# Previous greater element

Given an array of distinct elements, find previous greater element for every element. If previous greater element does not exist, print -1.

Examples:

```Input : arr[] = {10, 4, 2, 20, 40, 12, 30}
Output :         -1, 10, 4, -1, -1, 40, 40

Input : arr[] = {10, 20, 30, 40}
Output :        -1, -1, -1, -1

Input : arr[] = {40, 30, 20, 10}
Output :        -1, 40, 30, 20
```

Expected time complexity : O(n)

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to run two nested loops. The outer loop picks an element one by one. The inner loop, find the previous element that is greater.

## C++

 `// C++ program previous greater element ` `// A naive solution to print previous greater ` `// element for every element in an array. ` `#include ` `using` `namespace` `std; ` ` `  `void` `prevGreater(``int` `arr[], ``int` `n) ` `{  ` `    ``// Previous greater for first element never ` `    ``// exists, so we print -1. ` `    ``cout << ``"-1, "``; ` ` `  `    ``// Let us process remaining elements. ` `    ``for` `(``int` `i = 1; i < n; i++) { ` ` `  `        ``// Find first element on left side ` `        ``// that is greater than arr[i]. ` `        ``int` `j; ` `        ``for` `(j = i-1; j >= 0; j--) { ` `            ``if` `(arr[i] < arr[j]) { ` `            ``cout << arr[j] << ``", "``; ` `            ``break``; ` `            ``}              ` `        ``} ` ` `  `        ``// If all elements on left are smaller. ` `        ``if` `(j == -1) ` `        ``cout << ``"-1, "``; ` `    ``} ` `} ` `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 10, 4, 2, 20, 40, 12, 30 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``prevGreater(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program previous greater element ` `// A naive solution to print ` `// previous greater element  ` `// for every element in an array. ` `import` `java.io.*; ` `import` `java.util.*; ` `import` `java.lang.*; ` ` `  `class` `GFG ` `{ ` `static` `void` `prevGreater(``int` `arr[],  ` `                        ``int` `n) ` `{  ` `    ``// Previous greater for  ` `    ``// first element never ` `    ``// exists, so we print -1. ` `    ``System.out.print(``"-1, "``); ` ` `  `    ``// Let us process  ` `    ``// remaining elements. ` `    ``for` `(``int` `i = ``1``; i < n; i++) ` `    ``{ ` ` `  `        ``// Find first element on  ` `        ``// left side that is  ` `        ``// greater than arr[i]. ` `        ``int` `j; ` `        ``for` `(j = i-``1``; j >= ``0``; j--)  ` `        ``{ ` `            ``if` `(arr[i] < arr[j])  ` `            ``{ ` `            ``System.out.print(arr[j] + ``", "``); ` `            ``break``; ` `            ``}              ` `        ``} ` ` `  `        ``// If all elements on  ` `        ``// left are smaller. ` `        ``if` `(j == -``1``) ` `        ``System.out.print(``"-1, "``); ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = {``10``, ``4``, ``2``, ``20``, ``40``, ``12``, ``30``}; ` `    ``int` `n = arr.length; ` `    ``prevGreater(arr, n); ` `} ` `} `

## Python 3

 `# Python 3 program previous greater element ` `# A naive solution to print previous greater ` `# element for every element in an array. ` `def` `prevGreater(arr, n) : ` ` `  `    ``# Previous greater for first element never ` `    ``# exists, so we print -1. ` `    ``print``(``"-1"``,end ``=` `", "``) ` ` `  `    ``# Let us process remaining elements. ` `    ``for` `i ``in` `range``(``1``, n) : ` `        ``flag ``=` `0` ` `  `        ``# Find first element on left side ` `        ``# that is greater than arr[i]. ` `        ``for` `j ``in` `range``(i``-``1``, ``-``1``, ``-``1``) : ` `            ``if` `arr[i] < arr[j] : ` `                ``print``(arr[j],end ``=` `", "``) ` `                ``flag ``=` `1` `                ``break` ` `  `        ``# If all elements on left are smaller. ` `        ``if` `j ``=``=` `0` `and` `flag ``=``=` `0``: ` `            ``print``(``"-1"``,end ``=` `", "``) ` ` `  ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"` `: ` `    ``arr ``=` `[``10``, ``4``, ``2``, ``20``, ``40``, ``12``, ``30``] ` `    ``n ``=` `len``(arr) ` `    ``prevGreater(arr, n) ` ` `  `# This code is contributed by ANKITRAI1 `

