435. Non-overlapping Intervals 763. Partition Labels 56. Merge Intervals

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文章介绍了如何在给定区间列表中，通过排序和比较操作，最小化需要移除的区间数量以使得剩余区间互不重叠。提供了两种解决方案，包括使用快速排序对区间按起始位置排序，然后逐步合并重叠区间。

435. Non-overlapping Intervals

Given an array of intervals intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Record the number of non-crossing intervals from left to right.
Finally, subtract the number of non-crossing intervals from the total number of intervals, and that's the number of intervals to remove.

Time complexity: O(nlog n) with a Quicksort
Space complexity: O(n)

AC mysolution:

class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        intervals.sort(key = lambda x: (x[0],x[1]))
        result = 0
        right = intervals[0][1]

        for i in range(len(intervals) - 1):
            if right <= intervals[i + 1][0]:
                right =intervals[i + 1][1]
            
            elif right > intervals[i + 1][0] and right <= intervals[i + 1][1]:
                result += 1
            
            elif right > intervals[i + 1][0] and right > intervals[i + 1][1]:
                result += 1
                right = intervals[i + 1][1]
        
        return result

bro Ka solution:

class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        if not intervals:
            return 0
        
        intervals.sort(key=lambda x: x[0])  # 按照左边界升序排序
        count = 0  # 记录重叠区间数量
        
        for i in range(1, len(intervals)):
            if intervals[i][0] < intervals[i - 1][1]:  # 存在重叠区间
                intervals[i][1] = min(intervals[i - 1][1], intervals[i][1])  # 更新重叠区间的右边界
                count += 1
        
        return count

763. Partition Labels 分区标签

You are given a string s. We want to partition the string into as many parts as possible so that each letter appears in at most one part.

Note that the partition is done so that after concatenating all the parts in order, the resultant string should be s.

Return a list of integers representing the size of these parts

Time complexity: O(n)
Space complexity: O(1), using hash arrays is fixed size

class Solution:
    def partitionLabels(self, s: str) -> List[int]:
        last_occurrence = {}
        for i,l in enumerate(s):
            last_occurrence[l] = i
        
        result = []
        start = 0
        end = 0

        for i,l in enumerate(s):
            end = max(end, last_occurrence[l])
            if i == end:
                result.append(end - start + 1)
                start = i + 1
        
        return result

56. Merge Intervals

Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

AC:

class Solution:
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
        intervals.sort(key = lambda x: (x[0], x[1]))
        result = []
        left = intervals[0][0]
        right = intervals[0][1]

        for i in range(len(intervals) - 1):
            if right >= intervals[i + 1][0] and right < intervals[i + 1][1]:
                right = intervals[i + 1][1]

            elif right >= intervals[i + 1][0] and right >= intervals[i + 1][1]:
                continue
            
            else: # elif right < intervals[i + 1][0]:
                result.append([left,right])
                left = intervals[i + 1][0]
                right = intervals[i + 1][1]
        
        result.append([left,right])
        return result

AC simplified version:

class Solution:
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
        intervals.sort(key = lambda x: (x[0], x[1]))
        result = []
        left = intervals[0][0]
        right = intervals[0][1]

        for i in range(len(intervals) - 1):
            if right >= intervals[i + 1][0]: #and right < intervals[i + 1][1]:
                right = max(right, intervals[i + 1][1])

            #elif right >= intervals[i + 1][0] and right >= intervals[i + 1][1]:
            #    continue
            
            else: # elif right < intervals[i + 1][0]:
                result.append([left,right])
                left = intervals[i + 1][0]
                right = intervals[i + 1][1]
        
        result.append([left,right])
        return result

bro KA's solution:

class Solution:
    def merge(self, intervals):
        result = []
        if len(intervals) == 0:
            return result  # 区间集合为空直接返回

        intervals.sort(key=lambda x: x[0])  # 按照区间的左边界进行排序

        result.append(intervals[0])  # 第一个区间可以直接放入结果集中

        for i in range(1, len(intervals)):
            if result[-1][1] >= intervals[i][0]:  # 发现重叠区间
                # 合并区间，只需要更新结果集最后一个区间的右边界，因为根据排序，左边界已经是最小的
                result[-1][1] = max(result[-1][1], intervals[i][1])
            else:
                result.append(intervals[i])  # 区间不重叠

        return result