How to find the positions of the max and min numbers in the array?

Question

int A[5] = {2, 2, 1, 4, 1};    
int max = A[0], min = A[0];
int k,l;

for (i = 0; i < 5; i++) {
    if (A[i] >= max) {
        max = A[i];
        k = i;
    }
}

for (j = 0; j < 5; j++) {
    if (A[j] <= min) {
        min = A[j];
        l = j;
    }
}

printf("densely is: %d ", k);
printf("\n");
printf("loosely is: %d ", l);

Program prints position of densely ( max ) as 3 . It prints out position of loosely ( min ) as 4 . But the correct answer for the loosely ( min ) position should be 2 and 4 . In my case, the max number is 4 , so the position should be 3 (counted from 0 ). The min number is 1 , so the position should be 2 and 4 . Now, my code only shows one position for the min number: 4 . It should print both positions: 2 and 4 .

I want to print out all positions for the max and min numbers.

Answer 1

This is actually very good question. The key to the solution is to find the min and max value in the array A . That part has been done and works well. The only improvement here is to start from index 1 and remove = from >= and <= comparisons.
The missing part is to remember all indexes for min and max .

That is difficult to do in one pass. What you can do in one pass is to find both min and max .

Once you have min and max in the second pass you could mark all the index positions for max and min in additional index tables called maximums and minimums . (Since I want to print those values I have two separate loops for it.)

The solution is quite simple:

#include <stdio.h>

// Find all minimums and all maximums in the array A
// Find and remember where `max` and `min` values were in the array A
// The algorithm is symmetrical for minimum and maximum 

int main(void)
{             // 0  1  2  3  4 
    int A[5] = { 2, 2, 1, 4, 1};    

    int maximums[5] = {0};  // we keep max positions here
    int minimums[5] = {0};  // we keep min positions here

    int max = A[0];  // we assume max at index 0
    int min = A[0];  // we assume min at index 0

    for (int i = 1; i < 5; i++) {       // we can start from index 1 because of the above assumption
        if (A[i] > max) max = A[i];     // densely 
        if (A[i] < min) min = A[i];     // loosely
    }

    // Find the all elements equal to `max` and `min` 
    // and register them in `maximums` and `minimums` arrays

    // Print the results:
    printf("max value is: %d and it occurs at the following index(es):\n", max);         
    for (int i = 0; i < 5; i++) {

        if (A[i] == max){ 
            maximums[i] = 1;   // mark this index as maximum being here
            printf("%d ", i);
        }
    }

    printf("\nmin value is: %d and it occurs at the following index(es):\n", min );   
    for (int i = 0; i < 5; i++) {

        if (A[i] == min){ 
            minimums[i] = 1;   // mark this index as minimum being here
            printf("%d ", i);
        }    
    }

    // At this moment we not only printed the results 
    // but the array minimums and maximums remember where the `min` and `max` were.

    return 0;
}

Output:

max value is: 4 and it occurs at the following index(es):                                                                                       
3                                                                                                                                               
min value is: 1 and it occurs at the following index(es):                                                                                       
2 4