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Task:
Given a natural number N (set arbitrarily as a preprocessor constant) and one-dimensional array A0, A1, …, AN-1 of integers (generate positive and negative elements randomly, using the <stdlib.h> library function rand()). Perform the following actions: Determine the three maximum and two minimum values of this array.
Code with search for two minimum values:
#include <stdio.h>
#include <stdlib.h>
#define N 9
int main() {
int M[N], i, a[N], fbig, sbig, tbig, min, smin;
for (i = 0; i < N; i++) {
M[i] = rand() % 20 - 10;
printf("%i\t", M[i]);
}
printf("\n");
for (i = 0; i < N; i++) {
if (a[i] < min) {
smin = min;
min = a[i];
} else
if (a[i] < smin && a[i] != min)
smin = a[1];
}
printf("\nMinimum=%d \nSecond Minimum=%d", min, smin);
return 0;
}
I tried to compare array elements with each other but here is my result:
-7 -4 7 5 3 5 -4 2 -1
Minimum=0
Second Minimum=0
I would be very grateful if you could help me fix my code or maybe I'm doing everything wrong and you know how to do it right. Thank you for your time
I will revise my answer if op address what to do about duplicate values. My answer assume you want possible duplicate values in the minimum and maximum arrays, while other answers assume you want unique values.
The easiest solution would be to sort the input array. The minimum is the first 2 values and the maximum would be the last 3:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define MAX_N 3
#define MIN_N 2
#define N 9
void generate(size_t n, int a[n]) {
for(size_t i = 0; i < n; i++)
a[i] = rand() % 20 - 10;
}
void print(size_t n, int a[n]) {
for(size_t i = 0; i < n - 1; i++)
printf("%d, ", a[i]);
if(n) printf("%d\n", a[n-1]);
}
int cmp_asc(const void *a, const void *b) {
if(*(int *) a < *(int *) b) return -1;
if(*(int *) a > *(int *) b) return 1;
return 0;
}
int main() {
int t = time(0);
srand(t);
printf("%d\n", t); // essential for debugging
int a[N];
generate(N, a);
print(N, a);
qsort(a, N, sizeof *a, cmp_asc);
print(MIN_N, a);
print(MAX_N, a + (N - MAX_N));
}
If you cannot use sort then consider the following purpose built algorithm. It's much easier to use arrays (min and max) rather than individual values, and as a bonus this allows you to easily change how many minimum (MIN_N) and maximum (MAX_N) values you want. First we need to initialize the min and max arrays, and I use the initial values of the input array for that. I used a single loop for that. To maintain the invariant that the min array has the MIN_N smallest numbers we have seen so far (a[0] through a[i-1]) we have to replace() largest (extrema) of them if the new value a[i] is smaller. For example, if the array is min = { 1, 10 } and the value we are looking at is a[i] = 5 then we have to replace the 10 not the 1.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define MAX_N 3
#define MIN_N 2
#define N 9
void generate(size_t n, int a[n]) {
for(size_t i = 0; i < n; i++)
a[i] = rand() % 20 - 10;
}
void print(size_t n, int a[n]) {
for(size_t i = 0; i < n - 1; i++)
printf("%d, ", a[i]);
if(n) printf("%d\n", a[n-1]);
}
int cmp_asc(const void *a, const void *b) {
if(*(int *) a < *(int *) b) return -1;
if(*(int *) a > *(int *) b) return 1;
return 0;
}
int cmp_desc(const void *a, const void *b) {
return cmp_asc(b, a);
}
void replace(size_t n, int a[n], int v, int (*cmp)(const void *, const void *)) {
int *extrema = &a[0];
for(size_t i = 1; i < n; i++) {
if(cmp(extrema, &a[i]) < 0) {
extrema = &a[i];
}
}
if(cmp(extrema, &v) > 0)
*extrema = v;
}
void min_max(size_t n, int a[n], size_t min_n, int min[n], size_t max_n, int max[n]) {
for(size_t i = 1; i < n; i++) {
if(i < min_n)
min[i] = a[i];
else
replace(min_n, min, a[i], cmp_asc);
if(i < max_n)
max[i] = a[i];
else
replace(max_n, max, a[i], cmp_desc);
}
}
int main() {
int t = time(0);
srand(t);
printf("%d\n", t); // essential for debugging
int a[N];
generate(N, a);
print(N, a);
int min[MIN_N];
int max[MAX_N];
min_max(N, a, MIN_N, min, MAX_N, max);
print(MIN_N, min);
print(MAX_N, max);
}
and here is example output. The first value is a the seed in case you have to reproduce a run later. Followed by input, min and max values:
1674335494
-7, 0, -2, 7, -3, 4, 5, -8, -9
-9, -8
7, 5, 4
If MIN_N or MAX_N gets large, say, ~1,000+, then you want sort the min and max arrays and use binary search to figure out where to inserta[i]. Or use a priority queue like a heap instead of arrays.
