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This is a Medium difficulty Coding & Algorithms question, commonly asked during Technical Screen rounds at Meta.

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This question is commonly asked for Software Engineer candidates at Meta during technical interviews.

Solve two string algorithm tasks | Meta Coding Question

Quick Overview

Solve two string algorithm tasks evaluates algorithm design, data structures, correctness, complexity, edge cases, and implementation details in a realistic interview setting. A strong answer states assumptions, handles edge cases, explains trade-offs, and shows how to validate the result clearly.

Solve two string algorithm tasks

Company: Meta

Role: Software Engineer

Category: Coding & Algorithms

Difficulty: Medium

Interview Round: Technical Screen

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.

Quick Answer: Solve two string algorithm tasks evaluates algorithm design, data structures, correctness, complexity, edge cases, and implementation details in a realistic interview setting. A strong answer states assumptions, handles edge cases, explains trade-offs, and shows how to validate the result clearly.

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 ')'

Examples

Input: ('())',)

Expected Output: 1

Explanation: One ')' is unmatched at the front context; insert a single '(' to balance, giving "(())".

Input: ('(((',)

Expected Output: 3

Explanation: Three unmatched '(' remain; each needs a closing ')'.

Input: ('()',)

Expected Output: 0

Explanation: Already balanced.

Input: ('',)

Expected Output: 0

Explanation: Empty string is trivially valid.

Input: (')(',)

Expected Output: 2

Explanation: The leading ')' needs a '(' before it; the trailing '(' needs a ')' after it.

Input: ('()))((',)

Expected Output: 4

Explanation: Two stray ')' each force a '(' insertion (2), and two leftover '(' each need a ')' (2).

Input: (')))',)

Expected Output: 3

Explanation: Three ')' with no available '(' each force one '(' insertion.

Hints

Scan left to right tracking how many '(' are currently unmatched (open balance).
Every ')' that has no open '(' to pair with forces one insertion of '(' and does not change the balance; otherwise it consumes an open '('.
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

Examples

Input: ('ADOBECODEBANC', 'ABC')

Expected Output: 'BANC'

Explanation: The window "BANC" contains A, B, and C and is the shortest such window.

Input: ('a', 'a')

Expected Output: 'a'

Explanation: The single character satisfies t exactly.

Input: ('a', 'aa')

Expected Output: ''

Explanation: s has only one 'a' but t needs two, so no valid window exists.

Input: ('aa', 'aa')

Expected Output: 'aa'

Explanation: Both 'a' are needed; the whole string is the minimum window.

Input: ('abc', 'd')

Expected Output: ''

Explanation: 'd' never appears in s.

Input: ('xyzabc', 'cba')

Expected Output: 'abc'

Explanation: Order in t does not matter; "abc" covers a, b, c in the fewest characters.

Input: ('aaflslflsldkalskaaa', 'aaa')

Expected Output: 'aaa'

Explanation: The trailing run of three 'a' is the shortest window holding three 'a'.

Input: ('cabwefgewcwaefgcf', 'cae')

Expected Output: 'cwae'

Explanation: "cwae" is a length-4 window containing c, a, and e.

Hints

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

Quick Overview