SkylineWebZ

Under the former method, for every element we required to determine the highest element on the left and the right. Consequently, to lower the temporal complexity: First we compute and save the highest bar on the left and right for every element—say placed in arrays left[] and right[].Then iterate the array using the computed values to determine the water stored in this index—that is, the same as ( min(left[i], right[i]) – arr[i]). C++ C Java Python C# JavaScript Illustration

Given an array of n non-negative integers arr[] representing an elevation map where the width of each bar is 1, find how much water it can trap following rain. Examples Table of Content Observations: The problem’s fundamental idea runs like this: Brute force O(n2) time and O(1) Place: Space Go across every element in the array to locate the highest bars on both sides. Consider the lesser of the two heights. The quantity of water this array element can store defines the difference between the smaller height and the height of the current element. C++ C Java Python C# JavaScript Output 9

Problem Statement Given a string s, find the longest palindromic substring in s. A palindrome is a sequence of characters that reads the same backward as forward. Your task is to return the longest substring of s that is a palindrome. Example Input:s = “babad” Output:”bab” or “aba” (both are valid) Explanation:The string contains multiple palindromic substrings such as “bab”, “aba”, and “a”. The longest one is either “bab” or “aba”. Approach 1. Expand Around Center 2. Dynamic Programming 3. Brute Force Algorithm Expand Around Center Complexity Implementation C #include <stdio.h>#include <string.h>char* longestPalindrome(char* s) { int start = 0, maxLength = 0; int len = strlen(s); for (int i = 0; i < len; i++) { int l = i, r = i; while (l >= 0 && r < len && s[l] == s[r]) { if (r – l + 1 > maxLength) { start = l; maxLength = r – l + 1; } l–; r++; } l = i; r = i + 1; while (l >= 0 && r < len && s[l] == s[r]) { if (r – l + 1 > maxLength) { start = l; maxLength = r – l + 1; } l–; r++; } } s[start + maxLength] = ‘\0’; return s + start;}int main() { char s[] = “babad”; printf(“Longest Palindromic Substring: %s\n”, longestPalindrome(s)); return 0;} C++ #include <iostream>#include <string>using namespace std;string longestPalindrome(string s) { int start = 0, maxLength = 0, n = s.length(); for (int i = 0; i < n; i++) { int l = i, r = i; while (l >= 0 && r < n && s[l] == s[r]) { if (r – l + 1 > maxLength) { start = l; maxLength = r – l + 1; } l–; r++; } l = i, r = i + 1; while (l >= 0 && r < n && s[l] == s[r]) { if (r – l + 1 > maxLength) { start = l; maxLength = r – l + 1; } l–; r++; } } return s.substr(start, maxLength);}int main() { string s = “babad”; cout << “Longest Palindromic Substring: ” << longestPalindrome(s) << endl; return 0;} Java public class LongestPalindromicSubstring { public static String longestPalindrome(String s) { int start = 0, maxLength = 0; for (int i = 0; i < s.length(); i++) { int l = i, r = i; while (l >= 0 && r < s.length() && s.charAt(l) == s.charAt(r)) { if (r – l + 1 > maxLength) { start = l; maxLength = r – l + 1; } l–; r++; } l = i; r = i + 1; while (l >= 0 && r < s.length() && s.charAt(l) == s.charAt(r)) { if (r – l + 1 > maxLength) { start = l; maxLength = r – l + 1; } l–; r++; } } return s.substring(start, start + maxLength); } public static void main(String[] args) { String s = “babad”; System.out.println(“Longest Palindromic Substring: ” + longestPalindrome(s)); }} Python def longest_palindrome(s: str) -> str: start, max_length = 0, 0 for i in range(len(s)): l, r = i, i while l >= 0 and r < len(s) and s[l] == s[r]: if r – l + 1 > max_length: start, max_length = l, r – l + 1 l -= 1 r += 1 l, r = i, i + 1 while l >= 0 and r < len(s) and s[l] == s[r]: if r – l + 1 > max_length: start, max_length = l, r – l + 1 l -= 1 r += 1 return s[start:start + max_length]# Example usageprint(“Longest Palindromic Substring:”, longest_palindrome(“babad”)) C# using System;class LongestPalindromicSubstring { public static string LongestPalindrome(string s) { int start = 0, maxLength = 0; for (int i = 0; i < s.Length; i++) { int l = i, r = i; while (l >= 0 && r < s.Length && s[l] == s[r]) { if (r – l + 1 > maxLength) { start = l; maxLength = r – l + 1; } l–; r++; } l = i; r = i + 1; while (l >= 0 && r < s.Length && s[l] == s[r]) { if (r – l + 1 > maxLength) { start = l; maxLength = r – l + 1; } l–; r++; } } return s.Substring(start, maxLength); } static void Main() { string s = “babad”; Console.WriteLine(“Longest Palindromic Substring: ” + LongestPalindrome(s)); }} JavaScript function longestPalindrome(s) { let start = 0, maxLength = 0; for (let i = 0; i < s.length; i++) { let l = i, r = i; while (l >= 0 && r < s.length && s[l] === s[r]) { if (r – l + 1 > maxLength) { start = l; maxLength = r – l + 1; } l–; r++; } l = i; r = i + 1; while (l >= 0 && r < s.length && s[l] === s[r]) { if (r – l + 1 > maxLength) { start = l; maxLength = r – l + 1; } l–; r++; } } return s.substring(start, start + maxLength);}console.log(“Longest Palindromic Substring:”, longestPalindrome(“babad”));

