给定两个包含N 个元素的数组A[]和B[] ，任务是为数组B[]中的每个元素找到严格大于数组A[] 中存在的元素的元素。如果不存在值，则打印“null”。
注意：数组 A[] 中的值只能使用一次。
例子：
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
做法：思路是使用树集数据结构。但是由于树集不支持重复值，因此使用哈希图来存储元素的频率。
- 遍历数组 A[]。
- 将数组 A[] 中的元素添加到树集中。
- 更新它们在哈希图中的频率。
- 现在，对于数组 B[] 中的每个元素，使用树集的high()函数找到严格大于当前值的值。
- 现在，将这个数字在哈希图中的频率减少 1。
- 不断重复以上两步，直到数字出现的频率为0。如果为0，则表示该数字出现的所有元素都已用完。因此，从树集中删除该元素。
下面是上述方法的实现：
C++
// C++ program to find the values
// strictly greater than the element
// and present in the array
#include
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
settree;
// HashMap to store the frequencies
// of the values in array A
mapfreqMap;
// 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 Java
// 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 tree
= new TreeSet();
// HashMap to store the frequencies
// of the values in array A
HashMap freqMap
= new HashMap();
// 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);
}
} Python3
# Python program to find the values
# strictly greater than the element
# and present in the array
from typing import List
from bisect import bisect_right
# Function to find the values
# strictly greater than the element
# and present in the array
def 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 code
if __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输出：
2 77 77 77 89 89 5 100 null null 4 9时间复杂度： O(N * log(N))因为在树集中插入一个元素需要 log(N)。
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