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
[Naive Solution] - Nested Loops - O(n^2) Time and O(1) Space
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 <bits/stdc++.h>
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 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 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# 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);
}
}
<script>
// Javascript program previous greater element
// A naive solution to print
// previous greater element
// for every element in an array.
function prevGreater(arr,n)
{
// Previous greater for
// first element never
// exists, so we print -1.
document.write("-1, ");
// Let us process
// remaining elements.
for (let i = 1; i < n; i++)
{
// Find first element on
// left side that is
// greater than arr[i].
let j;
for (j = i-1; j >= 0; j--)
{
if (arr[i] < arr[j])
{
document.write(arr[j] + ", ");
break;
}
}
// If all elements on
// left are smaller.
if (j == -1)
document.write("-1, ");
}
}
// Driver Code
let arr=[10, 4, 2, 20, 40, 12, 30];
let n = arr.length;
prevGreater(arr, n);
// This code is contributed by avanitrachhadiya2155
</script>
<?php
// php program previous greater element
// A naive solution to print previous greater
// element for every element in an array.
function prevGreater(&$arr,$n)
{
// Previous greater for first element never
// exists, so we print -1.
echo( "-1, ");
// Let us process remaining elements.
for ($i = 1; $i < $n; $i++)
{
// Find first element on left side
// that is greater than arr[i].
for ($j = $i-1; $j >= 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,
[Expected Solution] - O(n) Time and O(n) Space
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 <bits/stdc++.h>
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 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<Integer> s = new Stack<Integer>();
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 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# 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
<script>
// Javascript program previous greater element
// An efficient solution to
// print previous greater
// element for every element
// in an array.
function prevGreater(arr,n)
{
// Create a stack and push
// index of first element
// to it
let s = [];
s.push(arr[0]);
// Previous greater for
// first element is always -1.
document.write("-1, ");
// Traverse remaining elements
for (let 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.length!=0 &&
s[s.length-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 (s.length==0)
document.write("-1, ");
else
document.write(s[s.length-1] + ", ");
s.push(arr[i]);
}
}
// Driver Code
let arr=[10, 4, 2, 20, 40, 12, 30];
let n = arr.length;
prevGreater(arr, n);
// This code is contributed by rag2127
</script>
Output-1, 10, 4, -1, -1, 40, 40,
Complexity Analysis:
- 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.
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