# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of efficient algorithm to find ` `// floor of every element ` `#include ` `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) { ` `            ``(arr[i] == v) ? 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); ` `    ``printPrevGreater(arr, n); ` `    ``return` `0; ` `} `

## Python3

 `# 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 `

Output:

```-1 8 6 7 12 11
```

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

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