Write the c++ code according to the following algorithm insert 1357 using forloop and then while loop
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Write the c++ code according to the following
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- explain working of this code.. explain its each line #include <bits/stdc++.h>using namespace std; // Function to change all duplicate elements to -1void eliminate_duplicates(int arr[] , int N){ int count=0, i, j; for(i=0; i<N; i++) { if(arr[i]!=-1) { for(j=i+1; j<N; j++) { if(arr[i]==arr[j]) { count++; arr[j]=-1; } } } } // Printing results cout<<endl<<"Final state of array is : "; for(int k=0; k<N; k++) { cout<<arr[k]<<" "; } N = N-count; cout<<endl<<"Final value of N is : "<<N;} // Driver Functionint main(){ int i,n ; cout<<"Enter array size: "; // Taking input for array size cin>>n; int a[n]; cout<<"Enter elements in array : "; // Taking input for array elements for(i=0; i<n; i++) { cin>>a[i]; } eliminate_duplicates(a,n); return 0;}Python Numpy Function to complete: def w14(v): Inputs: v: A numpy array of shape (N, 1) Returns: The L2 norm of v: norm = (sum_i^N v[i]^2)^(1/2) You MAY NOT use np.linalg.normplease convert the code to C language //C++ program to check if two arrays //are equal or not #include <bits/stdc++.h>using namespace std; bool similar_array(vector<int> arr1, vector<int> arr2){ //create teo different hash table where for each key //the hash function is h(arr[i])=arr[i] //we will use stl map as hash table and //will keep frequency stored //so say two keys arr[0] and arr[5] are mapping to //the same location, then the location will have value 2 //instead of the keys itself //if two hash tables are exactly same then //we can say that our arrays are similar map<int, int> hash1; map<int, int> hash2; //for each number for (int i = 0, j = 0; i < arr1.size(); i++, j++) { hash1[arr1[i]]++; hash2[arr2[i]]++; } //now check whether hash tables are exactly same or not for (auto it = hash1.begin(), ij = hash2.begin(); it != hash1.end() && ij != hash2.end(); it++, ij++) {…
- 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 40 M ㅏ void sequence::insert(const value_type& entry) if(used capacity) { resize((1.5* capacity) + 1); } if (current_index >= used) { current_index = 0; } for (int i = used; i > current_index; i--) { data[i] = data[i-1];Language: Python 3 • Autocomplete Ready O 1 v import ast lst = input(O lst = ast.literal_eval(lst) def binarysearch(lst,x,low,high): if low - high x: 10 11 Algorithm BinarylnsertionSort 12 Input/output: takes an integer array a = {a[0], ..., a[n – 1]} of size n 13 begin BinarylnsertionSort return binarysearch (lst, x, mid, high) 14 for i =1 to n val = a[i] p = BinarySearch(a, val, 0, i – 1) for j = i-1 to p alj + 1]= a[j] j= j-1 end for 15 else: 16 return mid 17 18 def BinaryInsertionSort(lst): 19 print (BinaryInsertionSort(lst)) 20 a[p] = val j=i+1 end for end BinarylnsertionSort Here, val = a[i] is the current value to be inserted at each step i into the already sorted part a[0], ..., ați – 1] of the array a. The binary search along that part returns the position p where the val will be inserted. After finding p, the data values in the subsequent positions j = i- 1, ..., p are sequentially moved one position up to i, ..., p+1 so that the value val can be inserted into the proper…#include<iostream>using namespace std;//function to sort elements of arrayvoid sort(int a[], int n){int i,j,temp;for(i=1;i<n;i++){for(j=0;j<n-i;j++){if(a[j]>a[j+1]){temp=a[j];a[j]=a[j+1];a[j+1]=temp;}}}}//function to search location of item using binary searchint binary_search(int a[],int n,int item){int beg,end,mid;beg=0;end=n-1;mid=(beg+end)/2;while((beg<=end)&&(a[mid]!