for min/max occurrences can be inefficient; a single pass to build min_idx and max_idx arrays (or maps) is needed.

Interview preparation tip

For the Hash Table Sorting String Greedy interview pattern, problems involving intervals and character occurrences often benefit from precomputing character ranges and then applying a greedy interval selection algorithm. Practice interval scheduling problems to internalize the optimal greedy strategies.