I am unable to calculate the largest index at which the maximum value appears in an array x[n] of n integers.

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Anybody help asap !!!
I am unable to calculate the largest index at which the maximum value appears in an array x[n] of n integers.

What I have tried:

int lastMax(int n, int x[n]) {
 if (n<=0) {
     return -1; }
    int max = x[0];
    int maxIndex = 0;
    for (int i = 1;i <n; i++) {
        if (x[i] > max) {
            maxIndex = i;
            max = x[i];
        }
    }
    return maxIndex;
}
/*** DO NOT MODIFY BELOW THIS LINE ***/

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

int main() {
    int n;
    scanf("%d", &n);
    int *x = NULL;
    if (n > 0) {
        x = (int *) malloc(n * sizeof(int));
        if (x != NULL) {
            for (int i = 0; i < n; ++i) {
                scanf("%d", &x[i]);
            }
        }
    }
    printf("%d\n", lastMax(n, x));
    if (x != NULL) {
        free(x);
    }
    return 0;
}

Posted 11-Jan-21 9:43am

Shri_hari

Updated 11-Jan-21 10:30am

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k5054 11-Jan-21 15:57pm

In what way does this not work?
Why "Do not modify below this line" ??
I think it would be helpful if you did something like:

int maxIndex = lastMax(n, x);
printf("Max value: x[%d] = %d\n", maxIndex, x[maxIndex]);

But you should probably test that maxIndex is not -1 first.

Shri_hari 11-Jan-21 22:29pm

It calculates the index of largest number but if the largest number is present at more than one position say 2 3 4 4 then I should get output as 3 and not 2.
But it is not doing that part properly ..!

2 solutions

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Richard MacCutchan · Answer 1 · 2021-01-11T10:01:00

Solution 1

Change the definition of your function to:

C++

int lastMax(int n, int* x) {  // x is a pointer to an array

Posted 11-Jan-21 10:01am

Richard MacCutchan

OriginalGriff · Answer 2 · 2021-01-11T10:30:00

To add to what Richard has said: the C Language Specification defines that the name of an array variable is a pointer to its first element.
So given this:

C++

int arr[10];

These two statements are equivalent:

C++

int *p1 = arr;

C++

int *p2 = &(arr[0]);

This means that it is legal to treat any pointer to a type as an indexable array of that type - so passing an array to a function is done by declaring the parameter as a pointer, and then using an index within the function to access individual elements;

C++

void PrintArray(int *arr, int n)
   {
   for (int i = 0; i < n; i++)
      {
      printf("%u\n", arr[i]);
      }
   }

Or as a pointer:

C++

void PrintArray(int *arr, int n)
   {
   while(n-- > 0)
      {
      printf("%u\n", *arr++);
      }
   }

In either case, you can hand the array and it's size, and the function will work:

C++

int myArray[10];
PrintArray(myArray, 10);