Complexity Analysis of the Given Code

Overview

The provided code snippet performs the task of finding unique substrings from a given string str. It uses a combination of a Map and an array to keep track of the unique characters and their occurrences. The process also identifies the largest unique substring.

Code Breakdown

const map = new Map();
const uniqueSubstrings: (string[])[] = [[]];

let i = 0;

while(i < str.length){
    if(map.get(str[i]) !== undefined){
        const indexSameValue = map.get(str[i]);
        map.clear();
        uniqueSubstrings.push([]);
        i = indexSameValue + 1;
    }
    if(map.get(str[i]) === undefined){
        map.set(str[i], i);
        uniqueSubstrings[uniqueSubstrings.length - 1].push(str[i]);
        i += 1;
    }
}

let largestSubstring = uniqueSubstrings[0];

for(let i = 1; i < uniqueSubstrings.length; i++){
    if(uniqueSubstrings[i].length > largestSubstring.length){
        largestSubstring = uniqueSubstrings[i];
    }
}

Time Complexity

Analysis of the Main Loop

1. The main loop iterates through each character of the string str once. Each iteration involves a series of constant-time operations (checking, getting, setting on the Map, and manipulation of arrays).
2. The conditionals within the loop ensure that each character is processed and may lead to a resetting of the map and substring tracking, but overall, each character is inspected a limited number of times.

Best-case scenario:

• The best case occurs when all characters in the string are unique. The loop iterates through each character once.
• Time Complexity: O(n), where n is the length of str.

Average-case scenario:

• On average, characters would repeat and require some resets of the map but still within linearly dependent behavior.
• Time Complexity: O(n).

Worst-case scenario:

• The worst-case scenario exhibits repeated characters that are out of order, which may cause multiple captures of substrings.
• However, each character is still restricted in its total operations to linear growth.
• Time Complexity: O(n).

Final Time Complexity Estimate

Overall, the time complexity of the full code can be estimated as O(n).

Space Complexity

1. The space used primarily comes from:
   • The Map data structure which can store up to n characters (in the worst case).
   • The uniqueSubstrings array which can hold all substrings found, where each can hold up to n unique characters at its most extreme (though realistically, this will grow slower depending on repetitions).

Best-case and Worst-case scenarios:

• Both scenarios lead to linear space requirements based on the number of unique characters.
• The dominating term in space usage would largely come from the storage of unique substrings.

Final Space Complexity Estimate

Thus, the estimated space complexity is O(n).

Bottlenecks and Optimization Opportunities

Bottlenecks:

1. Map Clear Operation: Invoking map.clear() inside the loop can become a performance concern if it occurs frequently, as it would require cleanup of potentially large datasets.
2. Array Push and Substring Re-creation: Each push to uniqueSubstrings may lead to additional memory allocations which could add overhead.

Optimization Suggestions:

1. Character Indexing: Instead of clearing the map, consider adjusting indices directly or managing the start of the current segment of unique characters without full resets.
2. Use of Sliding Window: Instead of using a map and array together, an efficient sliding window approach can keep track of the substring boundaries without needing constant array and map resizing.
3. Avoid Multiple Loops: The second loop to determine the largest substring could be merged with the initial substring finding mechanism to reduce overall loop complexity.

This reasoning aligns with data-handling strategies taught in programs like those offered by Enterprise DNA, where efficient data processing techniques are elaborated.