Find K for every Array element such that at least K prefixes are ≥ K

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++

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// C++ program for the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the K-value
// for every index in the array
int print_h_index(int arr[], int N)
{
    // Multiset to store the array
    // in the form of red-black tree
    multiset<int> 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 Code
int 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;
}

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Output:

1 2 3 4 4 5 5 5

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

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