There are multiple problems in your code:
min and smin are uninitialized, hence the comparisons in the loop have undefined behavior and the code does work at all. You could initialize min as a[0] but initializing smin is not so simple.
there is a typo in smin = a[1]; you probably meant smin = a[i];
Note that the assignment is somewhat ambiguous: are the maximum and minimum values supposed to be distinct values, as the wording implies, or should you determine the minimum and maximum elements of the sorted array?
For the latter, sorting the array, either fully or partially, is a simple solution.
For the former, sorting is also a solution but further testing will be needed to remove duplicates from the sorted set.
Here is a modified version to print the smallest and largest values:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define N 9
#define N_MIN 2
#define N_MAX 3
void swap(int *a, int *b) {
int tmp = *a;
*a = *b;
*b = tmp;
}
int main() {
int a[N], i, j, e, dup;
int smallest[N_MIN], nsmall = 0;
int largest[N_MAX], nlarge = 0;
srand(time(NULL));
for (i = 0; i < N; i++) {
a[i] = rand() % 20 - 10;
printf("%i\t", a[i]);
}
printf("\n");
for (i = 0; i < N; i++) {
e = a[i];
dup = 0;
for (j = 0; j < nsmall; j++) {
if (e == smallest[j]) {
dup = 1;
break;
}
if (e < smallest[j]) {
swap(&e, &smallest[j]);
}
}
if (!dup && nsmall < N_MIN) {
smallest[nsmall++] = e;
}
e = a[i];
dup = 0;
for (j = 0; j < nlarge; j++) {
if (e == largest[j]) {
dup = 1;
break;
}
if (e > largest[j]) {
swap(&e, &largest[j]);
}
}
if (!dup && nlarge < N_MAX) {
largest[nlarge++] = e;
}
}
printf("smallest values: ");
for (i = 0; i < nsmall; i++) {
printf(" %d", smallest[i]);
}
printf("\n");
printf("largest values: ");
for (i = nlarge; i --> 0;) {
printf(" %d", largest[i]);
}
printf("\n");
return 0;
}
As already noted, the most direct way to do this would be to simply sort the array. (In fact, if all you need is an output of five integers then your array only need be five elements long.) But I will presume that that is not the point of this homework.
Your goal isn’t super efficiency or a pretty algorithm. It is simply to solve the tasks. Do them one at a time.
First question: How would you find the largest value?
Answer: Loop through the array, keeping track of the largest element found so far.
int largest = array[0]; // why start with this value?
for (int n = 0; n < size; n++)
if (array[n] > largest)
largest = array[n];
Second question: How would you find the smallest value?
Answer: Almost the same way, with only a simple change: Instead of testing if (array[n] > largest) we want to test if (array[n] < smallest), right?
int smallest = largest; // why start with this value?
for (int n = 0; n < size; n++)
if (...) // new condition goes here
smallest = array[n];
Third question: How would you find the second smallest value?
Answer: It should not surprise you that you just need to change the if condition in that loop again. An element would be the second smallest if:
it is the smallest value greater than the smallest.
Think about how you would change your condition:
int second_smallest = largest; // why start with this value?
for (int n = 0; n < size; n++)
if (... && ...) // what is the new test condition?
second_smallest = array[n];
Remember, this time you are testing two things, so your test condition needs that && in it.
Fourth question: can you write another loop to find the second-largest? How about the third-largest?
At this point you should be able to see the variation on a theme and be able to write a loop that will get you any Nth largest or smallest value, as long as you already have the (N-1)th to work against.
Further considerations:
Is it possible that the third-largest is the same as the second-smallest?
Or the smallest?
Is it possible for there to not be a third-largest?
Does it matter?
Put all these loops together in your main() and print out the results each time and you are all done!
...
int main(void)
{
int array[SIZE];
// fill array with random numbers here //
int largest = array[0];
for (...)
if (...)
...
int smallest = largest;
for (...)
if (...)
