Find relative rank of each element in array

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++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find relative rank for
// each element in the array A[]
void findRank(int A[], int N)
{
    // Create Rank Array
    int rank[N] = {};

    // Stack to store numbers in
    // non-decreasing order from right
    stack<int> s;

    // Push last element in stack
    s.push(A[N - 1]);

    // Iterate from second last
    // element to first element
    for (int i = N - 2; i >= 0; i--) {

        // If current element is less
        // than the top of stack and
        // push A[i] in stack
        if (A[i] < s.top()) {

            s.push(A[i]);

            // Rank is stack size - 1
            // for current element
            rank[i] = s.size() - 1;
        }
        else {

            // Pop elements from stack
            // till current element is
            // greater than the top
            while (!s.empty()
                   && A[i] >= s.top()) {
                s.pop();
            }

            // Push current element in Stack
            s.push(A[i]);

            // Rank is stack size - 1
            rank[i] = s.size() - 1;
        }
    }

    // Print rank of all elements
    for (int i = 0; i < N; i++) {
        cout << rank[i] << " ";
    }
}

// Driver Code
int main()
{
    // Given array A[]
    int A[] = { 1, 2, 3, 5, 4 };

    int N = sizeof(A) / sizeof(A[0]);

    // Function call
    findRank(A, N);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
import java.lang.*;

class GFG{

// Function to find relative rank for
// each element in the array A[]
static void findRank(int[] A, int N)
{

    // Create Rank Array
    int[] rank = new int[N];

    // Stack to store numbers in
    // non-decreasing order from right
    Stack<Integer> s = new Stack<Integer>();

    // Push last element in stack
    s.add(A[N - 1]);

    // Iterate from second last
    // element to first element
    for(int i = N - 2; i >= 0; i--)
    {

        // If current element is less
        // than the top of stack and
        // push A[i] in stack
        if (A[i] < s.peek())
        {
            s.add(A[i]);

            // Rank is stack size - 1
            // for current element
            rank[i] = s.size() - 1;
        }
        else
        {

            // Pop elements from stack
            // till current element is
            // greater than the top
            while (!s.isEmpty() &&
                    A[i] >= s.peek())
            {
                s.pop();
            }

            // Push current element in Stack
            s.add(A[i]);

            // Rank is stack size - 1
            rank[i] = s.size() - 1;
        }
    }

    // Print rank of all elements
    for(int i = 0; i < N; i++)
    {
        System.out.print(rank[i] + " ");
    }
}

// Driver Code
public static void main(String[] args)
{

    // Given array A[]
    int A[] = { 1, 2, 3, 5, 4 };

    int N = A.length;

    // Function call
    findRank(A, N);
}
}

// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach

# Function to find relative rank for
# each element in the array A[]
def findRank(A, N):

    # Create Rank Array
    rank = [0] * N

    # Stack to store numbers in
    # non-decreasing order from right
    s = []

    # Push last element in stack
    s.append(A[N - 1])

    # Iterate from second last
    # element to first element
    for i in range(N - 2, -1, -1):

        # If current element is less
        # than the top of stack and
        # append A[i] in stack
        if (A[i] < s[-1]):
            s.append(A[i])

            # Rank is stack size - 1
            # for current element
            rank[i] = len(s) - 1

        else:

            # Pop elements from stack
            # till current element is
            # greater than the top
            while (len(s) > 0 and A[i] >= s[-1]):
                del s[-1]

            # Push current element in Stack
            s.append(A[i])

            # Rank is stack size - 1
            rank[i] = len(s) - 1

    # Print rank of all elements
    for i in range(N):
        print(rank[i], end = " ")

# Driver Code
if __name__ == '__main__':

    # Given array A[]
    A = [ 1, 2, 3, 5, 4 ]

    N = len(A)

    # Function call
    findRank(A, N)

# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
using System.Linq;

class GFG{

// Function to find relative rank for
// each element in the array A[]
static void findRank(int[] A, int N)
{

    // Create Rank Array
    int[] rank = new int[N];

    // Stack to store numbers in
    // non-decreasing order from right
    Stack<int> s = new Stack<int>();

    // Push last element in stack
    s.Push(A[N - 1]);

    // Iterate from second last
    // element to first element
    for(int i = N - 2; i >= 0; i--)
    {

        // If current element is less
        // than the top of stack and
        // push A[i] in stack
        if (A[i] < s.Peek())
        {
            s.Push(A[i]);

            // Rank is stack size - 1
            // for current element
            rank[i] = s.Count() - 1;
        }
        else
        {

            // Pop elements from stack
            // till current element is
            // greater than the top
            while (s.Count() != 0 &&
                   A[i] >= s.Peek())
            {
                s.Pop();
            }

            // Push current element in Stack
            s.Push(A[i]);

            // Rank is stack size - 1
            rank[i] = s.Count() - 1;
        }
    }

    // Print rank of all elements
    for(int i = 0; i < N; i++)
    {
        Console.Write(rank[i] + " ");
    }
}

// Driver Code
public static void Main()
{

    // Given array A[]
    int[] A = new int[] { 1, 2, 3, 5, 4 };

    int N = A.Length;

    // Function call
    findRank(A, N);
}
}

// This code is contributed by sanjoy_62

Javascript

class GFG
{
    // Function to find relative rank for
    // each element in the array A[]
    static findRank(A, N)
    {
        // Create Rank Array
        var rank = Array(N).fill(0);

        // Stack to store numbers in
        // non-decreasing order from right
        var s = Array();

        // Push last element in stack
        (s.push(A[N - 1]) > 0);

        // Iterate from second last
        // element to first element
        for (var i = N - 2; i >= 0; i--)
        {

            // If current element is less
            // than the top of stack and
            // push A[i] in stack
            if (A[i] < s[s.length-1])
            {
                (s.push(A[i]) > 0);

                // Rank is stack size - 1
                // for current element
                rank[i] = s.length - 1;
            }
            else
            {
                // Pop elements from stack
                // till current element is
                // greater than the top
                while (!(s.length == 0) && A[i] >= s[s.length-1])
                {
                    s.pop();
                }

                // Push current element in Stack
                (s.push(A[i]) > 0);

                // Rank is stack size - 1
                rank[i] = s.length - 1;
            }
        }

        // Print rank of all elements
        for (var i=0; i < N; i++)
        {
            console.log(rank[i] + " ");
        }
    }

    // Driver Code
    static main(args)
    {

        // Given array A[]
        var A = [1, 2, 3, 5, 4];
        var N = A.length;

        // Function call
        GFG.findRank(A, N);
    }
}
GFG.main([]);

// This code is contributed by aadityaburujwale.

Output:

3 2 1 0 0

Time Complexity: O(N)
Auxiliary Space: O(N)