For each element in 1st array count elements less than or equal to it in 2nd array

Given two unsorted arrays arr1[] and arr2[]. They may contain duplicates. For each element in arr1[] count elements less than or equal to it in array arr2[].

Source: Amazon Interview Experience | Set 354 (For SDE-2)

Examples:



Input : arr1[] = [1, 2, 3, 4, 7, 9]
        arr2[] = [0, 1, 2, 1, 1, 4]
Output : [4, 5, 5, 6, 6, 6]

Input : arr1[] = [5, 10, 2, 6, 1, 8, 6, 12]
        arr2[] = [6, 5, 11, 4, 2, 3, 7]
Output : [4, 6, 1, 5, 0, 6, 5, 7]

Naive Approach: Using two loops, outer loop for elements of array arr1[] and inner loop for elements of array arr2[]. Then for each element of arr1[], count elements less than or equal to it in arr2[].
Time complexity: O(m * n), considering arr1[] and arr2[] are of sizes m and n respectively.

Efficient Approach: Sort the elements of 2nd array, i.e., array arr2[]. Then perform a modified binary search on array arr2[]. For each element x of array arr1[], find the last index of the largest element smaller than or equal to x in sorted array arr2[].

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of For each element in 1st 
// array count elements less than or equal to it
// in 2nd array
#include <bits/stdc++.h>
  
using namespace std;
  
// function returns the index of largest element 
// smaller than equal to 'x' in 'arr'. For duplicates
// it returns the last index of occurrence of required
// element. If no such element exits then it returns -1. 
int binary_search(int arr[], int l, int h, int x)
{
    while (l <= h)
    {
        int mid = (l+h) / 2;
  
        // if 'x' is greater than or equal to arr[mid], 
        // then search in arr[mid+1...h]
        if (arr[mid] <= x)
            l = mid + 1;
  
        // else search in arr[l...mid-1]    
        else
            h = mid - 1;    
    }
      
    // required index
    return h;
}
  
// function to count for each element in 1st array,
// elements less than or equal to it in 2nd array
void countEleLessThanOrEqual(int arr1[], int arr2[], 
                             int m, int n)
{
    // sort the 2nd array
    sort(arr2, arr2+n);
      
    // for each element of 1st array
    for (int i=0; i<m; i++)
    {
        // last index of largest element 
        // smaller than or equal to x
        int index = binary_search(arr2, 0, n-1, arr1[i]);
          
        // required count for the element arr1[i]
        cout << (index+1) << " ";
    }
}
  
// Driver program to test above
int main()
{
    int arr1[] = {1, 2, 3, 4, 7, 9};
    int arr2[] = {0, 1, 2, 1, 1, 4};
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
    countEleLessThanOrEqual(arr1, arr2, m, n);
    return 0;

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of For each element in 1st 
// array count elements less than or equal to it
// in 2nd array
  
import java.util.Arrays;
  
class GFG
{
    // method returns the index of largest element 
    // smaller than equal to 'x' in 'arr'. For duplicates
    // it returns the last index of occurrence of required
    // element. If no such element exits then it returns -1. 
    static int binary_search(int arr[], int l, int h, int x)
    {
        while (l <= h)
        {
            int mid = (l+h) / 2;
       
            // if 'x' is greater than or equal to arr[mid], 
            // then search in arr[mid+1...h]
            if (arr[mid] <= x)
                l = mid + 1;
       
            // else search in arr[l...mid-1]    
            else
                h = mid - 1;    
        }
           
        // required index
        return h;
    }
       
    // method to count for each element in 1st array,
    // elements less than or equal to it in 2nd array
    static void countEleLessThanOrEqual(int arr1[], int arr2[], 
                                 int m, int n)
    {
        // sort the 2nd array
        Arrays.sort(arr2);
           
        // for each element of 1st array
        for (int i=0; i<m; i++)
        {
            // last index of largest element 
            // smaller than or equal to x
            int index = binary_search(arr2, 0, n-1, arr1[i]);
               
            // required count for the element arr1[i]
            System.out.print((index+1) + " ");
        }
    }
  
    // Driver method 
    public static void main(String[] args)
    {
        int arr1[] = {1, 2, 3, 4, 7, 9};
        int arr2[] = {0, 1, 2, 1, 1, 4};
          
        countEleLessThanOrEqual(arr1, arr2, arr1.length, arr2.length);
    }
}

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# python implementation of For each element in 1st 
# array count elements less than or equal to it
# in 2nd array
  