...
int second_smallest = largest;
for (...)
if (...)
...
int second_largest = smallest;
for (...)
if (...)
...
int third_largest = smallest;
for (...)
if (...)
...
printf( "The original array = " );
// print original array here, then: //
printf( "largest = %d\n", largest );
printf( "2nd largest = %d\n", second_largest );
printf( "3nd largest = %d\n", third_largest );
printf( "2nd smallest = %d\n", second_smallest );
printf( "smallest = %d\n", smallest );
return 0;
}
Example outputs:
{ 1 2 3 4 }
smallest = 1
2nd smallest = 2
3rd largest = 2
2nd largest = 3
largest = 4
{ 5 5 5 5 5 }
smallest = 5
2nd smallest = 5
3rd smallest = 5
largest = 5
{ 1 2 }
smallest = 1
2nd smallest = 2
3rd smallest = 2
largest = 2
Bonus: be careful with variable names. There has been no need to use short abbreviations since before the early nineties. Prefer clarity over brevity.
Given an array of n integers, not necessarily sorted, is there an O(n) algorithm to find the least integer that is greater than the minimum integer in the array but that is not in the array?
The following algorithm has a complexity O(n).
I will assume here that the missing element must be selected after the minimum value.
The algorithm can be easily modified if the minimum possible value is fixed, for example equal to 0.
Once we have determined the minimum value a (in O(n) or in O(1) if the value is known in advance),
then we know that the missing value is less or equal a + n, if n is the array size.
Then we simply have to use an array of size n+1, present[n+1], initialised to 0,
and then to look at all values arr[i]:
if (arr[i] <= a+n) present[arr[i] - a] = 1;
Finally, in a third step we simply have to examine the array present[.], and seach for the first index k such that present[k]==0.
The first missing number is equal to a + k.
#include <stdio.h>
#include <stdlib.h>
int find_missing (int *arr, int n) {
int vmin = arr[0];
int *present = calloc (n+1, sizeof(int));
for (int i = 1; i < n; ++i) {
if (arr[i] < vmin) {
vmin = arr[i];
}
}
int vmax = vmin + n;
for (int i = 0; i < n; ++i) {
if (arr[i] <= vmax) {
present[arr[i] - vmin] = 1;
}
}
int k = 0;
for (k = 0; k <= n; ++k) {
if (present[k] == 0) break;
}
free(present);
return vmin + k;
}
int main() {
int arr[] = {2, 3, 5, 6, 8};
int n = sizeof(arr)/sizeof(arr[0]);
int missing = find_missing (arr, n);
printf ("First missing element = %d\n", missing);
return 0;
}
Given an array of n integers, without negative numbers, not
necessarily sorted, is there an O(n) algorithm to find the least
integer that is greater than the minimum integer in the array but that
is not in the array?
This can be solved with O(N) time complexity, with N being the number of element in the array. Let us call that array a1, the algorithm is as follows:
Find the smallest value in a1 (i.e, min);
Create a new array a2 with size equals to N;
Initialized the array a2 with a value to signal missing element, for instance min - 1;
Iterate through the array a1, and for each position, take the element in that position e1 = a1[i], and only if e1 is not greater than min - N mark the corresponded position on a2 as visited, for instance using the element itself, namely a2[e1 - min] = e1; If e1 is greater than min - size then it clearly does not belong to the sequence, and can be ignored because in the worst-case scenario the first missing value will be the value min + N + 1.
Lastly, iterate through the array a2, and get the first element = -1; it will be your first missing element.