Problem Statement Given a string s, find the length of the longest substring without repeating characters. Constraints Examples Approach The problem can be solved efficiently using the Sliding Window Technique. Here’s how: Algorithm Code Implementations 1. C #include <stdio.h>#include <string.h>int lengthOfLongestSubstring(char *s) { int n = strlen(s); int max_len = 0, start = 0; int map[128] = {0}; for (int end = 0; end < n; end++) { if (map[s[end]] > 0) { start = (map[s[end]] > start) ? map[s[end]] : start; } map[s[end]] = end + 1; max_len = (max_len > end – start + 1) ? max_len : end – start + 1; } return max_len;}int main() { char str[] = “abcabcbb”; printf(“Length of Longest Substring: %d\n”, lengthOfLongestSubstring(str)); return 0;} 2. C++ #include <iostream>#include <unordered_map>using namespace std;int lengthOfLongestSubstring(string s) { unordered_map<char, int> map; int start = 0, max_len = 0; for (int end = 0; end < s.length(); end++) { if (map.find(s[end]) != map.end()) { start = max(map[s[end]] + 1, start); } map[s[end]] = end; max_len = max(max_len, end – start + 1); } return max_len;}int main() { string str = “abcabcbb”; cout << “Length of Longest Substring: ” << lengthOfLongestSubstring(str) << endl; return 0;} 3. Java import java.util.HashMap;public class LongestSubstring { public static int lengthOfLongestSubstring(String s) { HashMap<Character, Integer> map = new HashMap<>(); int start = 0, max_len = 0; for (int end = 0; end < s.length(); end++) { if (map.containsKey(s.charAt(end))) { start = Math.max(map.get(s.charAt(end)) + 1, start); } map.put(s.charAt(end), end); max_len = Math.max(max_len, end – start + 1); } return max_len; } public static void main(String[] args) { String str = “abcabcbb”; System.out.println(“Length of Longest Substring: ” + lengthOfLongestSubstring(str)); }} 4. Python def length_of_longest_substring(s: str) -> int: char_map = {} start, max_len = 0, 0 for end, char in enumerate(s): if char in char_map and char_map[char] >= start: start = char_map[char] + 1 char_map[char] = end max_len = max(max_len, end – start + 1) return max_len# Example usageprint(“Length of Longest Substring:”, length_of_longest_substring(“abcabcbb”)) 5. C# using System;using System.Collections.Generic;class LongestSubstring { public static int LengthOfLongestSubstring(string s) { Dictionary<char, int> map = new Dictionary<char, int>(); int start = 0, max_len = 0; for (int end = 0; end < s.Length; end++) { if (map.ContainsKey(s[end]) && map[s[end]] >= start) { start = map[s[end]] + 1; } map[s[end]] = end; max_len = Math.Max(max_len, end – start + 1); } return max_len; } static void Main(string[] args) { string str = “abcabcbb”; Console.WriteLine(“Length of Longest Substring: ” + LengthOfLongestSubstring(str)); }} 6. JavaScript Copy codefunction lengthOfLongestSubstring(s) { let charMap = new Map(); let start = 0, maxLen = 0; for (let end = 0; end < s.length; end++) { if (charMap.has(s[end]) && charMap.get(s[end]) >= start) { start = charMap.get(s[end]) + 1; } charMap.set(s[end], end); maxLen = Math.max(maxLen, end – start + 1); } return maxLen;}console.log(“Length of Longest Substring:”, lengthOfLongestSubstring(“abcabcbb”)); Time Complexity Space Complexity

