WAP to Print Next Greater Elemnt in Given UnSorted Array

Given an array, print the Next Greater Element (NGE) for every element. The Next greater Element for an element x is the first greater element on the right side of x in array. Elements for which no greater element exist, consider next greater element as -1.

Algorithm

1) Push the first element to stack.
2) Pick rest of the elements one by one and follow following steps in loop.
....a) Mark the current element as next.
....b) If stack is not empty, then pop an element from stack and compare it with next.
....c) If next is greater than the popped element, then next is the next greater element fot the popped element.
....d) Keep poppoing from the stack while the popped element is smaller than next. next becomes the next greater element for all such popped elements
....g) If next is smaller than the popped element, then push the popped element back.
3) After the loop in step 2 is over, pop all the elements from stack and print -1 as next element for them.
?
#include
#include
#define STACKSIZE 100

// stack structure
struct stack
{
int top;
int items[STACKSIZE];
};

// Stack Functions to be used by printNGE()
void push(struct stack *ps, int x)
{
if (ps->top == STACKSIZE-1)
{
printf("Error: stack overflow\n");
getchar();
exit(0);
}
else
{
ps->top += 1;
ps->items[ps->top] = x;
}
}

bool isEmpty(struct stack *ps)
{
return (ps->top == -1)? true : false;
}

int pop(struct stack *ps)
{
int temp;
if (ps->top == -1)
{
printf("Error: stack underflow \n");
getchar();
exit(0);
}
else
{
temp = ps->items[ps->top];
ps->top -= 1;
return temp;
}
}

/* prints element and NGE pair for all elements of
arr[] of size n */
void printNGE(int arr[], int n)
{
int i = 0;
struct stack s;
s.top = -1;
int element, next;

/* push the first element to stack */
push(&s, arr[0]);

// iterate for rest of the elements
for (i=1; i {
next = arr[i];

if (isEmpty(&s) == false)
{
// if stack is not empty, then pop an element from stack
element = pop(&s);

/* If the popped element is smaller than next, then
a) print the pair
b) keep popping while elements are smaller and
stack is not empty */
while (element < next)
{
printf("\n %d --> %d", element, next);
if(isEmpty(&s) == true)
break;
element = pop(&s);
}

/* If element is greater than next, then push
the element back */
if (element > next)
push(&s, element);
}

/* push next to stack so that we can find
next greater for it */
push(&s, next);
}

/* After iterating over the loop, the remaining
elements in stack do not have the next greater
element, so print -1 for them */
while(isEmpty(&s) == false)
{
element = pop(&s);
next = -1;
printf("\n %d --> %d", element, next);
}
}

/* Driver program to test above functions */
int main()
{
int arr[]= {11, 13, 21, 3};
int n = sizeof(arr)/sizeof(arr[0]);
printNGE(arr, n);
getchar();
return 0;
}

Output:

11 --> 13
13 --> 21
3 --> -1
21 --> -1

Time Complexity: O(n). The worst case occurs when all elements are sorted in decreasing order. If elements are sorted in decreasing order, then every element is processed at most 4 times.
a) Initialy pushed to the stack.
b) Popped from the stack when next element is being processed.
c) Pushed back to the stack because next element is smaller.
d) Popped from the stack in step 3 of algo.

Run Here https://ideone.com/z8Zpc

Approach & Solution Provided by Pchild