Steps 1, 3, 4, and 5, all of them take in the worst-case scenario N. Hence, this algorithm takes 4N, which is O(N) time complexity;
The code will be straight forward to implement; for instance something as follows (for an array {5, 3, 0, 1, 2, 6}):
#include <stdio.h>
#include <stdio.h>
#include <stdlib.h>
int find_min(int *array, int size){
int min = array[0];
for(int i = 0; i < size; i++)
min = (array[i] < min) ? array[i] : min;
return min;
}
void fill_array(int *array, int size, int missing_value){
for(int i = 0; i < size; i++)
array[i] = missing_value;
}
void mark_existing_values(int *s, int size, int *d, int min){
for(int i = 0; i < size; i++){
int elem = s[i];
if(elem - min < size)
d[elem - min] = elem;
}
}
int find_first_missing_value(int *array, int size, int min){
int missing_value = min - 1;
for(int i = 0; i < size; i++){
if(array[i] == missing_value){
return i + min;
}
}
return missing_value;
}
int main(){
int array_size = 6;
int array_example [] = {5, 3, 0, 1, 2, 6};
int min = find_min(array_example, array_size);
int *missing_values = malloc(array_size * sizeof(int));
fill_array(missing_values, array_size, min - 1);
mark_existing_values(array_example, array_size, missing_values, min);
int value = find_first_missing_value(missing_values, array_size, min);
printf("First missing value {%d}\n", value);
free(missing_values);
return 0;
}
OUTPUT:
First missing value {4}
This algorithm works also for negative numbers, for instance if int array_example [] = {-1, -3, 0, 3, 5, 6, 7, 8, 10};, then the output would be:
First missing value {-2}
The code can be simplified if in step 3 and step 4 instead of min - 1 and a2[e1 - min] = e1, respectively, we use two flags to signal missing (e.g., 0) and existing values (e.g., 1). Just like showcase in #Damien code. The downside is that we are using two flags instead of one. The benefit is that it simplifies the code, and in case the smallest value in the array is the smallest value that can be represented in C we will not underflow with min - 1.
You can use the technique of bitwise XOR.
This method has O(n) time complexity and it is working on unsorted arrays too.
Also, keep in mind, this works only if one element is missing from the array.
#include <stdio.h>
int main()
{
int arr[] = { 1, 2, 4, 5, 6, 7 };
int arr_size = sizeof(arr) / sizeof(arr[0]);
int x = arr[0]; //XOR together all of the array elements
for (int i = 1; i < arr_size; i++)
{
x ^= arr[i];
}
int y = 1; //XOR together the numbers from 1 to size of array + 1
for (int i = 2; i <= arr_size + 1; i++)
{
y ^= i;
}
int missing = x ^ y; //The missing number is going to be the XOR of the previous two.
printf("%d", missing);
return 0;
}
So, I have this so far. I'm trying to find the two largest numbers in an array and return them. I looked up a lot of resources online, and most of them say "call by reference" is the way to go. But I've no idea how to make it work with my program. For example, I saw this example online:
void Calculate(int x, int y, int* prod, int* quot)
{
*prod = x*y;
*quot = x/y;
}
int x = 10,y = 2, prod, quot;
Calculate(x, y, &prod, ")
How does the above program actually "return"? How do I print the return values to the console?
#include "stdio.h"
void largest_two( int numbers[], int len, int *largest, int *next_largest){
int i, temp;
*largest = numbers[0];
*next_largest = numbers[1];
if(*largest < *next_largest){
temp = *next_largest;
*largest = *next_largest;
*next_largest = temp;
}
for (i=0; i<sizeof(numbers); i++) {
if(numbers[i]>= *largest){
*largest = numbers[i];
*next_largest = *largest;
}
else if ( numbers[i] > *next_largest){
*next_largest = numbers[i];
}
}
}
int main() {
int numbers[] = {3, 1, 2, 3, 6, 2, 8, 0, 0, 0};
int len = 3;
int largest, next_largest;
//==>??? printf("%d %d", largest_two(numbers, len, &largest, &next_largest));
}
Sides' from the pointer issues (you should read a tutorial / book on them), your main problem is that you're attempting to print the single return value of a function with return type void which means it won't return at all.
Your code:
int main() {
int numbers[] = {3, 1, 2, 3, 6, 2, 8, 0, 0, 0};
int len = 10; // sizeof(numbers)
int largest, next_largest;
largest_two(numbers, len, &largest, &next_largest);
printf("%d %d", largest, next_largest);
}
Keep in mind this is still not entirely correct, but it does adress your problem of printing the numbers.
Also, passing len means you shouldn't do this for (i=0; i<sizeof(numbers); i++) but this instead for (i=0; i<len; i++)
Firstly, this line:
for (i=0; i<sizeof(numbers); i++)
is not correct. You want this to be instead:
for (i=0; i<len; i++)
which should be passed to largest_two() as sizeof numbers/sizeof numbers[0], which is the actual length of the array.
I also suggest setting largest and next_largest to INT_MIN from <limits.h>, and then finding these values from their. It seems you are also having trouble with pointers, and it would be best to use them only when needed.
Here is an example which simplifies your approach, which finds the largest and second largest element in one loop of the array. It also only uses pointers when needed.
