Floor of every element in same array

Given an array of integers, find the closest smaller or same element for every element. If all elements are greater for an element, then print -1. We may assume that the array has at least two elements.

Examples:

Input : arr[] = {10, 5, 11, 10, 20, 12}
Output : 10 -1 10 10 12 11
Note that there are multiple occurrences of 10, so floor of 10 is 10 itself.

Input : arr[] = {6, 11, 7, 8, 20, 12}
Output : -1 8 6 7 12 11



A simple solution is to run two nested loops. We pick an outer element one by one. For every picked element, we traverse remaining array and find closest greater element. Time complexity of this solution is O(n*n)

An better solution is to sort the array and create a sorted copy, then do binary search for floor. We traverse the array, for every element we search for the first occurrence of an element that is greater than or equal to given element. Once we find such an element, we check if next of it is also same, if yes, then there are multiple occurrences of the element, so we print the same element as output. Otherwise we print previous element in the sorted array. In C++, lower_bound() returns iterator to the first greater or equal element in a sorted array.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of efficient algorithm to find
// floor of every element
#include <bits/stdc++.h>
using namespace std;
  
// Prints greater elements on left side of every element
void printPrevGreater(int arr[], int n)
{
    // Create a sorted copy of arr[]
    vector<int> v(arr, arr + n);
    sort(v.begin(), v.end());
  
    // Traverse through arr[] and do binary search for
    // every element.
    for (int i = 0; i < n; i++) {
  
        // Floor of first element is -1 if there is only
        // one occurrence of it.
        if (arr[i] == v[0]) {
            (arr[i] == v[1]) ? cout << arr[i] : cout << -1;
            cout << " ";
            continue;
        }
  
        // Find the first element that is greater than or
        // or equal to given element
        auto it = lower_bound(v.begin(), v.end(), arr[i]);
  
        // If next element is also same, then there
        // are multiple occurrences, so print it
        if (it != v.end() && *(it + 1) == arr[i])
            cout << arr[i] << " ";
  
        // Otherwise print previous element
        else
            cout << *(it - 1) << " ";
    }
}
  
/* Driver program to test insertion sort */
int main()
{
    int arr[] = { 6, 11, 7, 8, 20, 12 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printPrevGreater(arr, n);
    return 0;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of efficient 
# algorithm to find floor of every element 
  
# Prints greater elements on left 
# side of every element 
def printPrevGreater(arr, n) :
  
    # Create a sorted copy of arr 
    v = arr.copy()
    v.sort()
      
  
    # Traverse through arr[] and do 
    # binary search for every element. 
    for i in range(n) : 
  
        # Floor of first element is -1 if 
        # there is only one occurrence of it. 
        if (arr[i] == v[0]) : 
            if (arr[i] == v[1]) :
                print(arr[i], end = " ")
                  
            else :
                print(-1, end = " "
                  
            continue
  
        # Find the first element that is greater
        # than or or equal to given element 
        if v.count(arr[i]) > 0:
            it = v[v.index(arr[i])]
        else :
            it = v[n - 1]
              
        # If next element is also same, then there 
        # are multiple occurrences, so print it 
        if (it != v[n - 1] and 
                  v[v.index(it) + 1] == arr[i]) : 
            print(arr[i], end = " "
  
        # Otherwise print previous element 
        else :
            print(v[v.index(it) - 1], end = " "
  
# Driver Code
if __name__ == "__main__"
  
    arr = [ 6, 11, 7, 8, 20, 12 ]
    n = len(arr) 
    printPrevGreater(arr, n) 
  
# This code is contributed by Ryuga

chevron_right


Output:

-1 8 6 7 12 11

Time Complexity : O(n Log n)
Auxiliary Space : O(n)



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : Ryuga