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irga5000 [103]
3 years ago
13

Merge arrays A and B into array C using merge sort algorithm. Give final sorted numbers of C using colored numbers without showi

ng intermediate steps. For example, your answer can be [1, 3, 3, 5, 7, 7, 9, 9, 13, 25] or [1, 3, 3, 5, 7, 7, 9,9, 13, 25), or another variation. void merge(int all, int aux[], int lo, int mid, int hi) / for (int k = lo; k<= hi; k++) aux [k] a[k]; int i lo, j = mid+1; for (int k lo; k<-hi; k++) if (i > mid) ak) else if (1 > hi) a[k] else if (small (aux(j), aux[1])) a[k] else ak aux[j++); aux (i++); aux [j++); /*++ aux (i++]; 1 j++
Computers and Technology
1 answer:
sweet-ann [11.9K]3 years ago
7 0

Answer:

Explanation:

Given code:-

#include<iostream>

using namespace std; //namespace

#define RED "\033[31m" /* Red */

#define BLUE "\033[34m" /* Blue */ //color definition

#define WHITE "\033[37m" /* White */

void mergeArrays(int arr1[], int arr2[], int n1, int n2, int arr3[])

//merger sort

{

int i = 0, j = 0, k = 0;

while (i<n1 && j <n2) //iterate over both array

{

if (arr1[i] < arr2[j]) //iff array 1 element is larger

{

arr3[k] = arr1[i]; //assign to a3

k++;

i++;

}

else

{

arr3[k] = arr2[j]; //else assign array 2 to a3

j++;

k++;

}

}

//say the array turns out to br of different sizes

//let us copy the remaining elemnt to array 3

while (i < n1)

{

arr3[k] = arr1[i];

i++;

k++;

}

while (j < n2)

{

arr3[k] = arr2[j];

k++;

j++;

}

}

int main(){

int A[4] = {3,5,7,9}; //array 1 assignment

int a_size = 4;

int b_size = 6;

int B[6] = {1,3,7,9,13,25}; //array 2 assignment

int C[10]; //array 3 declearation

int a_count = 0;

int b_count = 0; //counter for both array index

mergeArrays(A,B,a_size,b_size,C); //merger function applied;

cout<<WHITE<<"A: ";

for(int i = 0; i<4; i++){

cout<<RED<<A[i]<<" ";

}

cout<<endl<<WHITE<<"B: ";

for(int i = 0; i<6; i++){

cout<<BLUE<<B[i]<<" ";

}

cout<<endl<<WHITE<<"C: ";

for(int i = 0; i<10; i++){ //iterate over C

if(C[i] == A[a_count] && a_count < a_size){

cout<<RED<<C[i]<<" ";

// if element is common in array 1 print in red

a_count++;

}

else if (C[i] == B[b_count] && b_count < b_size){

//say element was similar in array 2,  print in blue

cout<<BLUE<<C[i]<<" ";

b_count++;

}

}

cout<<endl;

}

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What does the following loop do?int[] a = {6, 1, 9, 5, 12, 3};int len = a.length;int x = 0;for (int i = 1; i &lt; len; i++)if (a
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Answer:

Option 1: Finds the position of the largest value in a

Explanation:

Given the codes as follows:

  1.        int[] a = {6, 1, 9, 5, 12, 3};
  2.        int len = a.length;
  3.        int x = 0;
  4.        for (int i = 1; i < len; i++)
  5.        {
  6.            if (a[i] > a[x])
  7.                x = i;
  8.        }
  9.        System.out.println(x);

The code is intended to find a largest value in the array, a. The logic is as follows:

  • Define a variable to hold an index where largest value positioned. At the first beginning, just presume the largest value is held at index zero, x = 0. (Line 3)
  • Next, compare the value location in next index. If the value in the next index is larger, update the index-x to the next index value (Line 4 - 8). Please note the for-loop traverse the array starting from index 1. This is to enable the index-1 value can be compared with index-0 and then followed with index-2, index-3 etc.
  • After completion of the for-loop, the final x value will be the index where the largest value is positioned.
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