Distance from Next Greater element
Last Updated :
13 Apr, 2023
Given an array arr[] of size N, the task is to print the distance of every array element from its next greater element. For array elements having no next greater element, print 0.
Examples:
Input: arr[] = {73, 74, 75, 71, 69, 72, 76, 73}
Output: {1, 1, 4, 2, 1, 1, 0, 0}
Explanation:
The next greater element for 73 is 74, which is at position 1. Distance = 1 – 0 = 1
The next greater element for 74 is 75, which is at position 2. Distance = 2 – 1 = 1
The next greater element for 75 is 76, which is at position 6. Distance = 6 – 2 = 4
The next greater element for 71 is 72, which is at position 5. Distance = 5 – 3 = 2
The next greater element for 69 is 72, which is at position 5. Distance = 5 – 4 = 1
The next greater element for 72 is 76, which is at position 6. Distance = 6 – 5 = 1
No, next greater element for 76. Distance = 0
No, next greater element for 73. Distance = 0
Input: arr[] = {5, 4, 3, 2, 1}
Output: {0, 0, 0, 0, 0}
Naive Approach: The simplest approach is to traverse the array and for every array element, traverse to its right to obtain its next greater element and calculate the difference between the indices.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use Stack to find the next greater element.
Below are the steps:
- Maintain a Stack which will contain the elements in non-increasing order.
- Check if the current element arr[i]is greater than the element at the top of the stack.
- Keep popping all the elements from the stack one by one from the top, that are found to be smaller than arr[i] and calculate the distance for each of them as the difference of current index and the index of the popped element.
- Push the current element into the stack and repeat the above steps.
Below is the implementation of the above approach:
C++
#include<bits/stdc++.h>
using namespace std;
vector< int > mindistance(vector< int > arr)
{
int N = arr.size();
vector< int > ans(N);
int st = 0;
for ( int i = 0; i < N - 1; i++)
{
if (arr[i] < arr[i + 1])
{
ans[i] = 1;
}
else
{
st = i + 1;
while (st <= N - 1)
{
if (arr[i] < arr[st])
{
ans[i] = st - i;
break ;
}
else
{
st++;
}
}
}
}
return ans;
}
int main()
{
vector< int > arr = { 73, 74, 75, 71,
69, 72, 76, 73 };
vector< int > x = mindistance(arr);
cout << "[" ;
for ( int i = 0; i < x.size(); i++)
{
if (i == x.size() - 1)
cout << x[i];
else
cout << x[i] << ", " ;
}
cout << "]" ;
}
|
Java
import java.io.*;
class GFG{
public static int [] mindistance( int [] arr)
{
int N = arr.length;
int [] ans = new int [N];
int st = 0 ;
for ( int i = 0 ; i < N - 1 ; i++)
{
if (arr[i] < arr[i + 1 ])
{
ans[i] = 1 ;
}
else
{
st = i + 1 ;
while (st <= N - 1 )
{
if (arr[i] < arr[st])
{
ans[i] = st - i;
break ;
}
else
{
st++;
}
}
}
}
return ans;
}
public static void main(String[] args)
{
int arr[] = new int []{ 73 , 74 , 75 , 71 ,
69 , 72 , 76 , 73 };
int x[] = mindistance(arr);
System.out.print( "[" );
for ( int i = 0 ; i < x.length; i++)
System.out.print(x[i]+ ", " );
System.out.print( "]" );
}
}
|
Python3
def mindistance(arr, N):
if N < = 1 :
return [ 0 ]
ans = [ 0 for i in range (N)]
st = [ 0 ]
for i in range ( 1 , N):
while (st and arr[i] > arr[st[ - 1 ]]):
pos = st.pop()
ans[pos] = i - pos
st.append(i)
return ans
arr = [ 73 , 74 , 75 , 71 , 69 , 72 , 76 , 73 ]
N = len (arr)
print (mindistance(arr, N))
|
C#
using System;
class GFG{
public static int [] mindistance( int [] arr)
{
int N = arr.Length;
int [] ans = new int [N];
int st = 0;
for ( int i = 0; i < N - 1; i++)
{
if (arr[i] < arr[i + 1])
{
ans[i] = 1;
}
else
{
st = i + 1;
while (st <= N - 1)
{
if (arr[i] < arr[st])
{
ans[i] = st - i;
break ;
}
else
{
st++;
}
}
}
}
return ans;
}
public static void Main(String[] args)
{
int []arr = new int []{ 73, 74, 75, 71,
69, 72, 76, 73 };
int []x = mindistance(arr);
Console.Write( "[" );
for ( int i = 0; i < x.Length; i++)
Console.Write(x[i]+ ", " );
Console.Write( "]" );
}
}
|
Javascript
<script>
function mindistance(arr)
{
let N = arr.length;
let ans = [];
let st = 0;
for (let i = 0; i < N - 1; i++)
{
if (arr[i] < arr[i + 1])
{
ans[i] = 1;
}
else
{
st = i + 1;
while (st <= N - 1)
{
if (arr[i] < arr[st])
{
ans[i] = st - i;
break ;
}
else
{
st++;
}
}
}
}
return ans;
}
let arr = [ 73, 74, 75, 71,
69, 72, 76, 73 ];
let x = mindistance(arr);
document.write( "[" );
for (let i = 0; i < x.length; i++)
document.write(x[i]+ ", " );
document.write( "]" );
</script>
|
Output:
[1, 1, 4, 2, 1, 1, 0, 0]
Time Complexity: O(N) //since one traversal of the array is required to complete all operations hence overall time required by the algorithm is linear
Auxiliary Space: O(N)// an extra array is used and in the worst case all elements will be stored inside it the hence algorithm takes up linear space
Share your thoughts in the comments
Please Login to comment...