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Rank of remaining numbers in Array by replacing first and last with max and min alternatively

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  • Last Updated : 12 Jul, 2022

Given an array arr[ ] of size N, the task is to find the rank of the remaining element in an array after performing the given operation:

  • In each operation choose elements from both ends and delete them and insert the max of those values at the position of the left element and move one step towards the center from both ends and keep performing this operation.
  • In the next cycle keep performing the same operation but instead of max, insert min of the elements this time.
  • Perform this operation in alternate cycles until there is only one element left in the array.
  • The rank is the position of the remaining element in the original array when it is sorted in increasing order. (The elements with same values are considered only for the sorted order)

Examples:

Input: arr = {4, 5, 3, 56, 3, 24, 5, 6, 22, 4, 55, 50, 89}
Output: 6
Explanation: See the diagram given below

24 is th 6th smallest element. The elements with same values are considered once.

Input: N = {20, 4, 5, 35, 6, 22, 4, 34}
Output: 6

 

Approach: The solution is based on two pointer approach. Follow the steps mentioned below:

  • Take c as 1 which will indicate the number of iterations.
    • Take 2 pointers, start as zero and end as N-1.
    • Compare element at index s and e.
    • If c is odd then take the maximum of the element else take the minimum of element and store in an array.
    • Increment s, decrement e and repeat till s != e.
  • Increment c by 1
  • Repeat step 2 till the length of arr becomes 1.
  • Remove duplicate from arr[] and sort it, now find the rank of the remaining element.

Below is the implementation of the above approach.

C++




// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find rank of number in array
int rankOfNum(vector<int>& num)
{
 
  // Copying array to S
  vector<int> S = num;
 
  // c count no of iterations
  int c = 1;
  while (S.size() != 1) {
 
    // s is starting index
    int s = 0;
 
    // e is ending index
    int e = S.size() - 1;
 
    // Empty array to store
    // result of comparisons.
    vector<int> l;
 
    // loop till s <= e
    while (s <= e) {
 
      // In odd iterations take
      // maximum of element.
      if (c % 2 == 1)
        l.push_back(max(S[s], S[e]));
 
      // In even Iterations
      // take minimum of element.
      else {
        l.push_back(min(S[s], S[e]));
      }
      // Increment s by 1
      // and decrement e by 1
      s += 1;
      e -= 1;
    }
    // Assigning l to S
    S = l;
 
    // Increment iteration value by 1
    c += 1;
  }
 
  // Converting list into set and again to list
  // so that all duplicate will get removed
 
  set<int> setx;
 
  for (auto dt : num)
    setx.insert(dt);
 
  // Finding index of remained element
 
  int p = distance(setx.begin(), setx.find(S[0]));
 
  // Returning the rank of element
  return p + 1;
}
 
// Driver code
int main()
{
 
  // Original array
  vector<int> arr
    = { 4, 5, 3, 56, 3, 24, 5, 6, 22, 4, 55, 50, 89 };
 
  // Calling function
  int s = rankOfNum(arr);
 
  // Print its rank
  cout << s;
 
  return 0;
}
 
//     This code is contributed by rakeshsahni


Java




// Java code for the above approach
import java.util.*;
class GFG{
 
// Function to find rank of number in array
static int rankOfNum(Integer[] num)
{
 
  // Copying array to S
  List<Integer> S = Arrays.asList(num);
 
  // c count no of iterations
  int c = 1;
  while (S.size() != 1) {
 
    // s is starting index
    int s = 0;
 
    // e is ending index
    int e = S.size() - 1;
 
    // Empty array to store
    // result of comparisons.
    ArrayList<Integer> l = new ArrayList<Integer>();
 
    // loop till s <= e
    while (s <= e) {
 
      // In odd iterations take
      // maximum of element.
      if (c % 2 == 1)
        l.add(Math.max(S.get(s), S.get(e)));
 
      // In even Iterations
      // take minimum of element.
      else {
        l.add(Math.min(S.get(s), S.get(e)));
      }
      // Increment s by 1
      // and decrement e by 1
      s += 1;
      e -= 1;
    }
    // Assigning l to S
    S = l;
 
    // Increment iteration value by 1
    c += 1;
  }
 
  // Converting list into set and again to list
  // so that all duplicate will get removed
 
  HashSet<Integer> setx = new HashSet<Integer>();
 
  for (int dt : num)
    setx.add(dt);
 
  // Finding index of remained element
  List<Integer> l = new LinkedList<>(setx);
  Collections.sort(l);
  int p = l.indexOf(S.get(0));
 
  // Returning the rank of element
  return p + 1;
}
 
// Driver code
public static void main(String[] args)
{
 
  // Original array
  Integer[] arr
    = { 4, 5, 3, 56, 3, 24, 5, 6, 22, 4, 55, 50, 89 };
 
  // Calling function
  int s = rankOfNum(arr);
 
  // Print its rank
  System.out.print(s);
 
