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.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]);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]
- 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 number2.)Recurrences. a.)Provide a Sample code for the recurrences below. i.) T(n)=T(n/5)+T(3n/5)+n ii.)T(n)=T(2n/5)+T(3n/5)+n iii.) T(n)=4T(n/2)+n^3#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:)