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]
So the number of elements less than or equal to 
1 is 4, 2 is 5, 3 is 5, 4 is 6, 7 is 6 and 9 is 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]
So the number of elements less than or equal to 
5 is 4, 10 is 6, 2 is 1, 6 is 5, 1 is 0, 8 is 6, 
6 is 5 and 12 is 7 

Naive Approach:

Approach: The idea is to use 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[].



Algorithm :

  1. Traverse through the elements of the first array from start to end.
  2. For every element in the first array.
  3. Traverse through the elements in the second array and find the count of elements which are less than or equal to the element of the first array.
  4. Print the count for every index.

C++

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// C++ code for the above algorithm
#include <bits/stdc++.h>
using namespace std;
  
// 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)
{
    // Run two loops to count
    // First loop to traverse the first array
    // Second loop to traverse the second array
    for (int i = 0; i < m; i++) {
        int count = 0;
  
        // Traverse through second array
        for (int j = 0; j < n; j++)
            if (arr2[j] <= arr1[i])
                count++;
  
        cout << count << " ";
    }
}
  
// 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;
}

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Java

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import java.util.Arrays;
  
class GFG {
  
    // function to count for each
    // element in 1st array,
    // elements less than or equal
    // to it in 2nd array
    static int countEleLessThanOrEqual(
        int arr1[], int arr2[],
        int m, int n)
    {
        // Run two loops to count
        // First loop to traverse the first array
        // Second loop to traverse the second array
        for (int i = 0; i < m; i++) {
            int count = 0;
  
            // Traverse through second array
            for (int j = 0; j < n; j++)
                if (arr2[j] <= arr1[i])
                    count++;
            System.out.print(count + " ");
        }
        return m;
    }
  
    // 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);
    }
}
  
// This code is contributed by shivanisinghss2110

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Python3

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# Python3 code for the above algorithm
  
# function to count for each element in 1st array,
# elements less than or equal to it in 2nd array
def countEleLessThanOrEqual(arr1, arr2, m, n):
      
    # Run two loops to count 
    # First loop to traverse the first array
    # Second loop to traverse the second array
    for i in range(m):
          
        count = 0
        # Traverse through second array
        for j in range(n):
            if (arr2[j] <= arr1[i]):
                count+= 1
              
        print(count, end =" ")
  
# Driver program to test above
arr1 = [1, 2, 3, 4, 7, 9]
arr2 = [0, 1, 2, 1, 1, 4]
m = len(arr1)
n = len(arr2)
countEleLessThanOrEqual(arr1, arr2, m, n)
   
# This code is contributed by shubhamsingh10

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C#

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// C# implementation of For each element
using System;
  
class GFG {
  
    // function 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)
    {
        // Run two loops to count
        // First loop to traverse the first array
        // Second loop to traverse the second array
        for (int i = 0; i < m; i++) {
            int count = 0;
  
            // Traverse through second array
            for (int j = 0; j < n; j++)
                if (arr2[j] <= arr1[i])
                    count++;
            Console.Write((count) + " ");
        }
    }
  
    // 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 mohit kumar 29.

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Output:

4 5 5 6 6 6 

Complexity Analysis:

  • Time complexity: O(m * n).
    Considering arr1[] and arr2[] are of sizes m and n respectively.
  • Space Complexity: O(1).
    As no extra space is required

Efficient Solution:
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[]. The index of the largest element will give the count of elements.

Algorithm:

  1. Sort the second array.
  2. Traverse through the elements of the first array from start to end.
  3. For every element in the first array.
  4. Do a binary search on the second array and find the index of the largest element smaller than or equal to element of first array.
  5. The index of the largest element will give the count of elements. Print the count for every index.

C++

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// 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;
}

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Java

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// 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);
    }
}

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Python

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# 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

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C#

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// 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.

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PHP

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<?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.
?>

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Output:

4 5 5 6 6 6

Complexity Analysis:

  • Time Complexity: O(mlogn + nlogn).
    Considering arr1[] and arr2[] of sizes m and n respectively.
  • Space Complexity: O(1).
    As no extra space is required

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.

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