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Find K for every Array element such that at least K prefixes are ≥ K
• Difficulty Level : Hard
• Last Updated : 04 Dec, 2020

Given an array arr[] consisting of N non-negative integers, the task is to find an integer K for every index such that at least K integers in the array till that index are greater or equal to K.

Note: Consider 1 based indexing

Examples:

Input: arr[] = {3, 0, 6, 1, 5}
Output: K = {1, 1, 2, 2, 3}
Explanation:
At index 1, there is 1 number greater than or equal to 1 in the array i.e. 3. So K value for elements upto index 1 is 1.
At index 2, there is 1 number greater than or equal to 1 in the array i.e. 3. So K value for elements upto index 2 is 1.
At index 3, there are 2 numbers greater than or equal to 2 in the array i.e. 3 and 6. So K value for elements upto index 3 is 2.
At index 4, there are 2 numbers greater than or equal to 2 in the array i.e. 3 and 6. So K value for elements upto index 4 is 2.
At index 5, there are 3 numbers greater than or equal to 3 in the array i.e. 3, 6 and 5. So K value for elements up to index 5 is 3.

Input: arr[] = {9, 10, 7, 5, 0, 10, 2, 0}
Output: K = {1, 2, 3, 4, 4, 5, 5, 5}

Naive Approach:
The simplest approach is to find the value of K for all the elements of the array in the range [0, i], where i is the index of the array arr[], using the efficient approach used in the article whose link is given here

Time Complexity: O(N2
Space Complexity: O(N)

Efficient Approach:
The idea is to use Multiset(Red-Black Tree). Multiset stores the values in a sorted order which helps to check that if the current minimum value in the multiset is greater than or equal to its size. If yes, then the value of integer K will be the size of the multiset.

Below are the steps for the implementation:

1. Traverse the array from index 0 to N-1.
2. For each index, insert the element into the multiset and check if the smallest value in the multiset is less than the size of the multiset.
3. If true, then erase the starting element and print the size of the multiset.
4. If false then simply print the size of the multiset.
5. The size of the multiset is the required K value for every index i.

Below is the implementation of the above approach:

C++

 // C++ program for the above approach #include using namespace std; // Function to find the K-value// for every index in the arrayint print_h_index(int arr[], int N){    // Multiset to store the array    // in the form of red-black tree    multiset ms;     // Iterating over the array    for (int i = 0; i < N; i++) {         // Inserting the current        // value in the multiset        ms.insert(arr[i]);         // Condition to check if        // the smallest value        // in the set is less than        // it's size        if (*ms.begin()            < ms.size()) {             // Erase the smallest            // value            ms.erase(ms.begin());        }         // h-index value will be        // the size of the multiset        cout << ms.size() << " ";    }} // Driver Codeint main(){     // array    int arr[] = { 9, 10, 7, 5, 0,                10, 2, 0 };     // Size of the array    int N = sizeof(arr)            / sizeof(arr[0]);     // function call    print_h_index(arr, N);     return 0;}

Java

 // Java program for the above approachimport java.util.*; class GFG{     // Function to find the K-value// for every index in the arraystatic void print_h_index(int arr[], int N){         // Multiset to store the array    // in the form of red-black tree    List ms = new ArrayList();     // Iterating over the array    for(int i = 0; i < N; i++)    {                 // Inserting the current        // value in the multiset        ms.add(arr[i]);         // Condition to check if        // the smallest value        // in the set is less than        // it's size        int t = Collections.min(ms);        if (t < ms.size())        {             // Erase the smallest            // value            ms.remove(ms.indexOf(t));        }         // h-index value will be        // the size of the multiset        System.out.print(ms.size() + " ");    }} // Driver codepublic static void main(String[] args){         // Array    int arr[] = { 9, 10, 7, 5, 0,                10, 2, 0 };         // Size of the array    int N = arr.length;         // Function call    print_h_index(arr, N);}} // This code is contributed by offbeat

Python3

 # Python3 program for the above approach  # Function to find the K-value# for every index in the arraydef print_h_index(arr, N):     # Multiset to store the array    # in the form of red-black tree    ms = []      # Iterating over the array    for i in range(N):          # Inserting the current        # value in the multiset        ms.append(arr[i])        ms.sort()                 # Condition to check if        # the smallest value        # in the set is less than        # it's size        if (ms[0] < len(ms)):              # Erase the smallest            # value            ms.pop(0)          # h-index value will be        # the size of the multiset        print(len(ms), end = ' ')         # Driver Codeif __name__=='__main__':     # Array    arr = [ 9, 10, 7, 5, 0, 10, 2, 0 ]      # Size of the array    N = len(arr)      # Function call    print_h_index(arr, N) # This code is contributed by pratham76

C#

 // C# program for the above approachusing System;using System.Collections;using System.Collections.Generic; class GFG{     // Function to find the K-value// for every index in the arraystatic void print_h_index(int []arr, int N){         // Multiset to store the array    // in the form of red-black tree    ArrayList ms = new ArrayList();     // Iterating over the array    for(int i = 0; i < N; i++)    {                 // Inserting the current        // value in the multiset        ms.Add(arr[i]);         // Condition to check if        // the smallest value        // in the set is less than        // it's size        int t = int.MaxValue;        foreach(int x in ms)        {            if(x < t)            {                t = x;            }        }                 if (t < ms.Count)        {                         // Erase the smallest            // value            ms.Remove(t);        }         // h-index value will be        // the size of the multiset        Console.Write(ms.Count + " ");    }} // Driver codepublic static void Main(string[] args){         // Array    int []arr = { 9, 10, 7, 5, 0,                  10, 2, 0 };         // Size of the array    int N = arr.Length;         // Function call    print_h_index(arr, N);}} // This code is contributed by rutvik_56
Output:
1 2 3 4 4 5 5 5

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

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