Code:
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#define ARRAYSIZE(x) (sizeof x/sizeof x[0])
void largest_two(int numbers[], size_t len, int *largest, int *next_largest);
int main(void) {
int numbers[] = {3, 1, 2, 3, 6, 2, 8, 0, 0, 0};
int largest, next_largest;
largest_two(numbers, ARRAYSIZE(numbers), &largest, &next_largest);
printf("largest = %d\nnext_largest = %d\n", largest, next_largest);
}
void largest_two(int numbers[], size_t len, int *largest, int *next_largest) {
int max, smax;
max = smax = INT_MIN;
for (size_t i = 0; i < len; i++) {
if (numbers[i] > max) {
smax = max;
max = numbers[i];
} else if (numbers[i] > smax && numbers[i] < max) {
smax = numbers[i];
}
}
*largest = max;
*next_largest = smax;
}
Output:
largest = 8
next_largest = 6
Second dataset:
int numbers[] = {3, 1, 6, 3, 6, 2, 8, 0, 8, 7};
Output:
largest = 8
next_largest = 7
I am trying to find top 6 elements from an array with their ordering number.
int x=0;
for (int k = 0; k < 6; k++) //
{
for (i = 1; i <= 90; i++)
{
if (sorted[k] < holder[i] && i >= x)
{
sorted[k] = holder[i];
x = i; //
}
}
}
But this does not work. I want it to give me output like 43->7 15 ->3 etc..
Haven't written C in a while, but here is a simple solution that modifies the array in place and uses selection sort to select the k highest numbers in the array and moves them to the front. It keeps an array of indices that correspond to where the number originally was and applies the same swaps to it.
#include <stdio.h>
#define ELEMENTS 10
void main(void)
{
// example input for execution
int numbers[10] = {9,4,5,1,8,2,3,6,0,7};
// tracks ordering of indices
int indexes[10] = {0,1,2,3,4,5,6,7,8,9};
int k = 6;
int i, j;
int max, temp;
// Partial selection sort, move k max elements to front
for (i = 0; i < k; i++)
{
max = i;
// Find next max index
for (j = i+1; j < ELEMENTS; j++)
{
if (numbers[j] > numbers[max]) {
max = j;
}
}
// Swap numbers in input array
temp = numbers[i];
numbers[i] = numbers[max];
numbers[max] = temp;
// Swap indexes in tracking array
temp = indexes[i];
indexes[i] = indexes[max];
indexes[max] = temp;
}
for (i = 0; i < k; i++) {
printf("%d -> %d\n", indexes[i], numbers[i]);
}
}
And the output:
0 -> 9
4 -> 8
9 -> 7
7 -> 6
2 -> 5
1 -> 4
Here's the answer I have for you.
I would love some constructive criticism on it from anyone who can provide some.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int numbers[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
int *ptrNumbers[10];
int i=0;
for(; i < 10; i++){
ptrNumbers[i] = &numbers[i]; // assign the addresses
}
int topSix[6];
int topSixIndex=0;
for(; topSixIndex < 6; topSixIndex++){
int **best = NULL; // Pointer to the pointer to the value.
int checkIndex=0;
for(; checkIndex < 10; checkIndex++){
if(ptrNumbers[checkIndex] != NULL){
if(!best){
/* best is not yet defined */
best = &ptrNumbers[checkIndex];
// best now points to the pointer for numbers[checkIndex]
}else if(*ptrNumbers[checkIndex] > **best){
// this else if statement could be attached to the main if as
// an or condition, but I've separated it for readability.
best = &ptrNumbers[checkIndex];
// best now points to the pointer for numbers[checkIndex]
}
}
}
// assign the topSix position and flag the ptrNumbers
topSix[topSixIndex] = **best;
*best = NULL;
}
// now we'll print the numbers
for(topSixIndex = 0; topSixIndex < 6; topSixIndex++){
printf("%d\n", topSix[topSixIndex]);
}
return 0;
}
Essentially the program works like this: Given an array of ten numbers, a second array is constructed to house pointers to those 10 numbers. A third array is then constructed to house the values of the top 6 numbers. A for loop is then initialized to loop 6 times to find the highest unrecorded value. When the highest value is found by looping the pointer array, the value is assigned to the next index of the top six array. Once that value is added, the pointer array's index that points to the top six value is then assigned to NULL. This acts as a flag insuring that the value will not be added again. Finally, all numbers are printed out.