The goal is to identify the smallest positive number absent from an unsorted array arr[] including both positive and negative items. Note: You have original array modification power. Examples By Marking Indices – O(n) Time and O(1) Space The concept is to note the presence of elements inside the same array. Since 1 is the smallest positive integer, first we look to see whether 1 is there; should it be absent, the response is 1. All numbers that are either negative or greater than the array size are converted to 1 otherwise to maintain all numbers within a range of 1 to n. We next index the numbers using their values. To indicate the presence of a number x, we are adding n to the number existing at index x – 1. At last, we search the array for the first unmarked index to obtain the smallest missing positive value. C++ C Java Python C# JavaScript Output 3

First, all positive integers should be migrated to the left side of the array. We next traverse over this left segment and negating the value at index (x – 1) marks the occurrences of every number x. Finally, go over the left section once more and, looking for the first unmarked piece, determine the missing number. Alternate Approach] By Marking Indices – O(n) Time and O(1) Space The goal is to identify the smallest positive number absent from an unsorted array arr[] including both positive and negative items. Note: You have original array modification power. Illustrations: By Negating Array Elements – O(n) Time and O(1) Space Should our array consist entirely of positive numbers, we can record the presence of a number x by negating the number at index x – 1 in the array. We can thus find our first missing element by first looking for the first unmarked element in the same array. Moving all the positive integers to the left side of the array will help us to apply this method for an array comprising both positive and negative numbers. After this is finished, we may use the positive array technique on this left side—an array of positive numbers. C++ C Java Python C# JavaScript Output 3

We shall answer the most often requested interview question “Combination Sum II – Find all unique combinations” in this post. Find all unique combinations in candidates where the candidate numbers add to target given a set of candidate numbers (candidates) and a target number. Every number in the candidates could just be used once in the mix. Note: The set of solutions has to be devoid of repeating combinations. Some instances are: Solution: Disclaimer: Don’t jump directly to the solution, try it out yourself first. Solution 1: Using extra space and time complexity  Approach: Defining the Recursive Function: Sort the elements before beginning the recursive call; the ans should not be duplicated and should include the combinations in sorted sequence. We begin with the index 0; at index 0 we have n – 1 ways to choose the first element of our subsequence. See if our ds could benefit from the current index value. If so, include it into the ds and shift the index by one. As shifting the index skip the consecutive repeated elements will create duplicate sequences. After the call make sure to pop the element from the ds. Reduce the target by arr[i] and call the recursive call for f(idx + 1,target – 1,ds,ans).(Seeing the example repeatedly will help you to grasp). If arr[i] > target, then stop the recursive function since there is no need in checking; the index will be moving by 1 all the entries to its right will be in increasing order. Base state: Add the ds to the ans return anyhow the goal value is zero. Recursive call represented for the provided example below: Code C++ Java Python Output: [ [ 1 1 6 ][ 1 2 5 ][ 1 7 ][ 2 6 ] ] Time Complexity:O(2^n*k) Space Complexity:O(k*x)