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Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string. Note: You must not use any built-in BigInteger library or convert the inputs to integer directly. Example 1: Input: num1 = "2", num2 = "3" Output: "6" Example 2: Input: num1 = "123", num2 = "456" Output: "56088" Constraints: 1 <= num1.length, num2.length <= 200 num1 and num2 consist of digits only. Both num1 and num2 do not contain any leading zero, except the number 0 itself. Solution class Solution { public String multiply(String num1, String num2) { int[] products = new int[num1.length() + num2.length()]; for (int i = 0; i < num1.length(); i++) { for (int j = 0; j < num2.length(); j++) { int d1 = num1.charAt(i) - '0'; int d2 = num2.charAt(j) - '0'; products[i + j + 1] += d1 * d2; } } int carry = 0; for (int i = products.length - 1; i >= 0; i--) { int tmp = (products[i] + carry) % 10; carry = (products[i] + carry) / 10; products[i] = tmp; } StringBuilder sb = new StringBuilder(); for (int product: products) { if (product == 0 && sb.length() == 0) { continue; } sb.append(product); } if (sb.length() == 0) { sb.append("0"); } return sb.toString(); } }

Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer (similar to C/C++’s atoi function). The algorithm for myAtoi(string s) is as follows: Read in and ignore any leading whitespace. Check if the next character (if not already at the end of the string) is ‘-’ or ‘+’. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present. Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored. Convert these digits into an integer (i.e. “123” -> 123, “0032” -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2). If the integer is out of the 32-bit signed integer range [-231, 231 - 1], then clamp the integer so that it remains in the range. Specifically, integers less than -231 should be clamped to -231, and integers greater than 231 - 1 should be clamped to 231 - 1. Return the integer as the final result. Note: ...

Given two strings s and t of length N, find the maximum number of possible matching pairs in strings s and t after swapping exactly two characters within s. A swap is switching s[i] and s[j], where s[i] and s[j] denotes the character that is present at the ith and jth index of s, respectively. The matching pairs of the two strings are defined as the number of indices for which s[i] and t[i] are equal. Note: This means you must swap two characters at different indices. Signature int matchingPairs(String s, String t) Input ...

Given two string arrays word1 and word2, return true if the two arrays represent the same string, and false otherwise. A string is represented by an array if the array elements concatenated in order forms the string. Example 1: Input: word1 = ["ab", "c"], word2 = ["a", "bc"] Output: true Explanation: word1 represents string "ab" + "c" -> "abc" word2 represents string "a" + "bc" -> "abc" The strings are the same, so return true. Example 2: Input: word1 = ["a", "cb"], word2 = ["ab", "c"] Output: false Example 3: Input: word1 = ["abc", "d", "defg"], word2 = ["abcddefg"] Output: true Constraints: ...

1360. Number of Days Between Two Dates Write a program to count the number of days between two dates. The two dates are given as strings, their format is YYYY-MM-DD as shown in the examples. Example 1: Input: date1 = "2019-06-29", date2 = "2019-06-30" Output: 1 Example 2: Input: date1 = "2020-01-15", date2 = "2019-12-31" Output: 15 Constraints: The given dates are valid dates between the years 1971 and 2100. Solution import java.time.*; class Solution { public int daysBetweenDates(String date1, String date2) { return Math.abs(countDays(date1) - countDays(date2)); } int [] MONTHS = new int[] {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; int countDays(String date) { String[] parts = date.split("-"); Integer years = Integer.parseInt(parts[0]); Integer months = Integer.parseInt(parts[1]); Integer days = Integer.parseInt(parts[2]); for (int year = 1971; year < years; year++) { if (isLeapYear(year)) { days += 366; } else { days += 365; } } for (int month = 1; month < months; month++) { if (isLeapYear(years) && month == 2) { days += 29; } else { days += MONTHS[month]; } } return days; } boolean isLeapYear(int year) { return (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0); } public int daysBetweenDates2(String date1, String date2) { LocalDateTime time1 = LocalDate.parse(date1).atTime(LocalTime.NOON); LocalDateTime time2 = LocalDate.parse(date2).atTime(LocalTime.NOON); Duration duration = Duration.between(time1, time2); return (int) Math.abs(duration.toDays()); } }

1482. Minimum Number of Days to Make m Bouquets Given a string s, reverse only all the vowels in the string and return it. The vowels are ‘a’, ’e’, ‘i’, ‘o’, and ‘u’, and they can appear in both cases. Example 1: Input: s = "hello" Output: "holle" Example 2: Input: s = "leetcode" Output: "leotcede" Constraints: 1 <= s.length <= 3 * 105 s consist of printable ASCII characters. Solution class Solution { public String reverseVowels(String s) { Set<Character> vowels = new HashSet<>(); vowels.addAll(Arrays.asList('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U')); int lo = 0; int hi = s.length() - 1; char[] word = s.toCharArray(); while (lo < hi) { while (lo < hi && !vowels.contains(word[lo])) { lo++; } while (lo < hi && !vowels.contains(word[hi])) { hi--; } char c = word[lo]; word[lo] = word[hi]; word[hi] = c; lo++; hi--; } return new String(word); } }

1624. Largest Substring Between Two Equal Characters Given a string s, return the length of the longest substring between two equal characters, excluding the two characters. If there is no such substring return -1. A substring is a contiguous sequence of characters within a string. Example 1: Input: s = "aa" Output: 0 Explanation: The optimal substring here is an empty substring between the two 'a's. Example 2: Input: s = "abca" Output: 2 Explanation: The optimal substring here is "bc". Example 3: Input: s = "cbzxy" Output: -1 Explanation: There are no characters that appear twice in s. Example 4: Input: s = "cabbac" Output: 4 Explanation: The optimal substring here is "abba". Other non-optimal substrings include "bb" and "". Constraints: ...

244. Shortest Word Distance II Design a data structure that will be initialized with a string array, and then it should answer queries of the shortest distance between two different strings from the array. Implement the WordDistance class: WordDistance(String[] wordsDict) initializes the object with the strings array wordsDict. int shortest(String word1, String word2) returns the shortest distance between word1 and word2 in the array wordsDict. Example 1: Input ["WordDistance", "shortest", "shortest"] [[["practice", "makes", "perfect", "coding", "makes"]], ["coding", "practice"], ["makes", "coding"]] Output [null, 3, 1] Explanation WordDistance wordDistance = new WordDistance(["practice", "makes", "perfect", "coding", "makes"]); wordDistance.shortest("coding", "practice"); // return 3 wordDistance.shortest("makes", "coding"); // return 1 Constraints: ...

245. Shortest Word Distance III Given an array of strings wordsDict and two strings that already exist in the array word1 and word2, return the shortest distance between these two words in the list. Note that word1 and word2 may be the same. It is guaranteed that they represent two individual words in the list. Example 1: Input: wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "makes", word2 = "coding" Output: 1 Example 2: Input: wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "makes", word2 = "makes" Output: 3 Constraints: ...

833. Find And Replace in String To some string s, we will perform some replacement operations that replace groups of letters with new ones (not necessarily the same size). Each replacement operation has 3 parameters: a starting index i, a source word x and a target word y. The rule is that if x starts at position i in the original string S, then we will replace that occurrence of x with y. If not, we do nothing. ...