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Finding indices of the earliest and last occurrences of an element x in a sorted array arr[] with maybe duplicate elements is the goal. Example A naive method: Iteratively on the elements of an array, check given elements in an array and record first and last occurrence of the index of the discovered element. The following outlines the actions to carry out the aforementioned concept: Here is the application of the aforementioned strategy: C++ C Java Python3 C# PHP JavaScript Output A binary search-based effective method: 2. For the last occurrence of a number  C++ C Java Python3 C# PHP JavaScript Output

Given a sorted and rotated array arr[] of n unique entries, the goal is to locate the index of specified key in the array. Return -1 should the key not be found in the array. Examples Using Two Binary Searches For instance, the min in {4, 5, 6, 7, 0, 1, 2} is 0 at index 4. Find the pivot point—or index of the min element. And at index 1 the min of 50, 10, 20, 30, 40 equals 10. How to get the min’s index? We have covered in great length here. Once we identify the pivot, we may readily split the provided array into two sorted subarrays with the index of the min. For instance {{4, 5, 6, 7, 0, 1, 2} as {{4, 5, 6, 7}, {1, 2}} and {{50, 10, 20, 30, 40} as {{50}, {10, 20, 30, 40} }. The following are the arising cases: Should the provided key match minimal, we returnShould min index be zero, the entire array is sorted; hence, we use binary search over the complete array.How then should we choose the subarray in other circumstances? To call binary search for both sides, one basic concept is We can save one binary search even though the general time complexity will remain O(Log n). One intends to match the given key with the first element. For instance {{4, 5, 6, 7}, {1, 2}, and with a key = 6. We conduct binary search in the first subarray if key is more than equal to the first element; else, in the second. C++ Java Python C# JavaScript

The goal is to print the lexicographically next bigger permutation of an n-sized array arr[]. If no further permutation exists, then determine the lexicographically lowest permutation of the particular array. Let us better grasp the issue by first writing all permutations of [1, 2, 4] in lexicographical sequence. [1, 2, 4]; [1, 4, 2], [2, 1, 4], [2, 4, 1], [4, 1, 2] and [4, 2, 1] Should we provide any of the above—except from the last—as input, we must then determine the next one in order. Should we provide last as input, we must revert to the first one. Examples [Naive Approach] Generating All Permutations – O(n!*n*log(n!)) Time and O(n!) Space Our first basic thought is that we would first create and sort all potential permutations of a particular array. Once arranged, we find the current permutation in this list. The next permutation is just the subsequent arrangement in the sorted sequence. Show the first permutation—the smallest permutation—should the present arrangement be the last in the list. Note: This method will only be effective in an input array free of duplicate values. Please see the expected method to manage duplicates here. C++ Java Python Output 2 4 5 0 1 7 Time Complexity: O(n!nlog(n!). n is the total number of elements in the input sequence, so reflecting all conceivable permutation.Auxiliary Space: O(n!). Permutation storage [Expected Approach] Generating Only Next – O(n) Time and O(1) Space Notes on the next permutation: Apply the above observation by following the guidelines below: C++ C Java Python C# JavaScript

Fundamental JavaScript procedure, removal of members from an array is necessary for data manipulation, filtering, and transformation. This post will investigate several ways to effectively remove elements from an array, therefore improving your knowledge and capacity to manage arrays. 1. Using pop() method The last element of the array is deleted and returned using thepop() method. This operation shrinks the array’s length by 1. JavaScript Output 2. Using shift() Method The first element of the array is eliminated using the shift() technique, therefore shrinking the original array by 1.  JavaScript Output Removed element: shift Remaining elements: splice,filter,pop 3. Using splice() Method Any specific element from an array in JavaScript can be deleted with the JavaScript Array splice() technique. Furthermore, this ability allows one to add or remove multiple elements from an array. Syntax Javascript Output [ ‘C++ ‘, ‘ Java ‘, ‘ Go ‘, ‘ Prolog’ ] 4. Using filter() Method This approach generates a new array from a given array comprising just those elements satisfying a criteria defined by the argument function JavaScript. Output 5. Using Delete Operator Use the Delete Operator to remove elements from a JavaScript array. JavaScript Output Removed: true Remaining elements: lodash,remove,,reset 6. Using _.remove() Method The _.remove() method is used to remove all elements from the array that predicate returns True and returns the removed elements. Output 7. Using Array.prototype.slice() Method One might generate a new array excluding the elements you wish to delete by means of the slice() method While slice() returns a new array rather than changing the original one like splice() does,. Syntax JavaScript Output

With a sorted array arr[] of size n, the aim is to reorganize the array so that every unique element shows at the beginning in sorted sequence. Return also the length of this unique sorted subarray. Note: Since they have no effect on the outcome, the items following the unique ones can be in any sequence and have any value. Example Using Hash Set – Works for Unsorted Also – O(n) Time and O(n) Space C++ Java Python C# JavaScript Output Anticipated Method – O(n) Time and O(1) Space The array is sorted hence we do not have to have a hash set. Every incident of an element would be consecutive. We so mostly have to find whether the current element is exactly the same as the previous one or not. Step by step application: C++ C JAVA Python C# JavaScript Output 12345

