Solve two string algorithm tasks

Answer both parts. A) Parentheses correction: Given a string s consisting only of '(' and ')', output the minimum number of parentheses you must insert (at any positions) to make s a valid balanced parentheses string. Define validity in the usual way via matching pairs and proper nesting. Constraints: 1 <= |s| <= 100,000. Describe an O(n)-time algorithm with O(1) or O(n) extra space and briefly justify correctness. B) Shortest covering substring: Given two strings s and t (ASCII), return the shortest contiguous substring of s that contains every character of t at least as many times as it appears in t; if no such substring exists, return the empty string. If multiple answers have the same length, return any. Constraints: 1 <= |s| <= 200,000; 1 <= |t| <= 10,000. Provide the algorithm, time and space complexities, and key edge cases.

What a Strong Answer Covers

• A clear algorithm with the right data structures and enough pseudocode or code-level detail to implement it.
• A correctness argument that explains why the algorithm covers all required cases.
• Time and space complexity, plus at least one alternative approach when relevant.
• Focused tests for normal cases, edge cases, and failure modes.

Follow-up Questions

• How would the approach change if the input were streaming or too large for memory?
• What invariants would you assert in production code?
• Which tests would catch off-by-one, duplicate, or tie-breaking bugs?

Minimum Parentheses Insertions to Make Valid

Given a string s consisting only of '(' and ')', return the minimum number of parentheses you must insert (at any positions) to make s a valid balanced parentheses string. A string is valid if every '(' has a matching ')' that comes after it and the pairs are properly nested.

Examples:

• s = "())" -> 1 (insert one '(' at the front: "(())")
• s = "(((" -> 3 (insert three ')')
• s = "()" -> 0 (already valid)
• s = ")(" -> 2 (one '(' before and one ')' after)

Constraints: 1 <= |s| <= 100,000.

Constraints

• 1 <= |s| <= 100,000
• s consists only of the characters '(' and ')'

Hints

1. Scan left to right tracking how many '(' are currently unmatched (open balance).
2. Every ')' that has no open '(' to pair with forces one insertion of '(' and does not change the balance; otherwise it consumes an open '('.
3. At the end, every remaining unmatched '(' needs one inserted ')'. Answer = forced '(' insertions + leftover open balance.

Shortest Covering Substring (Minimum Window)

Given two strings s and t (ASCII), return the shortest contiguous substring of s that contains every character of t at least as many times as it appears in t. If no such substring exists, return the empty string "". If multiple answers share the minimum length, return any one of them.

Examples:

• s = "ADOBECODEBANC", t = "ABC" -> "BANC"
• s = "a", t = "a" -> "a"
• s = "a", t = "aa" -> "" (s does not have two 'a')

Constraints: 1 <= |s| <= 200,000; 1 <= |t| <= 10,000.

Constraints

• 1 <= |s| <= 200,000
• 1 <= |t| <= 10,000
• s and t consist of ASCII characters

Hints

1. Count the required frequency of each character in t. Use a sliding window with two pointers over s.
2. Expand the right pointer to include characters; track how many distinct required characters are currently satisfied ('formed').
3. When all required characters are satisfied, shrink from the left to minimize the window, recording the best length, then continue scanning.

Solution

Solution Alignment

The prompt asks for an implementation-level answer. The safest way to present it is to define the state, maintain clear invariants, then walk through complexity and tests.

Problem Restatement

Answer both parts. A) Parentheses correction: Given a string s consisting only of '(' and ')', output the minimum number of parentheses you must insert (at any positions) to make s a valid balanced parentheses string. Define validity in the usual way via matching pairs and proper nesting. Constraints: 1 <= |s| <= 100,000. Describe an O(n)-time algorithm with O(1) or O(n) extra space and briefly justify correctness. B) Shortest covering substring: Given two strings s and t (ASCII), return the shortest contiguous substring of s that contains every character of t at least as many times as it appears in t; if no such substring exists, return the empty string. If multiple answers have the same l...

Recommended Approach

Model the states explicitly and use BFS for unweighted shortest paths, Dijkstra for weighted non-negative paths, or topological DP for DAGs. Track visited states at the right granularity so cycles do not cause repeated work.

Correctness

The implementation should maintain an invariant after each loop or operation that directly matches the problem statement. At termination, that invariant implies the returned value has considered every valid candidate exactly once, or has preserved the required data-structure state after every API call.

Complexity

BFS is O(V + E) time and O(V) space for a standard graph. Expanded-state problems multiply those bounds by the number of state dimensions.

Edge Cases and Tests

Disconnected graph, source equals target, cycles, duplicate edges, unreachable target, and whether the answer counts nodes, edges, moves, or transfers.