The placement session has begun in a college. There is N number of students standing outside an interview room in a line. It is given that the person who goes first has higher chances of selection.
Each student has a number associated with them representing their problem-solving capability. The higher the capability the higher the chances of selection. Now every student wants to know the number of students ahead of him with higher problem-solving capability.
using structure in code
Input: 6(number of students) {4 , 9 , 5 , 3 , 2 , 10}
Output: {0 , 0 , 1 , 3 , 4 , 0}
#include <stdio.h>
#define N 6
void number_ahead(int n, int students[], int ahead[])
{
int i, j;
for (i=0; i<n; i++) {
ahead[i] = 0;
for (j=i-1; j>=0; j--) {
if (students[j] > students[i]) {
ahead[i]++;
}
}
}
}
int main() {
int i;
int students[N] = {4, 9, 5, 3, 2, 10};
int ahead[N];
number_ahead(N, students, ahead);
for (i=0; i<N; i++) {
printf("%d ", ahead[i]);
}
return 0;
}
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