## C#

 `// C# program previous greater element ` `// A naive solution to print ` `// previous greater element  ` `// for every element in an array. ` ` `  `using` `System;  ` `class` `GFG ` `{ ` `static` `void` `prevGreater(``int``[] arr,  ` `                        ``int` `n) ` `{  ` `    ``// Previous greater for  ` `    ``// first element never ` `    ``// exists, so we print -1. ` `    ``Console.Write(``"-1, "``); ` ` `  `    ``// Let us process  ` `    ``// remaining elements. ` `    ``for` `(``int` `i = 1; i < n; i++) ` `    ``{ ` ` `  `        ``// Find first element on  ` `        ``// left side that is  ` `        ``// greater than arr[i]. ` `        ``int` `j; ` `        ``for` `(j = i-1; j >= 0; j--)  ` `        ``{ ` `            ``if` `(arr[i] < arr[j])  ` `            ``{ ` `            ``Console.Write(arr[j] + ``", "``); ` `            ``break``; ` `            ``}              ` `        ``} ` ` `  `        ``// If all elements on  ` `        ``// left are smaller. ` `        ``if` `(j == -1) ` `        ``Console.Write(``"-1, "``); ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``int``[] arr = {10, 4, 2, 20, 40, 12, 30}; ` `    ``int` `n = arr.Length; ` `    ``prevGreater(arr, n); ` `} ` `} `

## PHP

 `= 0; ``\$j``--)  ` `    ``{ ` `            ``if` `(``\$arr``[``\$i``] < ``\$arr``[``\$j``])  ` `        ``{ ` `            ``echo``(``\$arr``[``\$j``]);  ` `        ``echo``( ``", "``); ` `            ``break``; ` `            ``}              ` `        ``} ` ` `  `        ``// If all elements on left are smaller. ` `        ``if` `(``\$j` `== -1) ` `        ``echo``(``"-1, "``); ` `    ``} ` `} ` ` `  `// Driver code ` `\$arr` `= ``array``(10, 4, 2, 20, 40, 12, 30); ` `\$n` `= sizeof(``\$arr``) ; ` `prevGreater(``\$arr``, ``\$n``); ` ` `  `//This code is contributed by Shivi_Aggarwal. ` `     `  `?> `

Output:

```-1, 10, 4, -1, -1, 40, 40
```

An efficient solution is to use stack data structure. If we take a closer look, we can notice that this problem is a variation of stock span problem. We maintain previous greater element in a stack.

## C++

 `// C++ program previous greater element ` `// An efficient solution to print previous greater ` `// element for every element in an array. ` `#include ` `using` `namespace` `std; ` ` `  `void` `prevGreater(``int` `arr[], ``int` `n) ` `{ ` `    ``// Create a stack and push index of first element  ` `    ``// to it ` `    ``stack<``int``> s; ` `    ``s.push(arr[0]); ` `     `  `    ``// Previous greater for first element is always -1. ` `    ``cout << ``"-1, "``; ` ` `  `    ``// Traverse remaining elements ` `    ``for` `(``int` `i = 1; i < n; i++) { ` ` `  `        ``// Pop elements from stack while stack is not empty  ` `        ``// and top of stack is smaller than arr[i]. We  ` `        ``// always have elements in decreasing order in a  ` `        ``// stack. ` `        ``while` `(s.empty() == ``false` `&& s.top() < arr[i]) ` `            ``s.pop(); ` ` `  `        ``// If stack becomes empty, then no element is greater ` `        ``// on left side. Else top of stack is previous ` `        ``// greater. ` `        ``s.empty() ? cout << ``"-1, "` `: cout << s.top() << ``", "``; ` ` `  `        ``s.push(arr[i]); ` `    ``} ` `} ` `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 10, 4, 2, 20, 40, 12, 30 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``prevGreater(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program previous greater element ` `// An efficient solution to ` `// print previous greater ` `// element for every element  ` `// in an array. ` `import` `java.io.*; ` `import` `java.util.*; ` `import` `java.lang.*; ` ` `  `class` `GFG ` `{ ` `static` `void` `prevGreater(``int` `arr[],  ` `                        ``int` `n) ` `{ ` `    ``// Create a stack and push  ` `    ``// index of first element  ` `    ``// to it ` `    ``Stack s = ``new` `Stack(); ` `    ``s.push(arr[``0``]); ` `     `  `    ``// Previous greater for  ` `    ``// first element is always -1. ` `    ``System.out.print(``"-1, "``); ` ` `  `    ``// Traverse remaining elements ` `    ``for` `(``int` `i = ``1``; i < n; i++)  ` `    ``{ ` ` `  `        ``// Pop elements from stack  ` `        ``// while stack is not empty  ` `        ``// and top of stack is smaller  ` `        ``// than arr[i]. We always have  ` `        ``// elements in decreasing order  ` `        ``// in a stack. ` `        ``while` `(s.empty() == ``false` `&&  ` `            ``s.peek() < arr[i]) ` `            ``s.pop(); ` ` `  `        ``// If stack becomes empty, then  ` `        ``// no element is greater on left  ` `        ``// side. Else top of stack is  ` `        ``// previous greater. ` `        ``if` `(s.empty() == ``true``)  ` `            ``System.out.print(``"-1, "``); ` `        ``else` `            ``System.out.print(s.peek() + ``", "``); ` ` `  `        ``s.push(arr[i]); ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``10``, ``4``, ``2``, ``20``, ``40``, ``12``, ``30` `}; ` `    ``int` `n = arr.length; ` `    ``prevGreater(arr, n); ` `} ` `} `

