Answer:
To do this you'll need to use malloc to assign memory to the pointers used. You'll also need to use free to unassign that memory at the end of the program using the free. Both of these are in stdlib.h.
#include <stdlib.h>
#include <stdio.h>
#define SIZE_X 3
#define SIZE_Y 4
int main(void){
int **matrix, i, j;
// allocate the memory
matrix = (int**)malloc(SIZE_X * sizeof(int*));
for(i = 0; i < SIZE_X; i++){
matrix[i] = (int *)malloc(SIZE_Y * sizeof(int));
}
// assign the values
for(i = 0; i < SIZE_X; i++){
for(j = 0; j < SIZE_Y; j++){
matrix[i][j] = SIZE_Y * i + j + 1;
}
}
// print it out
for(i = 0; i < SIZE_X; i++){
for(j = 0; j < SIZE_X; j++){
printf("%d, %d: %d\n", i, j, matrix[i][j]);
}
}
// free the memory
for(i = 0; i < SIZE_X; i++){
free(matrix[i]);
}
free(matrix);
return 0;
}
Incomplete question. The Options read;
A. The year purchased
B. Brian's classmates
C. The amount of memory
D. The type of operating system
Answer:
<u>B. Brian's classmates</u>
Explanation:
<em>Remember,</em> the question is concerned about <em>"the individuals"</em> in the data set, <u>so the year they purchased their computer, neither is the amount of memory of the computer and the type of operating system can be classified as individuals in the data set.</u>
Hence, we can correctly say, only Brian's classmates are the individuals in this data set.
Answer:
First we understand what is hash function.A hash function is mostly used in Hashmaps. It maps different keys to a set of values.There may occur a case when we have same key but different values.This case is called collision.So we have to use different collision handling techniques that are open addressing and separate chaining.
A perfect hash function maps key-value pair such that there are no collisions.
The answer is B becausePower brakes are a system of hydraulics used to slow down or stop most motor vehicles. It uses a combination of mechanical components to multiply the force applied to the brake pedal by the driver into enough force to actuate thebrakes and stop a vehicle that can weigh several tons.
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.
