Ceiling 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

Examples:

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

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

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 greater element. In C++ upper_bound() serves this purpose.

Below is the implementation of above approach.

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) {     if (n == 1) {         cout << "-1";         return;     }        // Create a sorted copy of arr[]     vector 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++) {            // Find the first element that is greater than         // the given element         auto it = upper_bound(v.begin(), v.end(), arr[i]);            // Since arr[i] also exists in array, *(it-1)         // will be same as arr[i]. Let us check *(it-2)         // is also same as arr[i]. If true, then arr[i]         // exists twice in array, so ceiling is same         // same as arr[i]         if ((it - 1) != v.begin() && *(it - 2) == arr[i]) {                // If next element is also same, then there             // are multiple occurrences, so print it             cout << arr[i] << " ";         }            else if (it != v.end())             cout << *it << " ";         else             cout << -1 << " ";     } }    /* Driver program to test insertion sort */ int main() {     int arr[] = {10, 5, 11, 10, 20, 12};     int n = sizeof(arr) / sizeof(arr);     printPrevGreater(arr, n);     return 0; }

Java

 // Java implementation of efficient algorithm to find // floor of every element import java.util.Arrays;    class GFG  {        // Prints greater elements on left side of every element     static void printPrevGreater(int arr[], int n)      {         if (n == 1)          {             System.out.println("-1");             return;         }            // Create a sorted copy of arr[]         int v[] = Arrays.copyOf(arr, arr.length);         Arrays.sort(v);            // Traverse through arr[] and do binary search for         // every element.         for (int i = 0; i < n; i++)          {                // Find the first element that is greater than             // the given element             int it = Arrays.binarySearch(v,arr[i]);             it++;                // Since arr[i] also exists in array, *(it-1)             // will be same as arr[i]. Let us check *(it-2)             // is also same as arr[i]. If true, then arr[i]             // exists twice in array, so ceiling is same             // same as arr[i]             if ((it - 1) != 0 && v[it - 2] == arr[i])             {                    // If next element is also same, then there                 // are multiple occurrences, so print it                 System.out.print(arr[i] + " ");             }             else if (it != v.length)                 System.out.print(v[it] + " ");             else                 System.out.print(-1 + " ");         }     }        // Driver code     public static void main(String[] args)      {         int arr[] = {10, 5, 11, 10, 20, 12};         int n = arr.length;         printPrevGreater(arr, n);     } }    // This code is contributed by // Rajnis09

Python3

 # Python implementation of efficient algorithm # to find floor of every element import bisect    # Prints greater elements on left side of every element def printPrevGreater(arr, n):        if n == 1:         print("-1")         return            # Create a sorted copy of arr[]     v = list(arr)     v.sort()        # Traverse through arr[] and do binary search for     # every element     for i in range(n):            # Find the location of first element that         # is greater than the given element         it = bisect.bisect_right(v, arr[i])            # Since arr[i] also exists in array, v[it-1]         # will be same as arr[i]. Let us check v[it-2]         # is also same as arr[i]. If true, then arr[i]         # exists twice in array, so ceiling is same         # same as arr[i]         if (it-1) != 0 and v[it-2] == arr[i]:                # If next element is also same, then there             # are multiple occurrences, so print it             print(arr[i], end=" ")                    elif it <= n-1:             print(v[it], end=" ")                    else:             print(-1, end=" ")       # Driver code if __name__ == "__main__":     arr = [10, 5, 11, 10, 20, 12]     n = len(arr)     printPrevGreater(arr, n)    # This code is contributed by  # sanjeev2552

C#

 // C# implementation of efficient algorithm  // to find floor of every element using System;        class GFG  {        // Prints greater elements on left side     // of every element     static void printPrevGreater(int []arr, int n)      {         if (n == 1)          {             Console.Write("-1");             return;         }            // Create a sorted copy of arr[]         int []v = new int[arr.GetLength(0)];         Array.Copy(arr, v, arr.GetLength(0));         Array.Sort(v);            // Traverse through arr[] and          // do binary search for every element.         for (int i = 0; i < n; i++)          {                // Find the first element that is             // greater than the given element             int it = Array.BinarySearch(v, arr[i]);             it++;                // Since arr[i] also exists in array, *(it-1)             // will be same as arr[i]. Let us check *(it-2)             // is also same as arr[i]. If true, then arr[i]             // exists twice in array, so ceiling is same             // same as arr[i]             if ((it - 1) != 0 && v[it - 2] == arr[i])             {                    // If next element is also same, then there                 // are multiple occurrences, so print it                 Console.Write(arr[i] + " ");             }             else if (it != v.Length)                 Console.Write(v[it] + " ");             else                 Console.Write(-1 + " ");         }     }        // Driver code     public static void Main(String[] args)      {         int []arr = {10, 5, 11, 10, 20, 12};         int n = arr.Length;         printPrevGreater(arr, n);     } }    // This code is contributed by 29AjayKumar

Output:

10 10 12 10 -1 20

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

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