void find_and replace (char A, char olde, char *newc) for (int i=0; i<5; i++) for (int j-0; j<3; j++) oldcljl) A[i] = newc[j]; if (A[i] The procedure is tested with the following four test cases. 1. oldc = "abc", newc 2. oldc = "cde", newc "bca", "abc" , newc "dab" "bcd" "cda" "bac" 3. oldc 66 newc 4. olde

void find_and replace (char A, char olde, char *newc) for (int i=0; i<5; i++) for (int j-0; j<3; j++) oldcljl) A[i] = newc[j]; if (A[i] The procedure is tested with the following four test cases. 1. oldc = "abc", newc 2. oldc = "cde", newc "bca", "abc" , newc "dab" "bcd" "cda" "bac" 3. oldc 66 newc 4. olde

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter15: Recursion

Section: Chapter Questions

Problem 8SA

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PROGRAM IN C++ Our BoardArray will store 5 entries maximum. Improve the following existing methods: bool add(entry* e) Improve the add method where you will return true if the entry has been successfully added (i.e. the entry is placed in the top 10), otherwise return false (i.e. if the entry did not make it to the top 10). bool remove(const char* person) Improve the remove method where you return true if an entry is successfully deleted (i.e. we have found an entry that matches the person), otherwise return false. Add the following additional methods: bool ban_country(const char* nation) Removes all the entries bearing the country equal to the given nation. Return true if at least one entry has been removed, otherwise return false. int country_wins(const char* nation) Returns the number of entries in the scoreboard who represents the given nation. int exemplary(int score) Returns how many entries in the scoreboard are greater than or equal to the given score. double…
Count consecutive summers def count_consecutive_summers(n): Like a majestic wild horse waiting for the rugged hero to tame it, positive integers can be broken down as sums of consecutive positive integers in various ways. For example, the integer 42 often used as placeholder in this kind of discussions can be broken down into such a sum in four different ways: (a) 3 + 4 + 5 + 6 + 7 + 8 + 9, (b) 9 + 10 + 11 + 12, (c) 13 + 14 + 15 and (d) 42. As the last solution (d) shows, any positive integer can always be trivially expressed as a singleton sum that consists of that integer alone. Given a positive integer n, determine how many different ways it can be expressed as a sum of consecutive positive integers, and return that count. The number of ways that a positive integer n can be represented as a sum of consecutive integers is called its politeness, and can also be computed by tallying up the number of odd divisors of that number. However, note that the linked Wikipedia de0inition…
include<cmath> using namespace std; int main(int argc, char const *argv[]){ int n; cin>>n; // since the divisors occur in pairs, it is enough to traverse till sqrt(n) for(int i=1;i<=sqrt(n);i++) { if(n%i==0) // check the divisibility { if(n/i==i) // if same number appears in the pair, for example for n =100, the number 10 occurs twice. It is enough to print // it once cout<<i<<" "; else // print the divisors cout<<i<<" "<<n/i<<" "; } } // The TC .
Count consecutive summers def count_consecutive_summers(n): Like a majestic wild horse waiting for someone to come and tame it, positive integers can be broken down as sums of consecutive positive integers in various ways. For example, the integer 42 often used as placeholder in this kind of discussions can be broken down into such a sum in four different ways: (a) 3 + 4 + 5 + 6 + 7 + 8 + 9, (b) 9 + 10 + 11 + 12, (c) 13 + 14 + 15 and (d) 42. As the last solution (d) shows, any positive integer can always be trivially expressed as a singleton sum that consists of that integer alone. Given a positive integer n, determine how many different ways it can be expressed as a sum of consecutive positive integers, and return that count. The count of how many different ways a positive integer n can be represented as a sum of consecutive integers is also called its politeness, and can be alternatively computed by counting how many odd divisors that number has. However, note that the linked…
please help me public static void main(String args[]){ int[] x={1,4,88,9,13,77,4,18}; for(int i=0;i<x.length;i++){ for( int j=i+1;j<x.length;j++){ if(x[i]==x[j]){ System.out.println("duplicates exists is : "+x[i]); }}} System.out.println("no duplication"); } The above segment of code is searching the duplicated values inside an array, where the complexity is O(n). Suggest another solution for the above problem with less time complexity.
