Floor of every element in same array

Last Updated : 21 Sep, 2023

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)

Algorithm:

Create a vector to store the result.
Loop through every element of the array from i = 0 to n-1.

a. Initialize the variable ‘closest’ as INT_MIN.

b. Loop through all elements of the array from j = 0 to n-1

i. If i and j are the same, continue to the next iteration of the loop

ii. If arr[j] is smaller than or equal to arr[i], update the variable closest with maximum of closest and arr[j]

c. If closest is still INT_MIN, push -1 to the result vector else push closest

3. Return the result vector
4. In the main function:

Create an array of integers arr[] of size n
Initialize n as the size of the array arr[]
Call the closestSmallerOrSame function and store the result in a vector called ‘result’
Loop through the result vector and print the elements

Below is the implementation of the approach:

C++

   // C++ program to find the closest smaller or same element
// for every element
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find closest smaller or same element for
// every element of array
vector<int> closestSmallerOrSame(int arr[], int n) {
    // Vector to store result
    vector<int> res;
 
    // Loop through every element of the array
    for (int i = 0; i < n; i++) {
          int closest = INT_MIN;
        // Loop through all elements to find closest
        // smaller or same element
        for (int j = 0; j < n; j++) {
            // if same leave it and continue
              if(i == j)
              continue;
           
            // If a smaller or same element is found, update
              // the closest variable as maximum
            if (arr[j] <= arr[i]) {
                closest = max(closest, arr[j]);
            }
        }
         
          // If no smaller or same element is found, add -1 to the result vector
          if(    closest == INT_MIN)
              res.push_back(-1);
          else // push the closest element to res for ith one
            res.push_back(closest);
    }
 
    // Return the result vector
    return res;
}
 
// Driver code
int main()
{
    // Sample input
    int arr[] = { 6, 11, 7, 8, 20, 12 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Find closest smaller or same element for every
    // element of the array
    vector<int> result = closestSmallerOrSame(arr, n);
 
    // Print the result
    for (int i = 0; i < result.size(); i++)
        cout << result[i] << " ";
    cout << endl;
 
    return 0;
}
 
  

Java

   // Java program to find the closest smaller or same element
// for every element
import java.util.*;
 
public class GFG
{
 
  // Function to find closest smaller or same element for
  // every element of array
  public static List<Integer> closestSmallerOrSame(int[] arr, int n) 
  {
 
    // List to store result
    List<Integer> res = new ArrayList<>();
 
    // Loop through every element of the array
    for (int i = 0; i < n; i++) {
      int closest = Integer.MIN_VALUE;
      // Loop through all elements to find closest
      // smaller or same element
      for (int j = 0; j < n; j++) {
        // if same leave it and continue
        if(i == j)
          continue;
 
        // If a smaller or same element is found, update
        // the closest variable as maximum
        if (arr[j] <= arr[i]) {
          closest = Math.max(closest, arr[j]);
        }
      }
 
      // If no smaller or same element is found, add -1 to the result list
      if( closest == Integer.MIN_VALUE)
        res.add(-1);
      else // push the closest element to res for ith one
        res.add(closest);
    }
 
    // Return the result list
    return res;
  }
 
  // Driver code
  public static void main(String[] args) {
    // Sample input
    int[] arr = { 6, 11, 7, 8, 20, 12 };
    int n = arr.length;
 
    // Find closest smaller or same element for every
    // element of the array
    List<Integer> result = closestSmallerOrSame(arr, n);
 
    // Print the result
    for (int i = 0; i < result.size(); i++)
      System.out.print(result.get(i) + " ");
    System.out.println();
  }
}
 
  

Python

   def closest_smaller_or_same(arr):
    n = len(arr)
    res = []
 
    # Loop through every element of the array
    for i in range(n):
        closest = float('-inf')
         
        # Loop through all elements to find closest smaller or same element
        for j in range(n):
            if i == j:  # if same, continue
                continue
 
            # If a smaller or same element is found, update the closest variable
            if arr[j] <= arr[i]:
                closest = max(closest, arr[j])
 