## Python3

 `# Python3 program to print previous greater element ` `# An efficient solution to print previous greater ` `# element for every element in an array. ` `import` `math as mt ` ` `  `def` `prevGreater(arr, n): ` ` `  `    ``# Create a stack and push index of  ` `    ``# first element to it ` `    ``s ``=` `list``(); ` `    ``s.append(arr[``0``]) ` `     `  `    ``# Previous greater for first element ` `    ``# is always -1. ` `    ``print``(``"-1, "``, end ``=` `"") ` ` `  `    ``# Traverse remaining elements ` `    ``for` `i ``in` `range``(``1``, n):  ` ` `  `        ``# Pop elements from stack while stack is  ` `        ``# not empty and top of stack is smaller  ` `        ``# than arr[i]. We always have elements in  ` `        ``# decreasing order in a stack. ` `        ``while` `(``len``(s) > ``0` `and` `s[``-``1``] < arr[i]): ` `            ``s.pop() ` ` `  `        ``# If stack becomes empty, then no element  ` `        ``# is greater on left side. Else top of stack  ` `        ``# is previous greater. ` `        ``if` `len``(s) ``=``=` `0``: ` `            ``print``(``"-1, "``, end ``=` `"") ` `        ``else``: ` `            ``print``(s[``-``1``], ``", "``, end ``=` `"") ` ` `  `        ``s.append(arr[i]) ` `     `  `# Driver code ` `arr ``=` `[ ``10``, ``4``, ``2``, ``20``, ``40``, ``12``, ``30` `] ` `n ``=` `len``(arr) ` `prevGreater(arr, n) ` ` `  `# This code is contributed by ` `# mohit kumar 29 `

## C#

 `// C# program previous greater element  ` `// An efficient solution to  ` `// print previous greater  ` `// element for every element  ` `// in an array.  ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG  ` `{  ` `static` `void` `prevGreater(``int` `[]arr,  ` `                        ``int` `n)  ` `{  ` `    ``// Create a stack and push  ` `    ``// index of first element  ` `    ``// to it  ` `    ``Stack<``int``> s = ``new` `Stack<``int``>();  ` `    ``s.Push(arr[0]);  ` `     `  `    ``// Previous greater for  ` `    ``// first element is always -1.  ` `    ``Console.Write(``"-1, "``);  ` ` `  `    ``// Traverse remaining elements  ` `    ``for` `(``int` `i = 1; i < n; i++)  ` `    ``{  ` ` `  `        ``// Pop elements from stack  ` `        ``// while stack is not empty  ` `        ``// and top of stack is smaller  ` `        ``// than arr[i]. We always have  ` `        ``// elements in decreasing order  ` `        ``// in a stack.  ` `        ``while` `(s.Count != 0 &&  ` `            ``s.Peek() < arr[i])  ` `            ``s.Pop();  ` ` `  `        ``// If stack becomes empty, then  ` `        ``// no element is greater on left  ` `        ``// side. Else top of stack is  ` `        ``// previous greater.  ` `        ``if` `(s.Count == 0)  ` `            ``Console.Write(``"-1, "``);  ` `        ``else` `            ``Console.Write(s.Peek() + ``", "``);  ` ` `  `        ``s.Push(arr[i]);  ` `    ``}  ` `}  ` ` `  `// Driver Code  ` `public` `static` `void` `Main(String[] args)  ` `{  ` `    ``int` `[]arr = { 10, 4, 2, 20, 40, 12, 30 };  ` `    ``int` `n = arr.Length;  ` `    ``prevGreater(arr, n);  ` `}  ` `}  ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```-1, 10, 4, -1, -1, 40, 40
```

Time Complexity: O(n). It seems more than O(n) at first look. If we take a closer look, we can observe that every element of array is added and removed from stack at most once. So there are total 2n operations at most. Assuming that a stack operation takes O(1) time, we can say that the time complexity is O(n).

Auxiliary Space: O(n) in worst case when all elements are sorted in decreasing order.

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.