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, the outer loop for elements of array arr1[] and an 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 that are less than or equal to the element of the first array.
4. Print the count for every index.

Implementation:

C++

// 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;
}

Java

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

Python3

# 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

C#

// 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.

Javascript

<script>

// JavaScript code for the above algorithm

// 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)
{
    // Run two loops to count
    // First loop to traverse the first array
    // Second loop to traverse the second array
    for (let i = 0; i < m; i++) {
        let count = 0;

        // Traverse through second array
        for (let j = 0; j < n; j++)
            if (arr2[j] <= arr1[i])
                count++;

        document.write(count + " ");
    }
}

// Driver program to test above
    let arr1 = [ 1, 2, 3, 4, 7, 9 ];
    let arr2 = [ 0, 1, 2, 1, 1, 4 ];
    let m = arr1.length;
    let n = arr2.length;
    countEleLessThanOrEqual(arr1, arr2, m, n);


// This code is contributed by Surbhi Tyagi.

</script>

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 the element of the first array.
5. The index of the largest element will give the count of elements. Print the count for every index.

C++

// 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;
}

Java

// 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);
    }
}

Python3

# 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,end=" ")

# 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

C#

// 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.

PHP

<?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.
?>

Javascript

<script>

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

    // 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.
       function binary_search(arr,l,h,x)
    {
        while (l <= h) {
            let mid = Math.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;
    }

    // Method 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
        arr2.sort(function(a,b){return a-b;});

        // for each element of 1st array
        for (let i = 0; i < m; i++) {
            // Last index of largest element
            // smaller than or equal to x
            let index = binary_search(
                arr2, 0, n - 1, arr1[i]);

            // Required count for the element arr1[i]
            document.write((index + 1) + " ");
        }
    }

    // Driver method
    let arr1=[1, 2, 3, 4, 7, 9 ];
    let arr2=[0, 1, 2, 1, 1, 4];
    countEleLessThanOrEqual(
            arr1, arr2, arr1.length,
            arr2.length);


// This code is contributed by patel2127

</script>

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

Another way of solving the problem is to sort the second array and use the upper_bound() inbuilt function for each value of first array.

Implementation:

C++

// 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 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++) {
        int x = upper_bound(arr2, arr2 + n, arr1[i]) - arr2;
        cout << x << " ";
    }
}

// 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;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// JAVA implementation of For each element in 1st
// array count elements less than or equal to it
// in 2nd array
import java.util.*;
class GFG
{
  public static int upper_bound(int arr[], int key)
  {
    int upperBound = 0;

    while (upperBound < arr.length)
    {

      // If current value is lesser than or equal to
      // key
      if (arr[upperBound] <= key)
        upperBound++;

      // This value is just greater than key
      else {
        return upperBound;
      }
    }
    return upperBound;
  }

  // function to count for each element in 1st array,
  // elements less than or equal to it in 2nd array
  public 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++) {
      int x = upper_bound(arr2, arr1[i]);
      System.out.print(x + " ");
    }
  }

  // Driver program to test above
  public static void main(String[] args)
  {
    int arr1[] = { 1, 2, 3, 4, 7, 9 };
    int arr2[] = { 0, 1, 2, 1, 1, 4 };
    int m = arr1.length;
    int n = arr2.length;
    countEleLessThanOrEqual(arr1, arr2, m, n);
  }
}

// This code is contributed by Taranpreet

C#

// C# implementation of For each element in 1st array count
// elements less than or equal to it in 2nd array

using System;
using System.Collections;

public class GFG {

    public static int upper_bound(int[] arr, int key)
    {
        int upperBound = 0;

        while (upperBound < arr.Length) {

            // If current value is lesser than or equal to
            // key
            if (arr[upperBound] <= key)
                upperBound++;

            // This value is just greater than key
            else {
                return upperBound;
            }
        }
        return upperBound;
    }

    // function to count for each element in 1st array,
    // elements less than or equal to it in 2nd array
    public 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++) {
            int x = upper_bound(arr2, arr1[i]);
            Console.Write(x + " ");
        }
    }

    static public void Main()
    {

        // Code
        int[] arr1 = { 1, 2, 3, 4, 7, 9 };
        int[] arr2 = { 0, 1, 2, 1, 1, 4 };
        int m = arr1.Length;
        int n = arr2.Length;
        countEleLessThanOrEqual(arr1, arr2, m, n);
    }
}

// This code is contributed by lokeshmvs21.

Output

4 5 5 6 6 6 

Time Complexity: O(n logn + m log n), where m and n represents the size of the given two arrays.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Another Approach: We can also use two pointers to find our solution.  first, sort both the arrays. after sorting put i pointer for arr1 and j  pointer for arr2 at the beginning. we will traverse over the elements of arr1 by using i pointer & inside this loop, we will go over each element of arr2 by j pointer. wherever arr2[j] <= arr1[i] we will increase j . if the condition fails print j. 

Note:- this will give answer in sorted manner

C++

// 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;

void countEleLessThanOrEqual(int arr1[], int arr2[], int m, int n)
{
    // sorting both the arrays
    sort(arr1, arr1 + m);
    sort(arr2, arr2 + n);

    // pointer for arr2
    int j = 0;

    // traversing  each element of 1st array
    for (int i = 0; i < m; i++)
    {
        // checking the condition for each element
        while (j < n && arr2[j] <= arr1[i])
            j++;

        cout << j << " ";
    }
}

// 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;
}
// This code is contributed by Naveen Shah

Java

// Java implementation of For each element in 1st
// array count elements less than or equal to it
// in 2nd array
import java.io.*;
import java.util.*;

class GFG
{
  static void countEleLessThanOrEqual(int[] arr1,
                                      int[] arr2, int m,
                                      int n)
  {
    Arrays.sort(arr1);
    Arrays.sort(arr2);

    // pointer for arr2
    int j = 0;

    // traversing  each element of 1st array
    for (int i = 0; i < m; i++)
    {

      // checking the condition for each element
      while (j < n && arr2[j] <= arr1[i])
        j++;

      System.out.print(j + " ");
    }
  }

  public static void main(String[] args)
  {
    int[] arr1 = { 1, 2, 3, 4, 7, 9 };
    int[] arr2 = { 0, 1, 2, 1, 1, 4 };
    int m = arr1.length;
    int n = arr2.length;

    countEleLessThanOrEqual(arr1, arr2, m, n);
  }
}

// This code is contributed by lokeshmvs21.

Output

4 5 5 6 6 6 

Time Complexity: O(nlogn + mlogm)
Auxiliary Space: O(1)

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