I am writing a c++ program, describing the partition of heat (temperatures) along a 1 dimensional bar. This partition is according to a recursive algorithm (which I will not include here). The bar consists of 100 points that can have a temperature. So, after each second there will be an array with 100 members (the temperatures of the 100 points). Since the algorithm is recursive, a new array (at t seconds) has to be calculated using the members of the former array (at t-1 seconds). How can I combine such arrays with the for-loops necessary for the recursive algorithm? Am I to make a matrix of dimension 100*t, growing as t increases? For large t this would use a lot of memory storage. I would rather let the program remember the former array (t-1 seconds) until the algorithm has been applied and then replace this array with the array of time t in order to produce the array of time t+1 a.s.o. till the final time tf. How can this be done?

Thanks!
One doesn’t have to know science for this question. It concern arrays, matrices and recursion in c++. I included the background just to indicate why it bothers me.

Do you really need recurrence? Iteration seems to be enough.
Just keep 2 arrays in the mem at once:
<type>[100] measurements1;
<type>[100] measurements2;
1. Measure values and store into measurements1.
2. Call the measurements procedure. It would read measurements1 and store output into measurements2.
3. Save measurements1 to somewhere (disk?).
4. Copy measurements2 to measurements1 (you could operate on pointers and simply replace the pointers).
5. goto 2.

You are given an arrayA[1..n] of numbers and an integer m<n. The numbers in the sub-arrays A[1..m] and A[m +1..n]are (individually) sorted. As example, consider the array [1,5,7,8,11,2,3,4,10] the underlined numbers indicate the two sub-arrays (here m = 5 ). Give an O(n) algorithm that sorts the array A using O(1) storage, i.e. can’t use a temporary array for storage.

yes with o(1) storage
there is algo.
compare first and m+1element if m+1 th element is small then swap else compare second with m+1 increment ptr of second arr if swapped on continuing this way we can get sorted arr with o(n) complexity

I have a Dell desktop computer. It came with 2 150g hard disks in a RAID array. One disk has failed.

1) How do I replace it?
2) Can I replace it with a bigger disk? Then, later, change out the other to a bigger disk too?

the first post is absolutely correct if it is set up as a mirrored raid 1 array. However, if by chance, it is a striped raid 0 array (which i doubt…I don’t think dell would ship this) you will lose all of your data because the 2 drives would be acting as one.

looking for a vendor recommendations and why ?

It all depends on the size requirements and speed of growth of your data/small business.
What is the size of your network? how many users?
What is your budget?
If your budget is limited you might just want to add another hard drive to your server and make it only for data.

Its a C program to be done in linux platform.

There are many ways to sort, including the qsort function provided by the C standard library. If you want to implement a sort algorithm yourself, selection sort is a good place to start. See below for my example of sorting an array of int :

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

const int N = 10;
const int MAX_VAL = 100;

void selectionSort(int *, int);
void printArray(const int *, int);
int random(int range);

int main(int argc, char *argv[]) {
    time_t t;
    int *intArray,i;

    /* allocate int array */
    intArray = (int *)malloc(N * sizeof(int));

    /* fill array */
    srand((unsigned)time(&t));
    for (i = 0; i < N; i++) {
        intArray[i] = random(MAX_VAL);
    }
   
    printf("unsorted : ");
    printArray(intArray,N);
    selectionSort(intArray,N);
    printf(" sorted : ");
    printArray(intArray,N);
    free(intArray);
   
    return 0;
}

void selectionSort(int *a, int n) {
    int i, j, min, t;

    for (i = 0; i < n; i++) {
        min = i;
        for (j = i+1; j < n; j++) {
            if (a[j] < a[min]) {
                min = j;
            }
        }
        t = a[min];
        a[min] = a[i];
        a[i] = t;
    }
}

void printArray(const int *a, int n) {
    int i;

    for (i = 0; i < n; i++) {
        printf("%2d ",a[i]);
    }
    puts("");
}

int random(int range) {
    float x = ((float)rand() / (float)RAND_MAX) * range;
    return (int)x;
}

#if 0

Sample run:

unsorted : 18 11 87 62 91 74 30 67 38 1
    sorted : 1 11 18 30 38 62 67 74 87 91

#endif