每个数组元素中另一个数组中最接近的较大元素

给定两个数组a []和b []，我们需要构建一个数组c []，使得c []的每个元素c [i]包含a []中的一个值，该值大于b [i]并且最接近到b [i]。如果a []不大于b [i]，则c [i]的值为-1。所有阵列的大小均相同。

例子：

Input : a[] = [ 2, 6, 5, 7, 0]  
        b[] = [1, 3, 2, 5, 8]
Output : c[] = [2, 5, 5, 7, -1]

Input : a[] = [ 2, 6, 5, 7, 0]   
        b[] = [0, 2, 3, 5, 1]
Output : c[] = [2, 5, 5, 6, 2]

天真的方法：对于b []中的每个元素，我们遍历整个a []并尝试找到最接近的更大元素，并保存每次搜索的结果。这将花费O(n ^ 2)的时间复杂度。

高效的方法：对数组a []进行排序，并对每个b [i]在排序后的数组a []中应用二进制搜索。对于这种方法，我们的时间复杂度将为O(nlogn)。
注意：对于最接近的较大元素，我们可以使用upper_bound()。

下面是实现。

C++

// CPP to find result from target array
// for closest element
#include 
using namespace std;
  
// Function for printing resultant array
void closestResult(int a[], int b[], int n)
{
    // change arr[] to vector
    vector vect(a, a + n);
  
    // sort vector for ease
    sort(vect.begin(), vect.end());
  
    // iterator for upper_bound
    vector::iterator up;
  
    // vector for result
    vector c;
  
    // calculate resultant array
    for (int i = 0; i < n; i++) {
  
        // check upper bound element
        up = upper_bound(vect.begin(), vect.end(), b[i]);
  
        // if no element found push -1
        if (up == vect.end())
            c.push_back(-1);
  
        // Else push the element
        else
            c.push_back(*up); // add to resultant
    }
  
    cout << "Result = ";
    for (auto it = c.begin(); it != c.end(); it++)
        cout << *it << " ";
}
  
// driver program
int main()
{
    int a[] = { 2, 5, 6, 1, 8, 9 };
    int b[] = { 2, 1, 0, 5, 4, 9 };
    int n = sizeof(a) / sizeof(a[0]);
    closestResult(a, b, n);
    return 0;
}

Java

// Java to find result from target array 
// for closest element
import java.util.*;
  
class GFG 
{
  
    // Function for printing resultant array
    static void closestResult(Integer[] a, 
                              int[] b, int n) 
    {
  
        // change arr[] to Set
        TreeSet vect = new TreeSet<>(Arrays.asList(a));
  
        // vector for result
        Vector c = new Vector<>();
  
        // calculate resultant array
        for (int i = 0; i < n; i++) 
        {
  
            // check upper bound element
            Integer up = vect.higher(b[i]);
  
            // if no element found push -1
            if (up == null)
                c.add(-1);
  
            // Else push the element
            else
                c.add(up); // add to resultant
        }
  
        System.out.print("Result = ");
        for (int i : c)
            System.out.print(i + " ");
    }
  
    // Driver Code
    public static void main(String[] args) 
    {
        Integer[] a = { 2, 5, 6, 1, 8, 9 };
        int[] b = { 2, 1, 0, 5, 4, 9 };
        int n = a.length;
  
        closestResult(a, b, n);
    }
}
  
// This code is contributed by
// sanjeev2552

Python3

# Python implementation to find result
# from target array for closest element
  
import bisect
  
# Function for printing resultant array
def closestResult(a, b, n):
      
    # sort list for ease
    a.sort()
  
    # list for result
    c = []
  
    # calculate resultant array
    for i in range(n):
  
        # check location of upper bound element
        up = bisect.bisect_right(a, b[i])
  
        # if no element found push -1
        if up == n:
            c.append(-1)
          
        # else puch the element
        else:
            c.append(a[up]) # add to resultant
      
    print("Result = ", end = "")
    for i in c:
        print(i, end = " ")
  
  
# Driver code
if __name__ == "__main__":
    a = [2,5,6,1,8,9]
    b = [2,1,0,5,4,9]
    n = len(a)
    closestResult(a, b, n)
  
# This code is contributed by
# sanjeev2552

输出：

Result = 5 2 1 6 5 -1