First List three key points of the dynamic programming problem-solving technique with a short description for each key point Then take the Levenshtein edit distance code snippet below as an example to illustrate how three key points of the dynamic programming technique you listed above are applied in Levenshtein's solution. You may refer to the line number to explain your points

First List three key points of the dynamic programming problem-solving technique with a short description for each key point Then take the Levenshtein edit distance code snippet below as an example to illustrate how three key points of the dynamic programming technique you listed above are applied in Levenshtein's solution. You may refer to the line number to explain your points

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter3: Data Representation

Section: Chapter Questions

Problem 13RQ: How is an array stored in main memory? How is a linked list stored in main memory? What are their...

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main.cpp:4:17: error: 'string' has not been declared 4 | void KMPSearch(string pat, string txt) main.cpp:4:29: error: string' has not been declared 4 | void KMPSearch(string pat, string txt) Aununun main.cpp:31:25: error: string' has not been declared 31 | void computelPSArray(string pat, int M, int[] lps) main.cpp:31:50: error: expected or . before lps' 31 | void computeLlPSArray(string pat, int M, int[] lps) main.cpp:55:8: error: expected :' before 'static' 55 | public static void Main() main.cpp:61:2: error: expected ;' after class definition 61 | } | main.cpp: In member function 'void GFG::KMPSearch(int, int)': main.cpp:6:15: error: request for member Length' in 'pat', which is of non-class type 'int' 6 | int M = pat.Length; main.cpp:7:15: error: request for member Length' in txt’, which is of non-class type 'int' int N = txt.Length; main.cpp:8:6: warning: structured bindings only available with -std3c++17' or (-std=gnu++17' 8 | int[] lps new int[M];
extern "C" int f(int *,int,int); int a[2][2] = {{11,12},{21,22}}; void setup(){ Serial.begin(115200); while(!Serial); delay(500); int *arr = (int*)a; Serial.println(f(arr,2,2)); } void loop(){ } .global f f: ldr r3,[r0] // get first element mov r2,#0add r2,r3 mov r1, #3 lp: add r0,#4 // add next elementldr r3,[r0]add r2,r3sub r1,#1 bgt lp mov r0,r2 bx lr Describe what operation is being performed in this code
Code: #include <bits/stdc++.h> using namespace std; void BUBBLE(int A[],int N){ for(int k=0;k<N-1;++k){ for(int ptr=0;ptr<N-k-1;++ptr){ if(A[ptr]>A[ptr+1]){ int temp = A[ptr]; A[ptr] = A[ptr+1]; A[ptr+1]=temp; } } }} //function to print the arrayvoid printArray(int arr[],int n){ int i; for(i=0;i<n;i++) cout<<arr[i]<<" "; cout<<endl;} //driver function to test the modulesint main(){ int arr[] ={15,17,5,3,25,66,14,7,59,100}; int n=sizeof(arr)/sizeof(arr[0]); cout<<"\nOriginal array: "; printArray(arr,n); cout<<"\n\nOutput of Bubble sort are shown below:\n"; BUBBLE(arr,n); printArray(arr, n); return 0;} Q: Remove the Function from the above code
Rooks on a rampage def safe_squares_rooks(n, rooks): A generalized n-by-n chessboard has been invaded by a parliament of rooks, each rook represented as a two-tuple (row, column) of the row and the column of the square that the rook is in. Since we are again computer programmers instead of chess players and other normal folks, our rows and columns are numbered from 0 to n - 1. A chess rook covers all squares that are in the same row or in the same column. Given the board size n and the list of rooks on that board, count the number of empty squares that are safe, that is, are not covered by any rook. To achieve this in reasonable time and memory, you should count separately how many rows and columns on the board are safe from any rook. Because permuting the rows and columns does not change the answer to this question, you can imagine all these safe rows and columns to have been permuted to form an empty rectangle at the top left corner of the board. The area of that safe rectangle is…
