查找每个给定 N 个区间右侧最近的非重叠区间的索引

给定一个包含N个区间的数组arr[] ，任务是计算与当前区间不重叠的每个给定N个区间右侧最近区间的索引。

例子：

Input: arr[] = {{3, 4}, {2, 3}, {1, 2}}
Output: -1 0 1
Explanation: For the interval arr[0], there exists no interval to the right of it. For interval arr[1], i.e, [2, 3], the interval [3, 4] is the closest interval to its right that does not overlap it. Hence the required answer for arr[1] is 0. For the interval arr[2], both arr[1] and arr[2] are on its right side and does not overlap with arr[2], but arr[1] is the closest among them.

Input: arr[] = {{1, 4}, {3, 4}, {2, 3}}
Output: -1 -1 1

编程需要懂一点英语

方法：给定的问题可以在二分搜索的帮助下使用贪心方法来解决。这个想法是按照起始点的递增顺序对区间进行排序，并为每个区间找到最接近右侧的区间。以下是要遵循的步骤：

创建一个向量对V ，将起点和每个区间的索引存储在arr[]中。
按起始点的升序对向量V进行排序。
遍历数组arr[]并为每个索引找到最右边的区间的索引，这样它就不会与使用下界函数的当前区间重叠。

下面是上述方法的实现：

C++

// C++ program of the above approach
#include 
using namespace std;
 
// Function to find the index of closest
// non overlapping interval to the right
// of each of the given N intervals
vector closestInterval(
    vector > arr,
    int N)
{
    // Vector to store starting value,
    // index pair of arr[]
    vector > V;
 
    // Loop to iterate arr[]
    for (int i = 0; i < N; i++) {
        V.push_back({ arr[i].first, i });
    }
 
    // Sort the vector V
    sort(V.begin(), V.end());
 
    // Stores the resultant vector
    vector res(N, -1);
 
    // Loop to iterate arr[]
    for (int i = 0; i < N; i++) {
        // Calculate the closest
        // interval to the right
        auto it = lower_bound(
            V.begin(), V.end(),
            make_pair(arr[i].second, 0));
 
        // If a valid interval
        // exist update res
        if (it != V.end()) {
            int idx = it - V.begin();
 
            // Update res
            res[i] = V[idx].second;
        }
    }
 
    // Return Answer
    return res;
}
 
// Driver Code
int main()
{
    vector > arr{ { 1, 4 },
                                 { 3, 4 },
                                 { 2, 3 } };
 
    for (auto x : closestInterval(arr, arr.size())) {
        cout << x << " ";
    }
 
    return 0;
}

Python3

# python3 program of the above approach
import bisect
 
# Function to find the index of closest
# non overlapping interval to the right
# of each of the given N intervals
def closestInterval(arr, N):
 
    # Vector to store starting value,
    # index pair of arr[]
    V = []
 
    # Loop to iterate arr[]
    for i in range(0, N):
        V.append([arr[i][0], i])
 
    # Sort the vector V
    V.sort()
 
    # Stores the resultant vector
    res = [-1 for _ in range(N)]
 
    # Loop to iterate arr[]
    for i in range(0, N):
        # Calculate the closest
        # interval to the right
        it = bisect.bisect_left(V, [arr[i][1], 0])
 
        # If a valid interval
        # exist update res
        if (it != len(V)):
            idx = it
 
            # Update res
            res[i] = V[idx][1]
 
    # Return Answer
    return res
 
# Driver Code
if __name__ == "__main__":
 
    arr = [[1, 4],
           [3, 4],
           [2, 3]]
 
    for x in closestInterval(arr, len(arr)):
        print(x, end=" ")
 
    # This code is contributed by rakeshsahni

输出

-1 -1 1

时间复杂度： O(N*log N)
辅助空间： O(1)