}
}
 
// This code is contributed by 29AjayKumar


Python3




# Python code to implement above approach
 
# Function to find rank of number in array
 
 
def rankOfNum(num):
 
    # Copying array to S
    S = num[:]
 
    # c count no of iterations
    c = 1
    while len(S) != 1:
 
        # s is starting index
        s = 0
 
        # e is ending index
        e = len(S) - 1
 
        # Empty array to store
        # result of comparisons.
        l = []
 
        # loop till s <= e
        while s <= e:
 
            # In odd iterations take
            # maximum of element.
            if c % 2 == 1:
                l.append(max(S[s], S[e]))
 
            # In even Iterations
            # take minimum of element.
            else:
                l.append(min(S[s], S[e]))
 
            # Increment s by 1
            # and decrement e by 1
            s += 1
            e -= 1
 
        # Assigning l to S
        S = l
 
        # Increment iteration value by 1
        c += 1
 
    # Converting list into set and again to list
    # so that all duplicate will get removed
    setx = list(set(num))
 
    # Sorting to get rank
    setx.sort()
 
    # Finding index of remained element
    p = setx.index(S[0])
 
    # Returning the rank of element
    return p + 1
 
 
if __name__ == "__main__":
 
    # Original array
    arr = [4, 5, 3, 56, 3, 24, 5, 6, 22, 4, 55, 50, 89]
 
    # Calling function
    s = rankOfNum(arr)
 
    # Print its rank
    print(str(s))


C#




// C# code for the above approach
using System;
using System.Linq;
using System.Collections.Generic;
class GFG {
 
  // Function to find rank of number in array
  static int rankOfNum(List<int> num)
  {
 
    // Copying array to S
    List<int> S = num;
 
    // c count no of iterations
    int c = 1;
    while (S.Count != 1) {
 
      // s is starting index
      int s = 0;
 
      // e is ending index
      int e = S.Count - 1;
 
      // Empty array to store
      // result of comparisons.
      List<int> l = new List<int>();
 
      // loop till s <= e
      while (s <= e) {
 
        // In odd iterations take
        // maximum of element.
        if (c % 2 == 1)
          l.Add(Math.Max(S[s], S[e]));
 
        // In even Iterations
        // take minimum of element.
        else {
          l.Add(Math.Min(S[s], S[e]));
        }
 
        // Increment s by 1
        // and decrement e by 1
        s += 1;
        e -= 1;
      }
      // Assigning l to S
      S = l;
 
      // Increment iteration value by 1
      c += 1;
 
    }
 
    // Converting list into set and again to list
    // so that all duplicate will get removed
    HashSet<int> setx = new HashSet<int>();
    foreach(var dt in num) setx.Add(dt);
 
 
    // Finding index of remained element
    List<int> setxList = setx.ToList();
    setxList.Sort();
 
    int p = setxList.IndexOf(S[0]);
 
    // Returning the rank of element
    return p + 1;
  }
 
  // Driver code
  public static void Main()
  {
 
    // Original array
    List<int> arr = new List<int>() {
      4, 5, 3, 56, 3, 24, 5, 6, 22, 4, 55, 50, 89
      };
 
    // Calling function
    int s = rankOfNum(arr);
 
    // Print its rank
    Console.Write(s);
  }
}
 
// This code is contributed by ukasp.


Javascript




<script>
       // JavaScript code for the above approach
 
       // Function to find rank of number in array
       function rankOfNum(num) {
 
           // Copying array to S
           let S = [...num]
 
           // c count no of iterations
           c = 1
           while (S.length != 1) {
 
               // s is starting index
               s = 0
 
               // e is ending index
               e = S.length - 1
 
               // Empty array to store
               // result of comparisons.
               l = []
 
               // loop till s <= e
               while (s <= e) {
 
                   // In odd iterations take
                   // maximum of element.
                   if (c % 2 == 1)
                       l.push(Math.max(S[s], S[e]))
 
                   // In even Iterations
                   // take minimum of element.
                   else {
                       l.push(Math.min(S[s], S[e]))
                   }
                   // Increment s by 1
                   // and decrement e by 1
                   s += 1
                   e -= 1
               }
               // Assigning l to S
               S = [...l]
 
               // Increment iteration value by 1
               c += 1
           }
            
           // Converting list into set and again to list
           // so that all duplicate will get removed
           let setx = new Set([...num])
 
           // Sorting to get rank
           t = [...setx].sort(function (a, b) { return a - b })
           // Finding index of remained element
 
           let p = t.indexOf(S[0])
            
           // Returning the rank of element
           return p + 1
 
       }
        
       // Original array
       arr = [4, 5, 3, 56, 3, 24, 5, 6, 22, 4, 55, 50, 89]
 
       // Calling function
       s = rankOfNum(arr)
 
       // Print its rank
       document.write((s))
 
 // This code is contributed by Potta Lokesh
   </script>


Output

6

Time Complexity: O(N)
Space Complexity: O(N), since n extra space has been taken


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