同一数组中每个元素的底面

给定一个整数数组，为每个元素找到最接近的较小或相同的元素。如果一个元素的所有元素都大，则打印-1。我们可以假设该数组至少包含两个元素。

例子：

Input : arr[] = {10, 5, 11, 10, 20, 12}
Output : 10 -1 10 10 12 11
Note that there are multiple occurrences of 10, so floor of 10 is 10 itself.

Input : arr[] = {6, 11, 7, 8, 20, 12}
Output : -1 8 6 7 12 11

一个简单的解决方案是运行两个嵌套循环。我们一个接一个地选择一个外部元素。对于每个拾取的元素，我们遍历剩余的数组并找到最接近的较大元素。该解决方案的时间复杂度为O(n * n)

更好的解决方案是对数组进行排序并创建排序后的副本，然后对地板进行二进制搜索。我们遍历数组，对于每个元素，我们搜索一个大于或等于给定元素的元素的首次出现。找到这样的元素后，我们检查它的下一个是否也相同，如果是，则存在该元素的多次出现，因此我们将相同的元素打印为输出。否则，我们将在已排序的数组中打印前一个元素。在C++中，lower_bound()将迭代器返回到排序数组中的第一个大于或等于第一个元素。

C++

// C++ implementation of efficient algorithm to find
// floor of every element
#include 
using namespace std;
  
// Prints greater elements on left side of every element
void printPrevGreater(int arr[], int n)
{
    // Create a sorted copy of arr[]
    vector v(arr, arr + n);
    sort(v.begin(), v.end());
  
    // Traverse through arr[] and do binary search for
    // every element.
    for (int i = 0; i < n; i++) {
  
        // Floor of first element is -1 if there is only
        // one occurrence of it.
        if (arr[i] == v[0]) {
            (arr[i] == v[1]) ? cout << arr[i] : cout << -1;
            cout << " ";
            continue;
        }
  
        // Find the first element that is greater than or
        // or equal to given element
        auto it = lower_bound(v.begin(), v.end(), arr[i]);
  
        // If next element is also same, then there
        // are multiple occurrences, so print it
        if (it != v.end() && *(it + 1) == arr[i])
            cout << arr[i] << " ";
  
        // Otherwise print previous element
        else
            cout << *(it - 1) << " ";
    }
}
  
/* Driver program to test insertion sort */
int main()
{
    int arr[] = { 6, 11, 7, 8, 20, 12 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printPrevGreater(arr, n);
    return 0;
}

Python3

# Python3 implementation of efficient 
# algorithm to find floor of every element 
  
# Prints greater elements on left 
# side of every element 
def printPrevGreater(arr, n) :
  
    # Create a sorted copy of arr 
    v = arr.copy()
    v.sort()
      
  
    # Traverse through arr[] and do 
    # binary search for every element. 
    for i in range(n) : 
  
        # Floor of first element is -1 if 
        # there is only one occurrence of it. 
        if (arr[i] == v[0]) : 
            if (arr[i] == v[1]) :
                print(arr[i], end = " ")
                  
            else :
                print(-1, end = " ") 
                  
            continue
  
        # Find the first element that is greater
        # than or or equal to given element 
        if v.count(arr[i]) > 0:
            it = v[v.index(arr[i])]
        else :
            it = v[n - 1]
              
        # If next element is also same, then there 
        # are multiple occurrences, so print it 
        if (it != v[n - 1] and 
                  v[v.index(it) + 1] == arr[i]) : 
            print(arr[i], end = " ") 
  
        # Otherwise print previous element 
        else :
            print(v[v.index(it) - 1], end = " ") 
  
# Driver Code
if __name__ == "__main__" : 
  
    arr = [ 6, 11, 7, 8, 20, 12 ]
    n = len(arr) 
    printPrevGreater(arr, n) 
  
# This code is contributed by Ryuga

输出：

-1 8 6 7 12 11

时间复杂度：O(n Log n)
辅助空间：O(n)