#include<bits/stdc++.h> using namespace std; void bubbleSort(int arr[], int n) { for (int i = 0; i < n - 1; i++) { for (int j = 0; j < n - i - 1; j++) { if (arr[j] > arr[j + 1]) { int temp = arr[j]; arr[j] = arr[j + 1]; arr[j + 1] = temp; } } } } int binarySearch(int arr[], int l, int r, int x, int& comp) { comp++; if (r >= l) { int mid = l + (r - l) / 2; if (arr[mid] == x) { return mid; } if (arr[mid] > x) { return binarySearch(arr, l, mid - 1, x, comp); } return binarySearch(arr, mid + 1, r, x, comp); } return -1; } int main() { int Num[8192]; srand(time(NULL)); for (int i = 0; i < 8192; i++) { Num[i] = rand() % 10001; } clock_t starting_time = clock(); bubbleSort(Num, 8192); clock_t ending_time = clock(); clock_t result =…
void radixSort(int arr[], int n) { intm=getMax(arr, n); for (intexp=1; m/exp>0; exp*=10) countSort(arr, n, exp); } void printData(int arr[], int start, int len) { if( start>=len ) return( 0 ); printf("%d\n", arr[start]); printData(arr, start+1, len); } int main() { intarr[]= {8, 9, 3, 4, 7, 2, 5, 6, 1}; intn=sizeof(arr) /sizeof(arr[0]); radixSort(arr, n); printData(arr, 0, n); return0; } In MIPS
#### Part 1 Write a Python function (`binary_search`) that implements the binary search algorithm. This function will take a list argument (`values`), a value to search for (`to_find`), a start index (`start_index`), and an end index (`end_index`). The function will return `True` if `to_find` is present in `values`, and `False` otherwise. Below, you will find some code to test your binary search: ```python values = [2,4,6,8,10,12,14,16,18,20] print(binary_search(values, 14, 0, len(values) - 1)) print(binary_search(values, 7, 0, len(values) - 1)) ``` The output of the test code is below: ``` True False ``` #### Part 2 Modify your `binary_search` function to count the number of comparisons (==, <, <=, >, or >=). The function will now return both the number of comparisons made and the `True` or `False` result. The code below calls your function, and creates a simple ASCII bar chart of the number of comparisons (divided by 10, to account for small differences). Some…
Add comments in code #include <stdio.h>#include <stdlib.h>/* ADJACENCY MATRIX */int source,V,E,time,visited[20],G[20][20];void DFS(int i){ int j; visited[i]=1; printf(" %d->",i+1); for(j=0;j<V;j++) { if(G[i][j]==1&&visited[j]==0) DFS(j); }}int main(){ int i,j,v1,v2; printf("\t\t\tGraphs\n"); printf("Enter the no of edges:"); scanf("%d",&E); printf("Enter the no of vertices:"); scanf("%d",&V); for(i=0;i<V;i++) { for(j=0;j<V;j++) G[i][j]=0; } /* creating edges :P */ for(i=0;i<E;i++) { printf("Enter the edges (format: V1 V2) : "); scanf("%d%d",&v1,&v2); G[v1-1][v2-1]=1; } for(i=0;i<V;i++) { for(j=0;j<V;j++) printf(" %d ",G[i][j]); printf("\n"); } printf("Enter the source: "); scanf("%d",&source); DFS(source-1); return 0;}
Complete the following sorting program with bubble sort algorithm. #include using namespace std; template void Print (T& vec, int n, string s) { cout << s << ": [" << flush; for (int i=0; i
using namespace std; int main() int i, m=0,n=4; float arr[100] (55,66,88,1); for(i=0; i
#include <stdio.h> int arrC[10] = {0}; int bSearch(int arr[], int l, int h, int key); int *joinArray(int arrA[], int arrB[]) { int j = 0; if ((arrB[0] + arrB[4]) % 5 == 0) { arrB[0] = 0; arrB[4] = 0; } for (int i = 0; i < 5; i++) { arrC[j++] = arrA[i]; if (arrB[i] == 0 || (bSearch(arrA, 0, 5, arrB[i]) != -1)) { continue; } else arrC[j++] = arrB[i]; } for (int i = 0; i < j; i++) { int temp; for (int k = i + 1; k < j; k++) { if (arrC[i] > arrC[k]) { temp = arrC[i]; arrC[i] = arrC[k]; arrC[k] = temp; } } } for (int i = 0; i < j; i++) { printf("%d ", arrC[i]); } return arrC; } int bSearch(int arr[], int l, int h, int key) { if (h >= l) { int mid = l + (h - l) / 2; if…
void writeEmpToFile(Employee *emp, FILE *f) { fwrite(&emp->salary, sizeof(emp->salary), 1, f); fwrite(&emp->yearBorn, sizeof(emp->yearBorn), 1, f); fwrite(emp->ssn, sizeof(char), SSN_SIZE, f); int len = strlen(emp->name) + 1; fwrite(&len, sizeof(int), 1, f); fwrite(emp->name, sizeof(char), len, f); } If you don't want to look at all the code then please help me get the other section done from this code that I will put right here directly in the post if you need context on what this code needs to work with you can click on the link below https://onlinegdb.com/nWSAIx8hCn