🚀Day 5. Maximize partitions in a String 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

Given a string s consisting of lowercase English alphabets, your task is to determine the maximum number of substrings that can be formed by partitioning s such that no two substrings share common characters.

Note: Sorry for uploading late, my exam is going on.

🔍 Example Walkthrough:

Example 1:

Input:

s = "acbbcc"

Output:

2

Explanation:

Possible partitions:

• "a" and "cbbcc"

• "ac" and "bbcc"

The maximum possible substrings with no common characters are 2.

Example 2:

Input:

s = "ababcbacadefegdehijhklij"

Output:

3

Explanation:

Partitioning at index 8 and 15 produces three substrings:

• "ababcbaca"

• "defegde"

• "hijhklij"

Each of these substrings contains unique characters within them.

Example 3:

Input:

s = "aaa"

Output:

1

Explanation:

Since the string consists of the same character, no partition can be performed. The entire string itself is a single valid substring.

Constraints:

• $1 \leq |s| \leq 10^5$

• $'a' \leq s[i] \leq 'z'$

🎯 My Approach:

Greedy Approach Using Last Occurrence

1. Find the last occurrence of each character and store it.

2. Traverse the string and maintain an interval that expands based on character occurrences.

3. When the current index reaches the last occurrence of the characters in that segment, count it as a partition.

Algorithm Steps:

1. Precompute the last occurrence of each character in O(N) time.

2. Traverse the string, maintaining an end pointer that marks the furthest last occurrence of encountered characters.

3. If the current index reaches end, increment the partition count.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(N), as we scan the string twice (once to find last occurrences, once for partitioning).

• Expected Auxiliary Space Complexity: O(1), since we use a fixed-size array (26 elements for lowercase letters).

📝 Solution Code

Code (C++)

class Solution {
public:
    int maxPartitions(string &s) {
        int ans = 0, end = 0;
        vector<int> last(26);
        for (int i = 0; i < s.size(); i++)
            last[s[i] - 'a'] = i;
        for (int i = 0; i < s.size(); i++) {
            end = max(end, last[s[i] - 'a']);
            if (i == end) ans++;
        }
        return ans;
    }
};

⚡ Alternative Approaches

📊 2️⃣ Interval Merging Approach

Algorithm Steps:

1. Preprocess: For each character, record its first and last occurrence in the string.

2. Interval Creation: Create intervals ([first, last]) for each character that appears.

3. Merge Intervals: Sort these intervals by their starting index and merge overlapping intervals. Each merged interval represents a valid partition.

class Solution {
public:
    int maxPartitions(string &s) {
        vector<pair<int, int>> intervals;
        vector<int> first(26, s.size()), last(26, -1);
        for (int i = 0; i < s.size(); i++) {
            int idx = s[i] - 'a';
            first[idx] = min(first[idx], i);
            last[idx] = i;
        }
        for (int i = 0; i < 26; i++) {
            if (last[i] != -1)
                intervals.push_back({first[i], last[i]});
        }
        sort(intervals.begin(), intervals.end());
        int count = 0, start = intervals[0].first, end = intervals[0].second;
        for (int i = 1; i < intervals.size(); i++) {
            if (intervals[i].first > end) {
                count++;
                start = intervals[i].first;
                end = intervals[i].second;
            } else {
                end = max(end, intervals[i].second);
            }
        }
        return count + 1;
    }
};

📝 Complexity Analysis:

• ✅ Time Complexity: O(N + K log K)

  • O(N) for scanning the string (where N is the string length)

  • O(K log K) for sorting intervals (with at most K = 26 intervals)

• ✅ Space Complexity: O(K) – Uses extra space for intervals (K is constant, at most 26).

🔹 Pros: Visualizes problem as merging intervals. 🔹 Cons: Requires sorting, making it slightly slower than the greedy approach.

✅ Why This Approach?

This method visualizes the problem as merging intervals, making it easier to understand the segmentation of the string, especially for conceptual clarity.

🔁 3️⃣ First & Last Occurrence Approach

Algorithm Steps:

1. Record Occurrences: Use two arrays to store the first and last occurrences of each character.

2. Greedy Partitioning: Traverse the string and update the partition end using the last occurrence of each character. When the current index equals the partition end, a partition is identified.

class Solution {
public:
    int maxPartitions(string &s) {
        vector<int> first(26, -1), last(26, -1);
        for (int i = 0; i < s.size(); i++) {
            int idx = s[i] - 'a';
            if (first[idx] == -1)
                first[idx] = i;
            last[idx] = i;
        }
        int count = 0, end = 0;
        for (int i = 0; i < s.size(); i++) {
            end = max(end, last[s[i] - 'a']);
            if (i == end)
                count++;
        }
        return count;
    }
};

📝 Complexity Analysis:

• ✅ Time Complexity: O(N) – Single pass for recording occurrences and one pass for partitioning.

• ✅ Space Complexity: O(1) – Fixed-size arrays (26 elements each) are used.

🔹 Pros: Fastest approach, minimal space usage. 🔹 Cons: Needs careful index management.

✅ Why This Approach?

It leverages the concept of last occurrence effectively to decide partitions with minimal extra space and linear runtime.

🆚 Comparison of Approaches

Approach	⏱️ Time Complexity	🗂️ Space Complexity	✅ Pros	⚠️ Cons
Greedy (Optimized)	🟢 O(N)	🟢 O(1)	Best performance, straightforward implementation.	Requires careful management of indices.
Interval Merging Approach	🟢 O(N + K log K)	🟢 O(K)	Intuitive by visualizing partitions as intervals.	Slightly more code due to sorting.
First & Last Occurrence Approach	🟢 O(N)	🟢 O(1)	Simple and efficient with minimal extra space.	Similar to the greedy approach.

✅ Best Choice?

• For best performance: Use the Greedy Approach (O(N), O(1)).

• For conceptual clarity: Use the Interval Merging Approach.

• For a structured solution: Use First & Last Occurrence Approach.

Code (Java)

class Solution {
    public int maxPartitions(String s) {
        int[] last = new int[26];
        for (int i = 0; i < s.length(); i++)
            last[s.charAt(i) - 'a'] = i;
        int count = 0, end = 0;
        for (int i = 0; i < s.length(); i++) {
            end = Math.max(end, last[s.charAt(i) - 'a']);
            if (i == end)
                count++;
        }
        return count;
    }
}

Code (Python)

class Solution:
    def maxPartitions(self, s: str) -> int:
        last = {c: i for i, c in enumerate(s)}
        count = end = 0
        for i, c in enumerate(s):
            end = max(end, last[c])
            if i == end:
                count += 1
        return count

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

⭐ If you find this helpful, please give this repository a star! ⭐

---

📍Visitor Count