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

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[].

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]

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

This problem is already discussed in the previous post. In this article, a more optimized linear time solution to the above problem is discussed. The approach discussed here works for arrays with values in small range. A range of values that can be used as index in an array.

The idea is to use an array to create a direct address table initially filled with zero, such that hash[i] gives the count of an element in the second array arr2[]. Now, calculate pre-sum of the hash array. Doing this, hash[i] will now give the count of elements less than or equal to in second array arr2[].

Now, traverse the first array and print hash[arr1[i]].

Below program illustrate the above approach:

C++

 // C++ program for each element in 1st  // array count elements less than or equal to it  // in 2nd array     #include   using namespace std;     #define MAX 100000     // Function for each element in 1st  // array count elements less than or equal to it  // in 2nd array  void countEleLessThanOrEqual(int arr1[], int m,                                  int arr2[], int n)  {      // Creating hash array initially      // filled with zero      int hash[MAX] = {0};             // Insert element of arr2[] to hash      // such that hash[i] will give count of      // element i in arr2[]      for (int i = 0; i < n; i++)          hash[arr2[i]]++;         // Presum of hash array      // such that hash[i] will give count of      // element less than or equals to i in arr2[]      for (int i=1; i

Java

 // Java program for each element  // in 1st array count elements  // less than or equal to it in  // 2nd array  import java.io.*;    class GFG  { static int MAX = 100000;    // Function for each element  // in 1st array count elements  // less than or equal to it  // in 2nd array  static void countEleLessThanOrEqual(int arr1[], int m,                                      int arr2[], int n) {      // Creating hash array initially     // filled with zero     int hash[] = new int[MAX];            // Insert element of arr2[] to     // hash such that hash[i] will     // give count of element i in arr2[]     for (int i = 0; i < n; i++)         hash[arr2[i]]++;        // Presum of hash array     // such that hash[i] will      // give count of element      // less than or equals to      // i in arr2[]     for(int i = 1; i < MAX; i++)     {         hash[i] = hash[i] +                    hash[i - 1];     }            // Traverse arr1[] and      // print hash[arr[i]]     for (int i = 0; i < m; i++)      {         System.out.print(hash[arr1[i]] + " ");     } }    // Driver code public static void main (String[] args)  {     int arr1[] = {1, 2, 3, 4, 7, 9};     int arr2[] = {0, 1, 2, 1, 1, 4};     int m, n;            m = arr1.length;     n = arr2.length;            countEleLessThanOrEqual(arr1, m, arr2, n); } }    // This code is contributed // by inder_verma

Python3

 # Python 3 program for each element in 1st  # array count elements less than or equal  # to it in 2nd array     MAX = 100000    # Function for each element in 1st  # array count elements less than or  # equal to it in 2nd array  def countEleLessThanOrEqual(arr1, m, arr2, n):            # Creating hash array initially      # filled with zero      hash = [0 for i in range(MAX)]             # Insert element of arr2[] to hash      # such that hash[i] will give count      # of element i in arr2[]      for i in range(n):         hash[arr2[i]] += 1        # Presum of hash array such that      # hash[i] will give count of element      # less than or equals to i in arr2[]      for i in range(1, MAX, 1):         hash[i] = hash[i] + hash[i - 1]             # Traverse arr1[] and print hash[arr[i]]      for i in range(m):         print(hash[arr1[i]], end = " ")         # Driver code  if __name__ == '__main__':     arr1 = [1, 2, 3, 4, 7, 9]      arr2 = [0, 1, 2, 1, 1, 4]      m = len(arr1)      n = len(arr2)            countEleLessThanOrEqual(arr1, m, arr2, n)         # This code is contributed by # Shashank_Sharma

C#

 // C# program for each element  // in 1st array count elements  // less than or equal to it in  // 2nd array  using System; public class GFG {     static int MAX = 100000;     // Function for each element  // in 1st array count elements  // less than or equal to it  // in 2nd array  static void countEleLessThanOrEqual(int []arr1, int m,                                      int []arr2, int n)  {      // Creating hash array initially      // filled with zero      int []hash = new int[MAX];             // Insert element of arr2[] to      // hash such that hash[i] will      // give count of element i in arr2[]      for (int i = 0; i < n; i++)          hash[arr2[i]]++;         // Presum of hash array      // such that hash[i] will      // give count of element      // less than or equals to      // i in arr2[]      for(int i = 1; i < MAX; i++)      {          hash[i] = hash[i] +                  hash[i - 1];      }             // Traverse arr1[] and      // print hash[arr[i]]      for (int i = 0; i < m; i++)      {          Console.Write(hash[arr1[i]] + " ");      }  }     // Driver code  public static void Main ()  {      int []arr1 = {1, 2, 3, 4, 7, 9};      int []arr2 = {0, 1, 2, 1, 1, 4};      int m, n;             m = arr1.Length;      n = arr2.Length;             countEleLessThanOrEqual(arr1, m, arr2, n);  }  }               // This code is contributed by Shikha Singh.

PHP



Output:

4 5 5 6 6 6

Another method:(Using vector and map)

 // C++ program for each element in 1st  // array count elements less than or equal to it  // in 2nd array  #include #include #include    // Function for each element in 1st  // array count elements less than or equal to it  // in 2nd array  void countLessThanOrEqual(const std::vector& vec1,                          const std::vector& vec2) {     std::map countOfVec2;     for (const auto& item : vec2) {         ++countOfVec2[item];     }        unsigned int prev = 0;     for (auto& pair : countOfVec2) {         pair.second += prev;         prev = pair.second;     }     // Traverse arr1[] and print result      for (const auto& item : vec1) {         unsigned int result = (--countOfVec2.upper_bound(item))->second;         std::cout << result << " ";     } }    // Driver code int main() {     std::vector arr1 = { 1, 2, 3, 4, 7, 9 };     std::vector arr2 = { 0, 1, 2, 1, 1, 4 };        countLessThanOrEqual(arr1, arr2);        return 0; }

Output:

4 5 5 6 6 6

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