C Exercises: Find the next greater elements in a given unsorted array
C Array: Exercise-43 with Solution
Write a program in C to print the next greatest elements in a given unsorted array. Elements for which no superior element exists, consider the next greatest element as -1.
Pictorial Presentation:

Sample Solution:
C Code:
#include <stdio.h>
void findNxtLrgElem (int *arr1, int arr1_size)
{
int nxtBgElem, i, j;
for(i = 0; i < arr1_size; i++)
{
for (j = i+1, nxtBgElem = -1; j < arr1_size; j++)
{
if (arr1[i] < arr1[j])
{
nxtBgElem = arr1[j];
break;
}
}
printf("Next bigger element of %d in the array is: %d\n", arr1[i], nxtBgElem);
}
}
void formBigElemArray (int *arr1, int arr1_size)
{
int nxtBgElem, i, j;
for(i = 0; i < arr1_size; i++)
{
for (j = i+1, nxtBgElem = -1; j < arr1_size; j++)
{
if (arr1[i] < arr1[j])
{
nxtBgElem = arr1[j];
break;
}
}
printf("%d ", nxtBgElem);
}
}
int main()
{
int i, arr1[]= {5, 3, 10, 9, 6, 13};
int ctr = sizeof(arr1) / sizeof(arr1[0]);
//------------- print original array ------------------
printf("The given array is : ");
for(i = 0; i < ctr; i++)
{
printf("%d ", arr1[i]);
}
printf("\n");
//-----------------------------------------------------
printf("\nNext Bigger Elements are:\n");
findNxtLrgElem(arr1, ctr);
printf("\nNext Bigger Elements Array:\n");
formBigElemArray(arr1, ctr);
return 0;
}
Sample Output:
The given array is : 5 3 10 9 6 13 Next Bigger Elements are: Next bigger element of 5 in the array is: 10 Next bigger element of 3 in the array is: 10 Next bigger element of 10 in the array is: 13 Next bigger element of 9 in the array is: 13 Next bigger element of 6 in the array is: 13 Next bigger element of 13 in the array is: -1 Next Bigger Elements Array: 10 10 13 13 13 -1
Flowchart:

C Programming Code Editor:
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Next: Write a program in C to find the two repeating elements in a given array.
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C Programming: Tips of the Day
C Programming - What is the argument for printf that formats a long?
Put an l (lowercased letter L) directly before the specifier.
unsigned long n; long m; printf("%lu %ld", n, m);
Ref : https://bit.ly/3dIwfkP
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