# function returns the index of largest element 
# smaller than equal to 'x' in 'arr'. For duplicates
# it returns the last index of occurrence of required
# element. If no such element exits then it returns -1
def bin_search(arr, n, x):
      
l = 0
h = n - 1
while(l <= h):
    mid = int((l + h) / 2)
    # if 'x' is greater than or equal to arr[mid], 
    # then search in arr[mid + 1...h]
    if(arr[mid] <= x):
    l = mid + 1;
    else:
    # else search in arr[l...mid-1]
    h = mid - 1
# required index
return h
  
# function to count for each element in 1st array,
# elements less than or equal to it in 2nd array
def countElements(arr1, arr2, m, n):
# sort the 2nd array
arr2.sort()
  
# for each element in first array
for i in range(m):
    # last index of largest element 
    # smaller than or equal to x
    index = bin_search(arr2, n, arr1[i])
    # required count for the element arr1[i]
    print(index + 1)
  
# driver program to test above function
arr1 = [1, 2, 3, 4, 7, 9]
arr2 = [0, 1, 2, 1, 1, 4]
m = len(arr1)
n = len(arr2)
countElements(arr1, arr2, m, n)
  
#This code is contributed by Aditi Sharma

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of For each element
// in 1st array count elements less than
// or equal to it in 2nd array
using System;
          
public class GFG 
{
      
    // method returns the index of
    // largest element smaller than
    // equal to 'x' in 'arr'. For 
    // duplicates it returns the last
    // index of occurrence of required
    // element. If no such element
    // exits then it returns -1. 
    static int binary_search(int []arr,
                     int l, int h, int x)
    {
        while (l <= h)
        {
            int mid = (l+h) / 2;
      
            // if 'x' is greater than or
            // equal to arr[mid], then 
            // search in arr[mid+1...h]
            if (arr[mid] <= x)
                l = mid + 1;
      
            // else search in
            // arr[l...mid-1] 
            else
                h = mid - 1; 
        }
          
        // required index
        return h;
    }
      
    // method to count for each element
    // in 1st array, elements less than
    // or equal to it in 2nd array
    static void countEleLessThanOrEqual(
                 int []arr1, int []arr2,
                           int m, int n)
    {
          
        // sort the 2nd array
        Array.Sort(arr2);
          
        // for each element of 1st array
        for (int i=0; i<m; i++)
        {
            // last index of largest
            // element smaller than or
            // equal to x
            int index = binary_search(
                   arr2, 0, n-1, arr1[i]);
              
            // required count for the
            // element arr1[i]
            Console.Write((index+1) + " ");
        }
    }
  
    // Driver method 
    public static void Main()
    {
        int []arr1 = {1, 2, 3, 4, 7, 9};
        int []arr2= {0, 1, 2, 1, 1, 4};
          
        countEleLessThanOrEqual(arr1,
           arr2, arr1.Length, arr2.Length);
    }
}
  
// This code is contributed by Sam007.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// php implementation of For each element
// in 1st array count elements less than 
// or equal to it in 2nd array
  
// function returns the index of largest
// element smaller than equal to 'x' in 
// 'arr'. For duplicates it returns the 
// last index of occurrence of required
// element. If no such element exits then
// it returns -1. 
function binary_search($arr, $l, $h, $x)
{
    while ($l <= $h)
    {
        $mid = (floor($l+$h) / 2);
  
        // if 'x' is greater than or 
        // equal to arr[mid], then 
        // search in arr[mid+1...h]
        if ($arr[$mid] <= $x)
            $l = $mid + 1;
  
        // else search in arr[l...mid-1] 
        else
            $h = $mid - 1; 
    }
      
    // required index
    return $h;
}
  
// function to count for each element
// in 1st array, elements less than or
// equal to it in 2nd array
function countEleLessThanOrEqual($arr1
                           $arr2, $m,$n)
{
    // sort the 2nd array
    sort($arr2); sort($arr2, $n);
      
    // for each element of 1st array
    for ($i = 0; $i < $m; $i++)
    {
        // last index of largest element 
        // smaller than or equal to x
        $index = binary_search($arr2, 0, 
                        $n-1, $arr1[$i]);
          
        // required count for the 
        // element arr1[i]
        echo ($index+1) , " ";
    }
}
  
// Driver program to test above
$arr1 = array(1, 2, 3, 4, 7, 9);
$arr2 = array(0, 1, 2, 1, 1, 4);
$m = sizeof($arr1) / sizeof($arr1[0]);
$n = sizeof($arr2) / sizeof($arr2[0]);
countEleLessThanOrEqual($arr1, $arr2, $m, $n);
   
//This code is contributed by nitin mittal.
?>

chevron_right



Output:

4 5 5 6 6 6

Time Complexity: O(mlogn + nlogn), considering arr1[] and arr2[] are of sizes m and n respectively.
This article is contributed by Ayush Jauhari. 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 write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : Sam007, nitin mittal