Problem Statement: 4Sum You are given an array of integers, nums, and an integer, target. Your task is to find all unique quadruplets [a,b,c,d][a, b, c, d][a,b,c,d] in the array such that: a+b+c+d=target a + b + c + d = \text{target}a+b+c+d=target Constraints: Approaches to Solve the Problem 1. Brute Force Approach 2. Optimized Approach Using Sorting and Two Pointers 3. Hash Table Approach (Alternate Method) Comparison of Approaches Approach Time Complexity Space Complexity Advantages Disadvantages Brute Force O(n4)O(n^4)O(n4) O(1)O(1)O(1) Simple to implement Inefficient for large arrays Sorting + Two Pointers O(n3)O(n^3)O(n3) O(1)O(1)O(1) Efficient, avoids duplicates easily Sorting modifies original array Hash Table O(n2)O(n^2)O(n2) O(n2)O(n^2)O(n2) Faster for large arrays High memory usage Key Observations C Implementation #include <stdio.h>#include <stdlib.h>void fourSum(int* nums, int numsSize, int target) { // Sort the array qsort(nums, numsSize, sizeof(int), cmp); for (int i = 0; i < numsSize – 3; i++) { if (i > 0 && nums[i] == nums[i – 1]) continue; for (int j = i + 1; j < numsSize – 2; j++) { if (j > i + 1 && nums[j] == nums[j – 1]) continue; int left = j + 1, right = numsSize – 1; while (left < right) { int sum = nums[i] + nums[j] + nums[left] + nums[right]; if (sum == target) { printf(“[%d, %d, %d, %d]\n”, nums[i], nums[j], nums[left], nums[right]); while (left < right && nums[left] == nums[left + 1]) left++; while (left < right && nums[right] == nums[right – 1]) right–; left++; right–; } else if (sum < target) { left++; } else { right–; } } } }}int cmp(const void* a, const void* b) { return (*(int*)a – *(int*)b);}int main() { int nums[] = {1, 0, -1, 0, -2, 2}; int target = 0; int numsSize = sizeof(nums) / sizeof(nums[0]); fourSum(nums, numsSize, target); return 0;} C++ Implementation #include <iostream>#include <vector>#include <algorithm>using namespace std;vector<vector<int>> fourSum(vector<int>& nums, int target) { vector<vector<int>> result; sort(nums.begin(), nums.end()); for (int i = 0; i < nums.size() – 3; i++) { if (i > 0 && nums[i] == nums[i – 1]) continue; for (int j = i + 1; j < nums.size() – 2; j++) { if (j > i + 1 && nums[j] == nums[j – 1]) continue; int left = j + 1, right = nums.size() – 1; while (left < right) { int sum = nums[i] + nums[j] + nums[left] + nums[right]; if (sum == target) { result.push_back({nums[i], nums[j], nums[left], nums[right]}); while (left < right && nums[left] == nums[left + 1]) left++; while (left < right && nums[right] == nums[right – 1]) right–; left++; right–; } else if (sum < target) { left++; } else { right–; } } } } return result;}int main() { vector<int> nums = {1, 0, -1, 0, -2, 2}; int target = 0; vector<vector<int>> result = fourSum(nums, target); for (const auto& quad : result) { cout << “[” << quad[0] << “, ” << quad[1] << “, ” << quad[2] << “, ” << quad[3] << “]\n”; } return 0;} Java Implementation import java.util.*;public class FourSum { public static List<List<Integer>> fourSum(int[] nums, int target) { List<List<Integer>> result = new ArrayList<>(); Arrays.sort(nums); for (int i = 0; i < nums.length – 3; i++) { if (i > 0 && nums[i] == nums[i – 1]) continue; for (int j = i + 1; j < nums.length – 2; j++) { if (j > i + 1 && nums[j] == nums[j – 1]) continue; int left = j + 1, right = nums.length – 1; while (left < right) { int sum = nums[i] + nums[j] + nums[left] + nums[right]; if (sum == target) { result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right])); while (left < right && nums[left] == nums[left + 1]) left++; while (left < right && nums[right] == nums[right – 1]) right–; left++; right–; } else if (sum < target) { left++; } else { right–; } } } } return result; } public static void main(String[] args) { int[] nums = {1, 0, -1, 0, -2, 2}; int target = 0; List<List<Integer>> result = fourSum(nums, target); for (List<Integer> quad : result) { System.out.println(quad); } }} Python Implementation def four_sum(nums, target): nums.sort() result = [] for i in range(len(nums) – 3): if i > 0 and nums[i] == nums[i – 1]: continue for j in range(i + 1, len(nums) – 2): if j > i + 1 and nums[j] == nums[j – 1]: continue left, right = j + 1, len(nums) – 1 while left < right: sum_ = nums[i] + nums[j] + nums[left] + nums[right] if sum_ == target: result.append([nums[i], nums[j], nums[left], nums[right]]) while left < right and nums[left] == nums[left + 1]: left += 1 while left < right and nums[right] == nums[right – 1]: right -= 1 left += 1 right -= 1 elif sum_ < target: left += 1 else: right -= 1 return result# Example usagenums = [1, 0, -1, 0, -2, 2]target = 0print(four_sum(nums, target)) C# Implementation using System;using System.Collections.Generic;class FourSum { public static IList<IList<int>> FourSumSolution(int[] nums, int target) { Array.Sort(nums); var result = new List<IList<int>>(); for (int i = 0; i < nums.Length – 3; i++) { if (i > 0 && nums[i] == nums[i – 1]) continue; for (int j = i + 1; j < nums.Length – 2; j++) { if (j > i + 1 && nums[j] == nums[j – 1]) continue; int left = j + 1, right = nums.Length – 1; while (left < right) { int sum = nums[i] + nums[j] + nums[left] + nums[right]; if (sum == target) { result.Add(new List<int> { nums[i], nums[j], nums[left], nums[right] }); while (left < right && nums[left] == nums[left + 1]) left++; while (left < right && nums[right] == nums[right – 1]) right–; left++; right–; } else if (sum < target) { left++; } else { right–; } } } } return result; } public static void Main(string[] args) { int[] nums = { 1, 0, -1, 0, -2, 2 }; int target = 0; var result = FourSumSolution(nums, target); foreach (var quad in result) { Console.WriteLine($”[{string.Join(“,