Correct answer will be upvoted else Multiple Downvoted. Computer science. pick a non-void adjacent substring of s that contains an equivalent number of 0's and 1's; flip all characters in the substring, that is, supplant all 0's with 1's, as well as the other way around; turn around the substring. For instance, think about s = 00111011, and the accompanying activity: Pick the initial six characters as the substring to follow up on: 00111011. Note that the number of 0's and 1's are equivalent, so this is a lawful decision. Picking substrings 0, 110, or the whole string would not be imaginable. Flip all characters in the substring: 11000111. Invert the substring: 10001111. Find the lexicographically littlest string that can be gotten from s after nothing or more tasks. Input The primary line contains a solitary integer T (1≤T≤5⋅105) — the number of experiments. Every one of the accompanying T lines contains a solitary non-void string — the input string s for…
Please define and initialize and array X of length 8. You need implement following and write outcomes with reasons; int X[ ] = { 1,2,3,4,5,6,7,8 }; 1.cout<< X+1; 2.cout<<&X+1; 3.cout<<2[X]; 4.cout<<&X[0]+1; 5.cout<<*(X+1); 6.cout<<*X+1; Also, practice and play with pointers using multiplication, division, subtraction and addition. You may also play with it your ways and find out reasoning. Write down those reasoning.
Debug the following program and answer the following questions. #include <stdio.h> typedef struct node { int value; struct node *next; } node; int ll_has_cycle(node *first) { node * head = first; while (head->next) { head = head->next; if (head == first) return 1; } return 0; } void test_ll_has_cycle(void) { int i,j; node nodes[5]; for(i=0; i < sizeof(nodes)/sizeof(node); i++) { nodes[i].next = NULL; nodes[i].value = i; } nodes[0].next = &nodes[1]; nodes[1].next = &nodes[2]; nodes[2].next = &nodes[1]; printf("Checking first list for cycles. There should be a cycle, ll_has_cycle says it has %s cycle\n", ll_has_cycle(&nodes[1])?"a":"no"); printf("Checking length-zero list for cycles. There should be none, ll_has_cycle says it has %s cycle\n",…
Q#2 Write a recursive function zeroCount ( int a[ ], int s, int e) that receives an array of integers a [], a start index s, and an end index e. The function should return the number of zeros in that array between s and e. int zeroCount ( int a[ ], int s, int e); Trace your function given the following array and function call. Draw your steps. int a[ ] = {1, 0, 0, 5}; int zeros = zeroCount(a, 0, 3); language c++
Using a source-level debugger, determine for what values of argumentsthe function Mystery returns a zero.#include <stdio.h>int Mystery(int a, int b, int c);int main(void){int sum = 0; // running sum of Mysteryfor (int i = 100; i > 0; i--) {for (int j = 1; j < i; j++) {for (int k = j; k < 100; k++)sum = sum + Mystery(i, j, k);}}}int Mystery(int a, int b, int c){int out;out = 3*a*a + 7*a - 5*b*b + 4*b + 5*c ;return out;}
What is wrong with the following code? int *p;. //Line 1int *q;. //Line 2 p = new int [5];. //Line 3 *p = 2; //Line 4 for (int i = 1; i < 5; i++). //Line 5p[i] = p[i-1] + i;. //Line 6 q = p;. //Line 7 delete [] p; //Line 8 for (int j = 0; j < 5; j++) //Line 9 cout << q[j] << " "; //Line 10 cout << endl; //Line 11
Create a function that takes a grid of # and -, where each hash (#) represents amine and each dash (-) represents a mine-free spot.Return a grid, where each dash is replaced by a digit, indicating the number ofmines immediately adjacent to the spot i.e. (horizontally, vertically, anddiagonally).Example of an input:[ ["-", "-", "-", "#", "#"],["-", "#", "-", "-", "-"],["-", "-", "#", "-", "-"],["-", "#", "#", "-", "-"],["-", "-", "-", "-", "-"] ]Example of the expected output:[ ["1", "1", "2", "#", "#"],["1", "#", "3", "3", "2"],["2", "4", "#", "2", "0"],["1", "#", "#", "2", "0"],["1", "2", "2", "1", "0"] ]Here is a tip. When checking adjacent positions to a specific position in the grid,the following table might assist you in determining adjacent indexes:NW position =current_row - 1current_col - 1N position =current_row - 1current_colNE position =current_row - 1current_col + 1W position =current_rowcurrent_col - 1Current position =current_rowcurrent_colE position…