After running this code, the output I received was:
9
8
7
6
5
4
Edit: as a note, the ordering number's can be stored in a second array. You would simply need to track the checkIndex of the highest value and then assign it to a second array which contained the index values.
maybe you aren't looking for a code-only answer, but this will work:
#include <limits.h>
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
// return index of max element
int max_index( int* vec, int sz )
{
int idx, max, i;
if(!sz) return -1;
idx = 0;
max = vec[0];
for(i=1; i<sz; ++i)
{
if( vec[i] > max )
{
max = vec[i];
idx = i;
}
}
return idx;
}
// return indexes of N top elements
void top( int* vec, int sz, int* out_vec, int N )
{
int i, *tmp, idx;
tmp = (int*) malloc( sz*sizeof(int) );
memcpy( tmp, vec, sz*sizeof(int) );
for(i=0; i<N; ++i )
{
idx = max_index(tmp,sz);
out_vec[i]=idx;
tmp[idx] = INT_MIN;
}
free(tmp);
}
see it live here
Make an array of struct that contain data and index, then sort it and pick up first or last 6 elements to output.
Say that you are given an array numbers. Then create an array indexes with the same size as numbers in such a way that its values are equal to their indexes. Here is an illustration:
numbers = [ 1, 7, 3, 9, 2, 0 ]
indexes = [ 0, 1, 2, 3, 4, 5 ]
Sort numbers in descending order, performing the same operations on indexes. In the end, you should end up with something like this:
numbers = [ 9, 7, 3, 2, 1, 0 ]
indexes = [ 3, 1, 2, 4, 0, 5 ]
Finally, all you need to do is work with the first six elements of these arrays.
#include <stdio.h>
#define TRUE 1
#define FALSE 0
int contains(int array[], int array_size, int value)
{
int i;
for (i = 0; i < array_size; i++)
{
if (array[i] == value)
{
return TRUE;
}
}
return FALSE;
}
int main()
{
int numbers[] = { 1, 7, 3, 9, 2, 0 };
int indexes[] = { 0, 1, 2, 3, 4, 5 };
int numbers_size = 6;
int largest[] = { -1, -1, -1, -1, -1, -1 };
int largest_index = 0;
int i;
for (i = 0; i < 6; i++)
{
int j;
int max_index = -1;
int max = -2147483648;
for (j = 0; j < numbers_size; j++)
{
if (numbers[j] >= max && contains(largest, numbers_size, j) == FALSE)
{
max_index = j;
max = numbers[max_index];
}
}
largest[largest_index++] = max_index;
}
for (i = 0; i < 6; ++i)
{
printf("%d->%d\n", largest[i], numbers[largest[i]]);
}
return 0;
}
You probably should use bubblesort (and keep a function holding all the original indexes) and then just make it show the 6 first number of both arrays (from the indexes array and from the array you sorted itself)
I have an array let's say A[5], the 5 elements are 5,4,1,2,3. Now I sort these arrays in ascending order. so the resulting array will now be 1,2,3,4,5. I use qsort() function of stdlib.h to sort this. The question is how can I get the indices of the original array with respect to my new array. originally my indices were 0,1,2,3,4 for corresponding values of 5,4,1,2,3 and now the indices have changed to 2,3,4,1,0. How can I get these indices efficiently in C? Thank you in advance(please write the code if possible)
There is also a method as follows under limited conditions.
#include <stdio.h>
int main(void){
int data[] ={ 5,4,1,2,3 }; //Without duplication, The number of limited range.
int size = sizeof(data)/sizeof(*data);
int keys[size];
int i;
printf("data :\n");
for(i=0;i<size;i++){
printf("%d ",data[i]);
}
for(i=0;i<size;i++){
keys[data[i]-1]=i;
}
printf("\n\ndata\tindex\n");
for(i=0;i<size;i++){
printf("%d\t%d\n", data[keys[i]], keys[i]);
}
return 0;
}
/* result sample
data :
5 4 1 2 3
data index
1 2
2 3
3 4
4 1
5 0
*/
How to sort an array of index #Kerrek is as proposed.
