Related Articles

Related Articles

Find just strictly greater element from first array for each element in second array
  • Last Updated : 19 Nov, 2020

Given two arrays A[] and B[] containing N elements, the task is to find, for every element in the array B[], the element which is just strictly greater than that element which is present in the array A[]. If no value is present, then print ‘null’.

Note: The value from the array A[] can only be used once. 

Examples:  

Input: A[] = {0, 1, 2, 3, 4}, B[] = {0, 1, 1, 2, 3} 
Output: 1 2 3 4 null 
Explanation: 
On iterating every element in the array B[]: 
The value which is strictly greater than 0 and present in the array A[] is 1. 
Similarly, the value which is strictly greater than 1 and present in the array A[] is 2. 
Similarly, the value which is strictly greater than 1 and present in the array A[] is 3 because 2 has already been used for the previous 1. 
Similarly, the value which is strictly greater than 2 and present in the array A[] is 4. 
Now, there is no value in the array which is greater than 3 because 4 has already been used for the previous 2. So, null is printed. 

Input: A[] = {0, 1, 6, 4, 0, 2, 4, 2, 4, 7}, B[] = {0, 1, 6, 4, 0, 2, 4, 2, 4, 7} 
Output: 1 2 7 6 2 4 null 4 null null 



Approach: The idea is to use the Tree set Data structure. But since a tree set doesn’t support duplicate values, a hashmap is used to store the frequency of the elements.  

  • Iterate through the array A[].
  • Add the elements in the array A[] into the tree set.
  • Update their frequencies in the hashmap.
  • Now, for every element in the array B[], find the value which is strictly greater than the current value by using the higher() function of the tree set.
  • Now, reduce the frequency of this number in the hash map by 1.
  • Keep repeating the above two steps until the frequency of the numbers become 0. If it is 0, then all the occurrences of that number have been used up for the elements. So, remove that element from the tree set.

Below is the implementation of the above approach: 

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find the values
// strictly greater than the element
// and present in the array
#include<bits/stdc++.h>
using namespace std;
 
// Function to find the values
// strictly greater than the element
// and present in the array
void operations(int n, long long A[],
                       long long B[])
{
     
    // Treeset to store the
    // values of the array A
    set<long long>tree;
     
    // HashMap to store the frequencies
    // of the values in array A
    map<long long, int>freqMap;
 
    // Iterating through the array
    // and add values in the treeset
    for(int j = 0; j < n; j++)
    {
        long long x = A[j];
        tree.insert(x);
        freqMap[x]++;
    }
 
    // Finding the strictly greater value
    // in the array A[] using "higher()"
    // function and also reducing the
    // frequency of that value because it
    // has to be used only once
    for(int j = 0; j < n; j++)
    {
        long long  x = B[j];
 
        // If the higher value exists
        if (tree.upper_bound(x) != tree.end())
        {
            cout << *tree.upper_bound(x) << " ";
             
            // If the frequency value is 1
            // then remove it from treeset
            // because it has been used
            // and its frequency becomes 0
            if (freqMap[*tree.upper_bound(x)] == 1)
            {
                tree.erase(*tree.upper_bound(x));
            }
             
            // Else, reducing the frequency
            // by 1
            else
            {
                freqMap[*tree.upper_bound(x)]--;
            }
        }
 
        // If the value is not present
        // then print null
        else
        {
            cout << "null ";
        }
    }
}
 
// Driver code
int main()
{
    int n = 12;
    long long A[] = { 9, 5, 100, 4, 89, 2,
                      0, 2, 89, 77, 77, 77 };
    long long B[] = { 0, 18, 60, 34, 50, 29,
                      4, 20, 48, 77, 2, 8 };
     
    operations(n, A, B);
}
 
// This code is contributed by Stream_Cipher

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find the values
// strictly greater than the element
// and present in the array
 
import java.io.*;
import java.util.*;
public class GFG {
 
    // Function to find the values
    // strictly greater than the element
    // and present in the array
    public static void operations(
        int n, long A[], long B[])
    {
 
        // Treeset to store the
        // values of the array A
        TreeSet<Long> tree
            = new TreeSet<Long>();
 
        // HashMap to store the frequencies
        // of the values in array A
        HashMap<Long, Integer> freqMap
            = new HashMap<Long, Integer>();
 
        // Iterating through the array
        // and add values in the treeset
        for (int j = 0; j < n; j++) {
            long x = A[j];
            tree.add(x);
 
            // Updating the frequencies
            if (freqMap.containsKey(x)) {
 
                freqMap.put(x, freqMap.get(x) + 1);
            }
            else {
 
                freqMap.put(x, 1);
            }
        }
 
        // Finding the strictly greater value
        // in the array A[] using "higher()"
        // function and also reducing the
        // frequency of that value because it
        // has to be used only once
        for (int j = 0; j < n; j++) {
            long x = B[j];
 
            // If the higher value exists
            if (tree.higher(x) != null) {
                System.out.print(tree.higher(x) + " ");
 
                // If the frequency value is 1
                // then remove it from treeset
                // because it has been used
                // and its frequency becomes 0
                if (freqMap.get(tree.higher(x)) == 1) {
                    tree.remove(tree.higher(x));
                }
 
                // Else, reducing the frequency
                // by 1
                else {
                    freqMap.put(
                        tree.higher(x),
                        freqMap.get(tree.higher(x))
                            - 1);
                }
            }
 
            // If the value is not present
            // then print null
            else {
                System.out.print("null ");
            }
        }
    }
 
    // Driver code
    public static void main(String args[])
    {
 
        int n = 12;
        long A[] = new long[] { 9, 5, 100, 4,
                                89, 2, 0, 2,
                                89, 77, 77, 77 };
        long B[] = new long[] { 0, 18, 60, 34,
                                50, 29, 4, 20,
                                48, 77, 2, 8 };
 
        operations(n, A, B);
    }
}

chevron_right


Output: 

2 77 77 77 89 89 5 100 null null 4 9



 

Time Complexity: O(N * log(N)) because the insertion of one element takes log(N) in a tree set.
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up
Recommended Articles
Page :