#
# * 435. Non-overlapping Intervals
# * Medium

# * Given a collection of intervals, find the minimum number of intervals you need 
# * to remove to make the rest of the intervals non-overlapping.

# * Example 1:

# Input: [[1,2],[2,3],[3,4],[1,3]]
# Output: 1
# Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.

# * Example 2:

# Input: [[1,2],[1,2],[1,2]]
# Output: 2
# Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.

# * Example 3:

# Input: [[1,2],[2,3]]
# Output: 0
# Explanation: You don't need to remove any of the intervals since they're already 
# non-overlapping.

# * Note:

#     You may assume the interval's end point is always bigger than its start point.
#     Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap 
#       each other.

#%%
class Solution:
    # * Solution 1
    def eraseOverlapIntervals(self, intervals: list) -> int:
        if len(intervals) == 0:
            return 0
            
        intervals.sort(key=lambda x : x[0])
        intervals.sort(key=lambda x : x[1])
        # print(intervals)

        w = intervals[0]
        result = []
        for interval in intervals:
            if interval[0] >= w[1]:
                result.append(w)
                w = interval
        
        result.append(w)
        return len(intervals) - len(result)

        



sol = Solution()
a1 = [[1,2], [2,3], [3,4], [1,3]]
r1 = sol.eraseOverlapIntervals(a1)
print(r1)

a1 = [[1,2], [1,2], [1,2]]
r1 = sol.eraseOverlapIntervals(a1)
print(r1)

a1 = [[1,2], [2,3]]
r1 = sol.eraseOverlapIntervals(a1)
print(r1)
# %%
