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)
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.
Implementation:
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
// 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
<?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. ?> |
Javascript
<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> |
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 <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
// 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
# 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 |
Javascript
<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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