=item)){if(item<a[mid]){end=mid-1;}else{beg=mid+1;}mid=(beg+end)/2;}if(item==a[mid]){return mid;}else{return -1;}}//function to calculate meanfloat mean(int a[],int n){int i;float sum=0;for(i=0;i<n;i++){sum=sum+a[i];}sum=sum/n;return sum;}//main function declarationint main(){int a[50],n,i,item,loc;cout<<"Enter no. of elements you wants: ";cin>>n;cout<<"Enter "<<n<<"array elements:"<<endl;for(i=0;i<n;i++){cin>>a[i];}cout<<"\nThe size of the array entered by user is: "<<n<<endl;cout<<"\nArray elements entered by user…
- / CPP program to demonstrate working of Unordered_map #include <bits/stdc++.h> using namespace std; int main() { unordered_map<int, int> um; um[1] = 2; um[4] = 5; um[2] = 3; um[8] = 5; um[3] = 6; cout << "Elements in unordered_map:\n"; for (auto it : um) cout << "[ " << it.first << ", " << it.second << "]\n"; return 0; } Convert this code to python and then compare between the two codes from where: the compiled size then try to execute it on a lot of data (for map or set) and track the time, State the, time, and size taken for STL and python.LSD Radix sort is used to sort the following list of data: [mak, kcl, pks, kih, brb, jyr, Iwk, scc, tea), how will the array elements look like after second pass? [ tea, brb, scc, kih, mak, Iwk, kcl, pks, jyr] [ mak, scc, kcl, tea, kih, pks, brb, Iwk, jyr] [ brb, jyr, kcl, kih, lwk, mak, pks, scc, tea] O [ mak, kcl, scc, tea, kih, pks, brb, lwk, jyr]arr[] = {64, 34, 25, 12, 22, 11, 90} n = 7 for i = 0 to i = n-2 for j= 0 to j = n-1-2 if arrj] > arr[j+1] output j, swap(arrlj], arrlj+1]);
- def selection_mean(numbs): for i in range (len (numbs)-1): # find the index of smalllest reamining element index_smalllest=i for j in range (1 + 1, len(numbs)): if numbs [j]< numbs [index_smalllest]: index_smalllest j #swap number [i] and numbs [index_smalllest] numbs [i] temp numbs [1] numbs [index_smalllest] numbs [index_smalllest] = temp numbs [10, 2, 78, 45, 32, 7, 11] print('UNSORTED: ', end= ') for num in numbs: print (str(num), end=' ') print() # Fix me # complete the code to sort only even numbers in ascending order #input = [10, 2, 78, 45, 32, 7, 11] #output= [2, 10, 32, 78] size = len(numbs) (numbs, 1, size + 1 ) selection_mean (numbs) print('SORTED: ', ') for num in number#include <bits/stdc++.h>using namespace std;int getMedian(int ar1[], int ar2[], int n){int j = 0;int i = n - 1;while (ar1[i] > ar2[j] && j < n && i > -1)swap(ar1[i--], ar2[j++]);sort(ar1, ar1 + n);sort(ar2, ar2 + n);return (ar1[n - 1] + ar2[0]) / 2;}// Driver Codeint main(){int ar1[] = { 1, 12, 15, 26, 38 };int ar2[] = { 2, 13, 17, 30, 45 };int n1 = sizeof(ar1) / sizeof(ar1[0]);int n2 = sizeof(ar2) / sizeof(ar2[0]);if (n1 == n2)cout << "Median is " << getMedian(ar1, ar2, n1);elsecout << "Doesn't work for arrays"<< " of unequal size";getchar();return 0;} PLEASE SOLVE THIS USING 'class' . Thanks a lot in advance:)ALGO2(A)// A is an integer array, index starting at 11:for i=1 to n-1 do2: for j=n to i+1 do3: if (A[j] < A[j-1]) then4: exchange A[j] with A[j-1] Which function best describes the worst-case running time behaviour of ALGO2 on an array of size n? Option 1) a*(n^2) + b*n +c, where a,b,c are constantsOption 2) a*n + b , where a and b are constantsOption 3) a * n * log(n) + b, where a,b are constantsOption 4) a*(n^3) + b*(n^2) + c*n + d, where a,b,c,d are constants