Find relative rank of each element in array

21 Aug, 2020
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 = 5Output: {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 = 2For i = 3, required sequence is {6, 9} Relative Rank = 1For 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 = 3For i = 1, required sequence is {2, 3, 5}, Relative Rank = 2For i = 2, required sequence is {3, 5}, Relative Rank = 1For i = 3, required sequence is {5}, Relative Rank = 0For 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 which in non-decreasing order from the right till 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++
Java
Python3
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 Codeint 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 program to implement// the above approachimport 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 Codepublic 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 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 Codeif __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# program to implement// the above approachusing 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 Codepublic 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
3 2 1 0 0

Time Complexity: O(N)Auxiliary Space: O(1)
mohit kumar 29
sanjoy_62
cpp-stack
Arrays
Competitive Programming
Greedy
Stack
Arrays
Greedy
Stack
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