Find just strictly greater element from first array for each element in second array - GeeksforGeeks

15 Feb, 2021
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++
Java
Python3
// 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 arrayvoid 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 codeint 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
// 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);    }}
# Python program to find the values# strictly greater than the element# and present in the arrayfrom typing import Listfrom bisect import bisect_right # Function to find the values# strictly greater than the element# and present in the arraydef operations(n: int, A: List[int], B: List[int]) -> None:     # Treeset to store the    # values of the array A    tree = set()     # HashMap to store the frequencies    # of the values in array A    freqMap = dict()     # Iterating through the array    # and add values in the treeset    for j in range(n):        x = A[j]        tree.add(x)        if x not in freqMap:            freqMap[x] = 0        freqMap[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 j in range(n):        x = B[j]         # If the higher value exists        sset = sorted(list(tree))        index = bisect_right(sset, x)        if index < len(tree):            print(sset[index], end=" ")             # If the frequency value is 1            # then remove it from treeset            # because it has been used            # and its frequency becomes 0            if (freqMap[sset[index]] == 1):                tree.remove(sset[index])             # Else, reducing the frequency            # by 1            else:                freqMap[sset[index]] -= 1         # If the value is not present        # then print null        else:            print("null", end=" ") # Driver codeif __name__ == "__main__":     n = 12    A = [9, 5, 100, 4, 89, 2, 0, 2, 89, 77, 77, 77]    B = [0, 18, 60, 34, 50, 29, 4, 20, 48, 77, 2, 8]     operations(n, A, B) # This code is contributed by sanjeev2552
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. 
Stream_Cipher
sanjeev2552
frequency-counting
Java-HashMap
java-treeset
Advanced Data Structure
Arrays
Hash
Tree
Arrays
Hash
Tree
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