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vivado [14]
3 years ago
14

Write a program that prompts the user to enter three words. The program will then sort the words in alphabetical order, and disp

lay them on the screen. The program must be able to support words with both uppercase and lowercase letters. You must use C style strings (string class).
Computers and Technology
1 answer:
Softa [21]3 years ago
3 0

Answer:

#include <iostream>

#include<bits/stdc++.h>

using namespace std;

bool sorter(string a, string b)

{

   return a<b;

}

int main(){

   string wordList[3];

   string word;

   for (int i = 0; i < 3; i++){

       cout<< "Enter word: ";

       cin>> word;

       wordList[i] = word;

   }

   sort(wordList, wordList+3, sorter);

   for (int i = 0; i < 3; i++){

       cout<< wordList[i]<< "\n";

}

}

Explanation:

The C++ source code prompts the user for string words that are appended to the wordList array. The array is sorted with the sort() function from the C++ bits library. The finally for loop statement prints out the items in the array.

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Which are technical and visual demands that need to be considered when planning a project? Choose three answers
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Given that ∠XQR = 180° and ∠LQM = 180°, which equation could be used to solve problems involving the relationships between ∠XQM
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C) 180 − (140 − 7a) = (70 − 3a)

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C) 180 − (140 − 7a) = (70 − 3a)

Explanation:

i got it wrong by clicking D on usatestprep

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Queues can be represented using linear arrays and have the variable REAR that point to the position from where insertions can be
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7 0
2 years ago
Suppose you have two arrays of ints, arr1 and arr2, each containing ints that are sorted in ascending order. Write a static meth
telo118 [61]
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.

public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}


So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.

A quick explanation:

We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.

The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.


4 0
3 years ago
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