grid47 Exploring patterns and algorithms
Sep 24, 2024
6 min read
Solution to LeetCode 436: Find Right Interval Problem
You are given an array of intervals, where each interval is represented by a pair of integers [start, end], and the start value is unique. For each interval, find the right interval. The right interval for an interval i is an interval j such that start[j] >= end[i] and the start[j] is minimized. If no right interval exists, return -1.
Problem
Approach
Steps
Complexity
Input: The input consists of an array of intervals, where each interval is represented as a pair of integers [start, end]. Each start value is unique.
Output: Return an array where each element is the index of the right interval for the corresponding input interval. If no right interval exists for a given interval, return -1.
Example: [0, -1, 1]
Constraints:
• The output array should have the same length as the input array.
Goal: The goal is to efficiently find the right interval for each given interval.
Steps:
• 1. Create a map of interval start values to their indices.
• 2. For each interval, use the map to find the first interval whose start value is greater than or equal to the end value of the current interval.
• 3. If such an interval is found, store its index. If not, store -1.
Goal: The solution should be efficient enough to handle large input sizes.
Steps:
• The length of the input intervals array is between 1 and 2 * 10^4.
• Each interval consists of two integers where start < end, and the start values are unique.
Assumptions:
• The intervals are sorted based on their start times or can be sorted to facilitate efficient searching.
• Input: Input: [[5, 8]]
• Explanation: There is only one interval, so there is no right interval, and the result is -1.
• Input: Input: [[10, 15], [6, 10], [2, 5]]
• Explanation: For the interval [10, 15], there is no right interval. For [6, 10], the right interval is [10, 15]. For [2, 5], the right interval is [6, 10].
• Input: Input: [[1, 4], [2, 5], [5, 9]]
• Explanation: For [1, 4], there is no right interval. For [2, 5], the right interval is [5, 9]. For [5, 9], there is no right interval.
Approach: The approach involves mapping each interval's start value to its index and then using a search strategy to efficiently find the right interval.
Observations:
• Using a map to store the start values of intervals allows for efficient lookups when searching for the right interval.
• By leveraging the map and lower bound search, we can minimize the time complexity compared to a brute force search.
Steps:
• 1. Create a map from interval start values to their indices.
• 2. Iterate over each interval and use the map's lower_bound function to find the smallest start value greater than or equal to the end value of the current interval.
• 3. If a matching interval is found, store its index; otherwise, store -1.
Empty Inputs:
• If the intervals array is empty, return an empty array.
Large Inputs:
• Ensure that the solution can handle up to 20,000 intervals efficiently.
Special Values:
• If an interval has the largest possible end value (10^6), ensure that the algorithm still works as expected.
Constraints:
• Handle cases where no right intervals exist for any interval.
vector<int> findRightInterval(vector<vector<int>>& iv) {
map<int, int> mk;
int n = iv.size();
for(int i =0; i < n; i++)
mk[iv[i][0]] = i;
vector<int> ans(n, -1);
for(int j =0; j < n; j++) {
auto i= iv[j];
if(mk.lower_bound(i[1]) != end(mk))
ans[j] = mk.lower_bound(i[1])->second;
}
return ans;
}