        # If no smaller or same element is found, append -1 to the result list
        if closest == float('-inf'):
            res.append(-1)
        else:  # append the closest element to res for the current one
            res.append(closest)
 
    # Return the result list
    return res
 
# Sample input
arr = [6, 11, 7, 8, 20, 12]
 
# Find closest smaller or same element for every element of the array
result = closest_smaller_or_same(arr)
 
# Print the result
print ' '.join(map(str, result))
 
     #This Code Is Contributed By Shubham Tiwari
 
  

C#

   using System;
using System.Collections.Generic;
 
public class Program {
    // Function to find the closest smaller or same element
    // for every element of the array
    public static List<int> ClosestSmallerOrSame(int[] arr,
                                                 int n)
    { // List to store result
        List<int> res = new List<int>();
        // Loop through every element of the array
        for (int i = 0; i < n; i++) {
            int closest = int.MinValue;
            // Loop through all elements to find closest
            // smaller or same element
            for (int j = 0; j < n;
                 j++) { // if same leave it and continue
                if (i == j)
                    continue;
                // If a smaller or same element is found,
                // update the closest variable as maximum
                if (arr[j] <= arr[i]) {
                    closest = Math.Max(closest, arr[j]);
                }
            }
            // If no smaller or same element is found, add
            // -1 to the result list
            if (closest == int.MinValue)
                res.Add(-1);
            else // push the closest element to res for ith
                 // one
                res.Add(closest);
        }
 
        return res;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 6, 11, 7, 8, 20, 12 };
        int n = arr.Length;
        // Find closest smaller or same element for every
        // element of the array
        List<int> result = ClosestSmallerOrSame(arr, n);
 
        foreach(var item in result)
        {
            Console.Write(item + " ");
        }
 
        Console.WriteLine();
 
        // This Code Is Contributed By Shubham Tiwari.
    }
}
 
  

Javascript

   // Function to find closest smaller or same element for every 
// element of array
function closestSmallerOrSame(arr) {
    const n = arr.length;
    const res = [];
 
    // Loop through every element of the array
    for (let i = 0; i < n; i++) {
        let closest = Number.MIN_SAFE_INTEGER;
 
        // Loop through all elements to find closest smaller or 
        // same element
        for (let j = 0; j < n; j++) {
            // If same element, skip and continue to the next element
            if (i === j) {
                continue;
            }
 
            // If a smaller or same element is found, update the 
            // closest variable
            if (arr[j] <= arr[i]) {
                closest = Math.max(closest, arr[j]);
            }
        }
 
        // If no smaller or same element is found, add -1 to the 
        // result vector
        if (closest === Number.MIN_SAFE_INTEGER) {
            res.push(-1);
        } else {
            // Push the closest element to res for the ith element
            res.push(closest);
        }
    }
 
    // Return the result vector
    return res;
}
 
// Driver code
const arr = [6, 11, 7, 8, 20, 12];
const result = closestSmallerOrSame(arr);
 
// Print the result
console.log(result.join(" "));
 
  

Output

-1 8 6 7 12 11

Time Complexity: O(N*N) as two nested loops are executing. Here, N is size of the input array.

Space Complexity: O(1) as no extra space has been used. Note here res vector space is ignored as it is the resultnt vector.

A better solution is to sort the array and create a sorted copy, then do a 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 the next of it is also the 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.

Implementation:

C++

   // 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;
}
 
  

Java

   // Java implementation of efficient algorithm to find floor
// of every element
import java.io.*;
import java.util.*;
 
class GFG {
 
  // Function to count the occurences of a target number.
  static int count(int[] arr, int target)
  {
    int count = 0;
    for (int i = 0; i < arr.length; i++) {
      if (arr[i] == target) {
        count++;
      }
    }
    return count;
  }
 
  // Function to find index of an element
  static int index(int[] arr, int target)
  {
    int index = -1;
    for (int i = 0; i < arr.length; i++) {
      if (arr[i] == target) {
        return i;
      }
    }
    return index;
  }
 
