Hello Everyone,
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 the next greater element as -1.
Examples:
For an array, the rightmost element always has the next greater element as -1.
For an array that is sorted in decreasing order, all elements have the 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:
// 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 Code
int main()
{
int arr[] = {11, 13, 21, 3};
int n = sizeof (arr)/ sizeof (arr[0]);
printNGE(arr, n);
return 0;
}
Output
11 – 13 13 – 21 21 – -1 3 – -1
Time Complexity: O(N2)
Auxiliary Space: O(1)
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 the stack and print -1 as the next element for them.
Below is the implementation of the above approach:
// 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++)
{
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;
}
Output
11 → 13 13 → 21 3 → -1 21 → -1
Time Complexity: O(N)
Auxiliary Space: 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:
// 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 all
elements 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 Code
int main()
{
int arr[] = { 11, 13, 21, 3 };
int n = sizeof (arr) / sizeof (arr[0]);
// Function call
printNGE(arr, n);
return 0;
}
Output
11 —> 13 13 —> 21 21 —> -1 3 —> -1
Time Complexity: O(N)
Auxiliary Space: O(N)