Find relative rank of each element in array
Last Updated :
03 Jan, 2023
Given an array A[] of N integers, the task is to find the relative rank for each element in the given array.
The relative rank for each element in the array is the count of elements which is greater than the current element in the Longest Increasing Subsequence from the current element.
Examples:
Input: A[] = {8, 16, 5, 6, 9}, N = 5
Output: {1, 0, 2, 1, 0}
Explanation:
For i = 0, required sequence is {8, 16} Relative Rank = 1.
For i = 1, Since all elements after 16 are smaller than 16, Relative Rank = 0.
For i = 2, required sequence is {5, 6, 9} Relative Rank = 2
For i = 3, required sequence is {6, 9} Relative Rank = 1
For i = 4, required sequence is {9} Relative Rank = 0
Input: A[] = {1, 2, 3, 5, 4}
Output: {3, 2, 1, 0, 0}
Explanation:
For i = 0, required sequence is {1, 2, 3, 5}, Relative Rank = 3
For i = 1, required sequence is {2, 3, 5}, Relative Rank = 2
For i = 2, required sequence is {3, 5}, Relative Rank = 1
For i = 3, required sequence is {5}, Relative Rank = 0
For i = 4, required sequence is {4}, Relative Rank = 0
Naive Approach: The idea is to generate the longest increasing subsequence for each element and then, the relative rank for each element is the (length of LIS – 1).
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use a Stack and store the elements in non-decreasing order from the right to each element(say A[i]) then the rank for each A[i] is the (size of stack – 1) till that element. Below is the illustration of the same:
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void findRank( int A[], int N)
{
int rank[N] = {};
stack< int > s;
s.push(A[N - 1]);
for ( int i = N - 2; i >= 0; i--) {
if (A[i] < s.top()) {
s.push(A[i]);
rank[i] = s.size() - 1;
}
else {
while (!s.empty()
&& A[i] >= s.top()) {
s.pop();
}
s.push(A[i]);
rank[i] = s.size() - 1;
}
}
for ( int i = 0; i < N; i++) {
cout << rank[i] << " " ;
}
}
int main()
{
int A[] = { 1, 2, 3, 5, 4 };
int N = sizeof (A) / sizeof (A[0]);
findRank(A, N);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
import java.lang.*;
class GFG{
static void findRank( int [] A, int N)
{
int [] rank = new int [N];
Stack<Integer> s = new Stack<Integer>();
s.add(A[N - 1 ]);
for ( int i = N - 2 ; i >= 0 ; i--)
{
if (A[i] < s.peek())
{
s.add(A[i]);
rank[i] = s.size() - 1 ;
}
else
{
while (!s.isEmpty() &&
A[i] >= s.peek())
{
s.pop();
}
s.add(A[i]);
rank[i] = s.size() - 1 ;
}
}
for ( int i = 0 ; i < N; i++)
{
System.out.print(rank[i] + " " );
}
}
public static void main(String[] args)
{
int A[] = { 1 , 2 , 3 , 5 , 4 };
int N = A.length;
findRank(A, N);
}
}
|
Python3
def findRank(A, N):
rank = [ 0 ] * N
s = []
s.append(A[N - 1 ])
for i in range (N - 2 , - 1 , - 1 ):
if (A[i] < s[ - 1 ]):
s.append(A[i])
rank[i] = len (s) - 1
else :
while ( len (s) > 0 and A[i] > = s[ - 1 ]):
del s[ - 1 ]
s.append(A[i])
rank[i] = len (s) - 1
for i in range (N):
print (rank[i], end = " " )
if __name__ = = '__main__' :
A = [ 1 , 2 , 3 , 5 , 4 ]
N = len (A)
findRank(A, N)
|
C#
using System;
using System.Collections.Generic;
using System.Linq;
class GFG{
static void findRank( int [] A, int N)
{
int [] rank = new int [N];
Stack< int > s = new Stack< int >();
s.Push(A[N - 1]);
for ( int i = N - 2; i >= 0; i--)
{
if (A[i] < s.Peek())
{
s.Push(A[i]);
rank[i] = s.Count() - 1;
}
else
{
while (s.Count() != 0 &&
A[i] >= s.Peek())
{
s.Pop();
}
s.Push(A[i]);
rank[i] = s.Count() - 1;
}
}
for ( int i = 0; i < N; i++)
{
Console.Write(rank[i] + " " );
}
}
public static void Main()
{
int [] A = new int [] { 1, 2, 3, 5, 4 };
int N = A.Length;
findRank(A, N);
}
}
|
Javascript
class GFG
{
static findRank(A, N)
{
var rank = Array(N).fill(0);
var s = Array();
(s.push(A[N - 1]) > 0);
for ( var i = N - 2; i >= 0; i--)
{
if (A[i] < s[s.length-1])
{
(s.push(A[i]) > 0);
rank[i] = s.length - 1;
}
else
{
while (!(s.length == 0) && A[i] >= s[s.length-1])
{
s.pop();
}
(s.push(A[i]) > 0);
rank[i] = s.length - 1;
}
}
for ( var i=0; i < N; i++)
{
console.log(rank[i] + " " );
}
}
static main(args)
{
var A = [1, 2, 3, 5, 4];
var N = A.length;
GFG.findRank(A, N);
}
}
GFG.main([]);
|
Time Complexity: O(N)
Auxiliary Space: O(N)
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