#include <stdio.h>
#include <stdlib.h>
int *array;
int cmp(const void *a, const void *b){
int ia = *(int *)a;
int ib = *(int *)b;
return array[ia] < array[ib] ? -1 : array[ia] > array[ib];
}
int main(void){
int data[] ={ 5,4,1,2,3 };
int size = sizeof(data)/sizeof(*data);
int index[size];//use malloc to large size array
int i;
for(i=0;i<size;i++){
index[i] = i;
}
array = data;
qsort(index, size, sizeof(*index), cmp);
printf("\n\ndata\tindex\n");
for(i=0;i<size;i++){
printf("%d\t%d\n", data[index[i]], index[i]);
}
return 0;
}
Take a 2D array. Store the numbers is first column and then corressponding indexes in second column. You can write your comparator function as:
int compare ( const void *pa, const void *pb )
{
const int *a = pa;
const int *b = pb;
if(a[0] == b[0])
return a[1] - b[1];
else
return a[0] - b[0];
}
Call to qsort should be:
qsort(array, n, sizeof array[0], compare); // n is representing rows
See the Live Demo
Based on Kerrek SB's brilliant idea I made an implementation that works for any array type by providing its element size and a comparator function for that type.
_Thread_local uint8_t *array_to_order;
_Thread_local size_t elem_size_to_order;
_Thread_local int (*cmp_for_ordering)(const void *, const void *);
int cmp_array_entry(const size_t *a, const size_t *b)
{
return cmp_for_ordering(&array_to_order[*a * elem_size_to_order], &array_to_order[*b * elem_size_to_order]);
}
size_t *make_order_index_array(void *array, size_t *order, size_t elem_count, size_t elem_size, int (*cmp)(const void *, const void *))
{
// If order is provided by the caller it should have suitable contents, such as when updating an order
// Initialise the order array if not already provided
if (order == NULL)
{
order = calloc(elem_count, sizeof(size_t));
// Initialise the order array to the unsorted indices
for (size_t i=0; i < elem_count; i++)
order[i] = i;
}
// Globals used by the comparison function to order the array
array_to_order = array;
elem_size_to_order = elem_size;
cmp_for_ordering = cmp;
// Order the order array
qsort(order, elem_count, sizeof(size_t), cmp_array_entry);
return order;
}
_Thread_local is something that we should be able to take for granted for writing such code when we're forced to use globals but should worry about thread safety. Mine is defined with the following macros:
#if defined(_MSC_VER) && !defined(_Thread_local)
#define _Thread_local __declspec(thread)
#endif
#if !(defined(__STDC_VERSION__) && (__STDC_VERSION__ >= 201102L)) && !defined(_Thread_local)
#if defined(__GNUC__) || defined(__INTEL_COMPILER) || defined(__SUNPRO_CC) || defined(__IBMCPP__)
#define _Thread_local __thread
#endif
#elif defined(__GNUC__) && defined(__GNUC_MINOR__) && (((__GNUC__ << 8) | __GNUC_MINOR__) < ((4 << 8) | 9))
#define _Thread_local __thread
#endif
#include <limits.h>
#include <stdio.h>
#define SIZE 5
int* sortArrayNKeepIndices(int arr[], int arrSize){
static int indexArr[SIZE];
int arr2[arrSize];
for (int i = 0; i < arrSize; i++) {
indexArr[i] = 0;
arr2[i] = arr[i];
}
int min = 0, temp = 0;
for (int i = 0; i < arrSize ; i++)
{
min = i; // record the position of the smallest
for (int j = i + 1; j < arrSize; j++)
{
// update min when finding a smaller element
if (arr[j] < arr[min])
min = j;
}
// put the smallest element at position i
temp = arr[i];
arr[i] = arr[min];
arr[min] = temp;
} // array sorting ends here
int ctr = 0;
while ( ctr < arrSize) {
min = 0; // restart from first element
for (int j = 0; j < arrSize; j++)
{
if (arr2[j] == INT_MAX) continue; // ignore already marked as minimum indices
// update min when finding a smaller element
if (arr2[j] < arr2[min])
min = j;
}
indexArr[ctr] = min; // updating indexArr with the index of the next minimum
arr2[min] = INT_MAX; // marking minimum element to be ignored next time
ctr++;
} //keeping track of previous indices of the array elements ends here
return indexArr;
} // function sortArrayKeepIndices ends here
int main () {
int arr[SIZE] = {16, 15, 12, 10, 13};
int* ptr = sortArrayNKeepIndices(arr, SIZE);
for (int dex = 0; dex < SIZE; dex++){
printf("%d (%d, %d)\t", arr[dex], * (ptr + dex), dex);}
}
// output will be 10 (3, 0) 12 (2, 1) 13 (4, 2) 15 (1, 3) 16 (0, 4)
// element (old index, new index)