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 the array. Elements for which no greater element exist, consider next greater element as -1.
Examples:
- For an array, the rightmost element always has the next greater element as -1.
- For an array which is sorted in decreasing order, all elements have next greater element as -1.
- 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
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 the outer loop. If a greater element is found then that element is printed as next, otherwise -1 is printed.
Below is the implementation of the above approach:
C++
// Simple C++ program to print// next greater elements in a// given array#include<iostream>using namespace std;
/* 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;
}
}
cout << arr[i] << " -- "
<< next << endl;
}
}// Driver Codeint main()
{ int arr[] = {11, 13, 21, 3};
int n = sizeof(arr)/sizeof(arr[0]);
printNGE(arr, n);
return 0;
}// This code is contributed // by Akanksha Rai(Abby_akku) |
C
// Simple C program to print next greater elements// in a given array#include<stdio.h>/* prints element and NGE pair for all elements ofarr[] 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);
return 0;
} |
Java
// Simple Java program to print next // greater elements in a given arrayclass 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 listdef 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 functionarr = [11,13,21,3]
printNGE(arr)# This code is contributed by Sunny Karira |
C#
// Simple C# program to print next // greater elements in a given arrayusing System;
class GFG
{ /* 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;
}
}
Console.WriteLine(arr[i] + " -- " + next);
}
}
// driver code
public static void Main()
{
int []arr= {11, 13, 21, 3};
int n = arr.Length;
printNGE(arr, n);
}
}// This code is contributed by Sam007 |
PHP
<?php// Simple PHP program to print next// greater elements in a given array/* prints element and NGE pair for all elements of arr[] of size n */
function printNGE($arr, $n)
{ for ($i = 0; $i < $n; $i++)
{
$next = -1;
for ($j = $i + 1; $j < $n; $j++)
{
if ($arr[$i] < $arr[$j])
{
$next = $arr[$j];
break;
}
}
echo $arr[$i]." -- ". $next."\n";
}
} // Driver Code
$arr= array(11, 13, 21, 3);
$n = count($arr);
printNGE($arr, $n);
// This code is contributed by Sam007?> |
Output
11 -- 13 13 -- 21 21 -- -1 3 -- -1
Time Complexity: O(n2). The worst case occurs when all elements are sorted in decreasing order.
Method 2 (Using Stack)
- Push the first element to stack.
- Pick rest of the elements one by one and follow the following steps in loop.
- Mark the current element as next.
- If stack is not empty, compare top element of stack with next.
- If next is greater than the top element,Pop element from stack. next is the next greater element for the popped element.
- Keep popping from the stack while the popped element is smaller than next. next becomes the next greater element for all such popped elements
- Finally, push the next in the stack.
- After the loop in step 2 is over, pop all the elements from stack and print -1 as next element for them.
Below image is a dry run of the above approach:
Below is the implementation of the above approach:
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 allelements 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++)
{
if (s.empty()) {
s.push(arr[i]);
continue;
}
/* if stack is not empty, then
pop an element from stack.
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 (s.empty() == false
&& s.top() < arr[i])
{
cout << s.top()
<< " --> " << arr[i] << endl;
s.pop();
}
/* push next to stack so that we can find
next greater for it */
s.push(arr[i]);
}
/* 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 code */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 <stdbool.h>#include <stdio.h>#include <stdlib.h>#define STACKSIZE 100// stack structurestruct 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 ofarr[] 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 code */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 stackpublic 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);
}
}
// Driver Code
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 codearr = [11, 13, 21, 3]
printNGE(arr)# This code is contributed by Sunny Karira |
C#
using System;
// c# program to print next// greater element using stackpublic class NGE {
public class stack {
public int top;
public int[] items = new int[100];
// Stack functions to be used by printNGE
public virtual void push(int x)
{
if (top == 99) {
Console.WriteLine("Stack full");
}
else {
items[++top] = x;
}
}
public virtual int pop()
{
if (top == -1) {
Console.WriteLine("Underflow error");
return -1;
}
else {
int element = items[top];
top--;
return element;
}
}
public virtual bool Empty
{
get { return (top == -1) ? true : false; }
}
}
/* prints element and NGE pair for
all elements of arr[] of size n */
public 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.Empty == 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) {
Console.WriteLine(element + " --> "
+ next);
if (s.Empty == 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.Empty == false) {
element = s.pop();
next = -1;
Console.WriteLine(element + " -- " + next);
}
}
// Driver Code
public static void Main(string[] args)
{
int[] arr = new int[] { 11, 13, 21, 3 };
int n = arr.Length;
printNGE(arr, n);
}
}// This code is contributed by Shrikant13 |
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.
