How to find and display, for example, three maximum or minimum elements of an array, in the C programming language? For example, an array

char mas[] = { 8, 7, 3, 9, 5, 2};

and to output (3 max.) 9 8 7.

  • one
    you need to look for a maximum (minimum) 3 times each time checking that the previous maximum is not equal to the current one - ampawd
  • @Vladimir If the array contains several coincident maxima, such as {9, 9, 8, 7, 6}, then do you need to output 9, 9, 8 or 9, 8, 7? - Vlad from Moscow
  • @ampawd It's not a fact that it works for {5, 5, 5, 1, 2, 3} . By the way, what is the correct result in this case 5, 5, 5 or 5, 3, 2 ? - andy.37
  • one
    The simplest (not the most efficient) 1) sort 2) take 3 first / last elements (if necessary - different) - andy.37
  • @ andy.37 is more interesting to solve this problem in linear time, if you sort it, you can calculate it, but its constant is big - ampawd

2 answers 2

You can define a function that will find any number of maximal elements in an integer array. To do this, you just need to pass to the function an array of the required dimension for storing maximum values.

The function returns the number of maximum elements, since in general you can request to find more maximum elements than the total number of elements is in the array, or, for example, when an empty array is passed to you.

Below is a demo program.

 #include <stdlib.h> #include <stdio.h> #include <string.h> #include <time.h> size_t max_elements(const int *a, size_t n, int *max, size_t m) { size_t k = 0; for (size_t i = 0; i < n; i++) { size_t j = 0; while (j < k && !(a[i] > max[j])) j++; if (j == k ) { if ( k < m ) max[k++] = a[i]; } else { memmove(max + j + 1, max + j, ( k < m ? k - j: k - j - 1 ) * sizeof(int)); max[j] = a[i]; if (k < m) ++k; } } return k; } #define N1 10 #define M 3 int main() { int a[N1]; int max[M]; srand((unsigned int)time(NULL)); for (size_t i = 0; i < N1; i++) a[i] = rand() % (2 * N1); for (size_t i = 0; i < N1; i++) printf("%d ", a[i]); printf("\n"); size_t k = max_elements(a, N1, max, M); if (k != 0) { printf("%zu max elements are ", k); for (size_t i = 0; i < k; i++) printf("%d ", max[i]); printf("\n"); } }

Its output to the console might look like this:

 13 17 8 9 14 10 2 15 3 19 3 max elements are 19 17 15

Similarly, you can find the minimum items. All you need to do is change the condition in this cycle.

 while (j < k && !(a[i] > max[j])) j++; ^^^^^^^^^^^^^^^^

    As an option -

     void max3(int * a, int N) { int i, retry; for(i = 0; i < N-3; ++i) { for(retry = 1; retry;) { retry = 0; for(int j = N-1; j >= N-3; --j) { if (a[i] > a[j]) { int tmp = a[i]; a[i] = a[j]; a[j] = tmp; retry = 1; break; } } } } } int main(int argc, const char * argv[]) { int a[] = { 3, 5, 7, 1, 9, 2, 8, 0, 1, 9 }; max3(a,sizeof(a)/sizeof(a[0])); for(int j = sizeof(a)/sizeof(a[0])-3; j < sizeof(a)/sizeof(a[0]); ++j) { printf("%3d",a[j]); } printf("\n"); }

    About efficiency - well, O (n), because 3 is a constant :)