# Next Greater Element

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.

Examples:
a) For any array, rightmost element always has next greater element as -1.
b) For an array which is sorted in decreasing order, all elements have next greater element as -1.
c) For the input array [4, 5, 2, 25}, the next greater elements for each element are as follows.

```Element       NGE
4      -->   5
5      -->   25
2      -->   25
25     -->   -1
```

d) For the input array [13, 7, 6, 12}, the next greater elements for each element are as follows.

```  Element        NGE
13      -->    -1
7       -->     12
6       -->     12
12     -->     -1
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Method 1 (Simple)
Use two loops: The outer loop picks all the elements one by one. The inner loop looks for the first greater element for the element picked by outer loop. If a greater element is found then that element is printed as next, otherwise -1 is printed.

Thanks to Sachin for providing following code.

## C

```
// Simple C program to print next greater elements
// in a given array
#include<stdio.h>

/* prints element and NGE pair for all elements of
arr[] of size n */
void printNGE(int arr[], int n)
{
int next, i, j;
for (i=0; i<n; i++)
{
next = -1;
for (j = i+1; j<n; j++)
{
if (arr[i] < arr[j])
{
next = arr[j];
break;
}
}
printf("%d -- %dn", arr[i], next);
}
}

int main()
{
int arr[]= {11, 13, 21, 3};
int n = sizeof(arr)/sizeof(arr[0]);
printNGE(arr, n);
getchar();
return 0;
}
```

## Java

```// Simple Java program to print next
// greater elements in a given array

class Main
{
/* prints element and NGE pair for
all elements of arr[] of size n */
static void printNGE(int arr[], int n)
{
int next, i, j;
for (i=0; i<n; i++)
{
next = -1;
for (j = i+1; j<n; j++)
{
if (arr[i] < arr[j])
{
next = arr[j];
break;
}
}
System.out.println(arr[i]+" -- "+next);
}
}

public static void main(String args[])
{
int arr[]= {11, 13, 21, 3};
int n = arr.length;
printNGE(arr, n);
}
}
```

## Python

```
# Function to print element and NGE pair for all elements of list
def printNGE(arr):

for i in range(0, len(arr), 1):

next = -1
for j in range(i+1, len(arr), 1):
if arr[i] < arr[j]:
next = arr[j]
break

print(str(arr[i]) + " -- " + str(next))

# Driver program to test above function
arr = [11,13,21,3]
printNGE(arr)

# This code is contributed by Sunny Karira
```

Output:

```11 -- 13
13 -- 21
21 -- -1
3 -- -1```

Time Complexity: O(n^2). The worst case occurs when all elements are sorted in decreasing order.

Method 2 (Using Stack)
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 for the popped element.
….d) Keep popping 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.

## C++

```
// A Stack based C++ program to find next
// greater element for all array elements.
#include<bits/stdc++.h>
using namespace std;

/* prints element and NGE pair for all
elements of arr[] of size n */
void printNGE(int arr[], int n)
{
stack<int> s;

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

// iterate for rest of the elements
for (int i=1; i<n; i++)
{
int next = arr[i];

if (s.empty() == false)
{
// if stack is not empty, then
// pop an element from stack
int element = s.top();
s.pop();

/* 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)
{
cout << element << " --> " << next
<< endl;
if (s.empty() == true)
break;
element = s.top();
s.pop();
}

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

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

/* After iterating over the loop, the remaining
elements in stack do not have the next greater
element, so print -1 for them */
while (s.empty() == false)
{
cout << s.top() << " --> " << -1 << endl;
s.pop();
}
}

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

## C

```// A Stack based C program to find next greater element
// for all array elements.
#include<stdio.h>
#include<stdlib.h>
#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 overflown");
getchar();
exit(0);
}
else
{
ps->top += 1;
int top = ps->top;
ps->items [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
{
int top = ps->top;
temp = ps->items [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<n; 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;
}
```

## Java

```//Java program to print next
//greater element using stack

public class NGE
{
static class stack
{
int top;
int items[] = new int[100];

// Stack functions to be used by printNGE
void push(int x)
{
if (top == 99)
{
System.out.println("Stack full");
}
else
{
items[++top] = x;
}
}

int pop()
{
if (top == -1)
{
System.out.println("Underflow error");
return -1;
}
else
{
int element = items[top];
top--;
return element;
}
}

boolean isEmpty()
{
return (top == -1) ? true : false;
}
}

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

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

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

if (s.isEmpty() == false)
{

// if stack is not empty, then
// pop an element from stack
element = s.pop();

/* 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)
{
System.out.println(element + " --> " + next);
if (s.isEmpty() == true)
break;
element = s.pop();
}

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

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

/* After iterating over the loop, the remaining
elements in stack do not have the next greater
element, so print -1 for them */
while (s.isEmpty() == false)
{
element = s.pop();
next = -1;
System.out.println(element + " -- " + next);
}
}

public static void main(String[] args)
{
int arr[] = { 11, 13, 21, 3 };
int n = arr.length;
printNGE(arr, n);
}
}

// Thanks to Rishabh Mahrsee for contributing this code
```

## Python

```# Python program to print next greater element using stack

# Stack Functions to be used by printNGE()
def createStack():
stack = []
return stack

def isEmpty(stack):
return len(stack) == 0

def push(stack, x):
stack.append(x)

def pop(stack):
if isEmpty(stack):
print("Error : stack underflow")
else:
return stack.pop()

'''prints element and NGE pair for all elements of
arr[] '''
def printNGE(arr):
s = createStack()
element = 0
next = 0

# push the first element to stack
push(s, arr[0])

# iterate for rest of the elements
for i in range(1, len(arr), 1):
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 :
print(str(element)+ " -- " + str(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
print(str(element) + " -- " + str(next))

# Driver program to test above functions
arr = [11, 13, 21, 3]
printNGE(arr)

# This code is contributed by Sunny Karira
```

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) Initially 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.

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.

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