  // Prints greater elements on left
  // side of every element
  static void printPrevGreater(int[] arr, int n)
  {
    // Create a sorted copy of arr
    int[] v = new int[n];
    for (int i = 0; i < n; i++) {
      v[i] = arr[i];
    }
 
    Arrays.sort(v);
    int it = 0;
 
    // 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]) {
        System.out.print(
          ((arr[i] == v[1]) ? arr[i] : -1) + " ");
        continue;
      }
 
      // Find the first element that is greater
      // than or or equal to given element
      if (count(arr, arr[i]) > 0) {
        it = v[index(v, 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]
          && v[index(v, it) + 1] == arr[i]) {
        System.out.print(arr[i] + " ");
      }
      // Otherwise print previous element
      else {
        System.out.print(v[index(v, it) - 1] + " ");
      }
    }
  }
 
  public static void main(String[] args)
  {
    int[] arr = { 6, 11, 7, 8, 20, 12 };
    int n = arr.length;
 
    printPrevGreater(arr, n);
  }
}
 
// This code is contributed by lokeshmvs21.
 
  

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
 
  

C#

   // C# implementation of efficient algorithm to find floor
// of every element
using System;
using System.Collections;
 
public class GFG {
 
  // Function to count the occurences of a target number.
  static int count(int[] arr, int target)
  {
    int count = 0;
    for (int i = 0; i < arr.Length; i++) {
      if (arr[i] == target) {
        count++;
      }
    }
    return count;
  }
 
  // Function to find index of an element
  static int index(int[] arr, int target)
  {
    int index = -1;
    for (int i = 0; i < arr.Length; i++) {
      if (arr[i] == target) {
        return i;
      }
    }
    return index;
  }
 
  // Prints greater elements on left
  // side of every element
  static void printPrevGreater(int[] arr, int n)
  {
    // Create a sorted copy of arr
    int[] v = new int[n];
    for (int i = 0; i < n; i++) {
      v[i] = arr[i];
    }
 
    Array.Sort(v);
    int it = 0;
 
    // 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]) {
        Console.Write(
          ((arr[i] == v[1]) ? arr[i] : -1) + " ");
        continue;
      }
 
      // Find the first element that is greater
      // than or equal to given element
      if (count(arr, arr[i]) > 0) {
        it = v[index(v, 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]
          && v[index(v, it) + 1] == arr[i]) {
        Console.Write(arr[i] + " ");
      }
      // Otherwise print previous element
      else {
        Console.Write(v[index(v, it) - 1] + " ");
      }
    }
  }
 
  static public void Main()
  {
 
    // Code
    int[] arr = { 6, 11, 7, 8, 20, 12 };
    int n = arr.Length;
 
    printPrevGreater(arr, n);
  }
}
 
// This code is contributed by lokeshmvs21.
 
  

Javascript

   <script>
 
// JavaScript implementation of efficient algorithm to find
// floor of every element
 
// Prints greater elements on left side of every element
function printPrevGreater(arr, n)
{
    // Create a sorted copy of arr[]
    let v = [...arr]
    v.sort((a, b) => a - b);
 
    // Traverse through arr[] and do binary search for
    // every element.
    for (let 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]) ? 
            document.write(arr[i]) : document.write(-1);
            document.write(" ");
            continue;
        }
 
        // Find the first element that is greater than or
        // or equal to given element
        if (v.includes(arr[i]))
            it = v[v.indexOf(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] && (v[v.indexOf(it) + 1] == arr[i]))
            document.write(arr[i] + " ");
 
        // Otherwise print previous element
        else
            document.write(v[v.indexOf(it) - 1] + " ");
    }
}
 
function lower_bound(arr, val){
 
     
}
 
/* Driver program to test insertion sort */
 
    let arr = [ 6, 11, 7, 8, 20, 12 ];
    let n = arr.length;
    printPrevGreater(arr, n);
 
 
// This code is contributed by _saurabh_jaiswal
 
</script>
 
  

Output

-1 8 6 7 12 11

Complexity Analysis:

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

Ceiling of every element in same array

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