找到出现在每个数组元素左侧的最近值

给定一个大小为N的数组 arr[] ，任务是为每个数组元素找到数组中存在于其左侧的最近的不等值。如果没有找到这样的元素，则打印-1

例子：

Input: arr[] = { 2, 1, 5, 8, 3 }
Output: -1 2 2 5 2
Explanation:
[2], it is the only number in this prefix. Hence, answer is -1. 
[2, 1], the closest number to 1 is 2
[2, 1, 5], the closest number to 5 is 2
[2, 1, 5, 8], the closest number to 8 is 5 
[2, 1, 5, 8, 3], the closest number to 3 is 2

Input: arr[] = {3, 3, 2, 4, 6, 5, 5, 1}
Output: -1 -1 3 3 4 4 4 2
Explanation:
[3], it is the only number in this prefix. Hence, answer is -1. 
[3, 3], it is the only number in this prefix. Hence, answer is -1
[3, 3, 2], the closest number to 2 is 3
[3, 3, 2, 4], the closest number to 4 is 3 
[3, 3, 2, 4, 6], the closest number to 6 is 4
[3, 3, 2, 4, 6, 5], the closest number to 5 is 4
[3, 3, 2, 4, 6, 5, 5], the closest number to 5 is 4
[3, 3, 2, 4, 6, 5, 5, 1], the closest number to 1 is 2

朴素的方法：最简单的想法是遍历给定的数组，对于每个^第i个元素，找到索引i左侧不等于arr[i]的最近元素。
时间复杂度： O(N^2)
辅助空间： O(1)

有效的方法：

这个想法是将给定数组的元素插入到 Set 中，这样插入的数字会被排序，然后对于一个整数，找到它的位置并将它的下一个值与前一个值进行比较，然后打印出两者中更接近的值。
完成以下步骤来解决问题：

• 初始化一组整数S以按排序顺序存储元素。
• 使用变量i遍历数组arr[] 。
• 现在，找到比 arr[i]更小和更大的最近值，分别说 X和Y。
  • 如果找不到 X ，则打印Y。
  • 如果找不到 Y ，则打印Z。
  • 如果找不到 X和Y ，则打印“-1” 。
• 之后，如果abs(X – arr[i])小于abs(Y – arr[i]) ，则将 arr[i]添加到 Set S并打印X。否则，打印Y 。
• 对每个元素重复上述步骤。

下面是上述方法的实现：

C++14

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the closest number on
// the left side of x
void printClosest(set& streamNumbers, int x)
{
 
    // Insert the value in set and store
    // its position
    auto it = streamNumbers.insert(x).first;
 
    // If x is the smallest element in set
    if (it == streamNumbers.begin())
    {
        // If count of elements in the set
        // equal to 1
        if (next(it) == streamNumbers.end())
        {
 
            cout << "-1 ";
            return;
        }
 
        // Otherwise, print its
        // immediate greater element
        int rightVal = *next(it);
        cout << rightVal << " ";
        return;
    }
 
    // Store its immediate smaller element
    int leftVal = *prev(it);
 
    // If immediate greater element does not
    // exists print it's immediate
    // smaller element
    if (next(it) == streamNumbers.end()) {
        cout << leftVal << " ";
        return;
    }
 
    // Store the immediate
    // greater element
    int rightVal = *next(it);
 
    // Print the closest number
    if (x - leftVal <= rightVal - x)
        cout << leftVal << " ";
    else
        cout << rightVal << " ";
}
 
// Driver Code
int main()
{
 
    // Given array
    vector arr = { 3, 3, 2, 4, 6, 5, 5, 1 };
 
    // Initialize set
    set streamNumbers;
 
    // Print Answer
    for (int i = 0; i < arr.size(); i++) {
 
        // Function Call
        printClosest(streamNumbers, arr[i]);
    }
 
    return 0;
}

输出

-1 -1 3 3 4 4 4 2 

时间复杂度： O(N * log(N))
辅助空间： O(N)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live