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Problem solution · Python

Maximum Number of Non Overlapping Substrings

Maximum Number of Non Overlapping Substrings: a Python solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Sliding window or two pointers
Source
Kamyu LeetCode Solutions
Length
74 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For Maximum Number of Non Overlapping Substrings, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 74 lines of Python from the credited upstream file maximum-number-of-non-overlapping-substrings.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximum Number of Non Overlapping Substrings · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n)# space: O(1) class Solution(object):    def maxNumOfSubstrings(self, s):        """        :type s: str        :rtype: List[str]        """        def find_right_from_left(s, first, last, left):            right, i = last[ord(s[left])-ord('a')], left            while i <= right:                if first[ord(s[i])-ord('a')] < left:                    return -1                right = max(right, last[ord(s[i])-ord('a')])                i += 1            return right                    first, last = [float("inf")]*26, [float("-inf")]*26        for i, c in enumerate(s):            first[ord(c)-ord('a')] = min(first[ord(c)-ord('a')], i)            last[ord(c)-ord('a')] = max(last[ord(c)-ord('a')], i)        result = [""]        right = float("inf")        for left, c in enumerate(s):            if left != first[ord(c)-ord('a')]:                continue            new_right = find_right_from_left(s, first, last, left)            if new_right == -1:                continue            if left > right:                result.append("")            right = new_right            result[-1] = s[left:right+1]        return result  # Time:  O(n)# space: O(1)class Solution2(object):    def maxNumOfSubstrings(self, s):        """        :type s: str        :rtype: List[str]        """        def find_right_from_left(s, first, last, left):            right, i = last[ord(s[left])-ord('a')], left            while i <= right:                if first[ord(s[i])-ord('a')] < left:                    return -1                right = max(right, last[ord(s[i])-ord('a')])                i += 1            return right         first, last = [float("inf")]*26, [float("-inf")]*26        for i, c in enumerate(s):            first[ord(c)-ord('a')] = min(first[ord(c)-ord('a')], i)            last[ord(c)-ord('a')] = max(last[ord(c)-ord('a')], i)        intervals = []        for c in xrange(len(first)):            if first[c] == float("inf"):                continue            left, right = first[c], find_right_from_left(s, first, last, first[c])            if right != -1:                intervals.append((right, left))        intervals.sort()  # Time: O(26log26)        result, prev = [], -1        for right, left in intervals:            if left <= prev:                continue            result.append(s[left:right+1])            prev = right        return result

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