- Initially pushed to the stack.
- Popped from the stack when next element is being processed.
- Pushed back to the stack because the next element is smaller.
- Popped from the stack in step 3 of the algorithm.
How to get elements in the same order as input?
The above approach may not produce output elements in the same order as the input. To achieve the same order, we can traverse the same in reverse order
Below is the implementation of the above approach:
C++
// A Stack based C++ program to find next// greater element for all array elements// in same order as input.#include <bits/stdc++.h>using namespace std;
/* prints element and NGE pair for allelements of arr[] of size n */void printNGE(int arr[], int n)
{ stack<int> s;
unordered_map<int, int> mp;
/* push the first element to stack */
s.push(arr[0]);
// iterate for rest of the elements
for (int i = 1; i < n; i++)
{
if (s.empty()) {
s.push(arr[i]);
continue;
}
/* if stack is not empty, then
pop an element from stack.
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 (s.empty() == false
&& s.top() < arr[i])
{
mp[s.top()] = arr[i];
s.pop();
}
/* push next to stack so that we can find
next smaller for it */
s.push(arr[i]);
}
/* After iterating over the loop, the remaining
elements in stack do not have the next smaller
element, so print -1 for them */
while (s.empty() == false)
{
mp[s.top()] = -1;
s.pop();
}
for (int i = 0; i < n; i++)
cout << arr[i] << " ---> "
<< mp[arr[i]] << endl;
}// Driver Codeint main()
{ int arr[] = { 11, 13, 21, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
printNGE(arr, n);
return 0;
} |
Java
// A Stack based Java program to find next// greater element for all array elements// in same order as input.import java.util.Stack;
class NextGreaterElement {
static int arr[] = { 11, 13, 21, 3 };
/* prints element and NGE pair for all
elements of arr[] of size n */
public static void printNGE()
{
Stack<Integer> s = new Stack<>();
int nge[] = new int[arr.length];
// iterate for rest of the elements
for (int i = arr.length - 1; i >= 0; i--)
{
/* if stack is not empty, then
pop an element from stack.
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 */
if (!s.empty())
{
while (!s.empty()
&& s.peek() <= arr[i])
{
s.pop();
}
}
nge[i] = s.empty() ? -1 : s.peek();
s.push(arr[i]);
}
for (int i = 0; i < arr.length; i++)
System.out.println(arr[i] +
" --> " + nge[i]);
}
/* Driver Code */
public static void main(String[] args)
{
// NextGreaterElement nge = new
// NextGreaterElement();
printNGE();
}
} |
C#
// A Stack based C# program to find next// greater element for all array elements// in same order as input.using System;
using System.Collections.Generic;
class GFG {
private int[] arr = new int[] { 11, 13, 21, 3 };
/* prints element and NGE pair for all
elements of arr[] of size n */
private void printNGE()
{
Stack<int> s = new Stack<int>();
int[] nge = new int[arr.Length];
// iterate for rest of the elements
for (int i = arr.Length - 1; i >= 0; i--)
{
/* if stack is not empty, then
pop an element from stack.
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 */
if (s.Count > 0)
{
while (s.Count > 0
&& s.Peek() <= arr[i])
{
s.Pop();
}
}
nge[i] = s.Count == 0 ? -1 : s.Peek();
s.Push(arr[i]);
}
for (int i = 0; i < arr.Length; i++)
{
Console.WriteLine(arr[i] + " --> " + nge[i]);
}
}
// Driver Code
public static void Main(string[] args)
{
GFG nge = new GFG();
nge.printNGE();
}
} |
Output
11 ---> 13 13 ---> 21 21 ---> -1 3 ---> -1
Please see for optimized solution for printing in same order.
Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.