使范围不重叠所需的最小移除量

给定具有起始值和结束值的范围列表，任务是找到需要删除的最小范围数，以使其余范围不重叠。

例子：

Input : input = {{1, 2}, {4, 7}, {3, 8}}
Output : 1
Explanation: Removal of {3, 8} makes {{1, 2} and {4, 7}} non-overlapping.

Input : input = {{ 10, 20 }, { 10, 20 } , { 10, 20 }}
Output : 2
Explanation: Removal of [10, 20] makes the remaining ranges non-overlapping.

Input : input = {{1, 2}, {5, 10}, {18, 35}, {40, 45}}
Output : 0
Explanation:All ranges are already non-overlapping.

方法：

• 按范围的起始值对范围进行排序。
• 遍历范围并检查是否有任何范围的起点小于上一个范围的终点(即存在重叠)。
• 删除终点更大的范围。

  下面的代码是上述方法的实现：

  C++

  #include 
  using namespace std;
    
  int minRemovals(vector >& ranges)
  {
    
      int size = ranges.size(), rem = 0;
    
      if (size <= 1)
          return 0;
    
      // Sort by minimum starting point
      sort(ranges.begin(), ranges.end(), 
          [](const vector& a, const vector& b) 
                      { return a[0] < b[0]; });
    
      int end = ranges[0][1];
      for (int i = 1; i < ranges.size(); i++) {
    
          // If the current starting point is less than
          // the previous interval's ending point 
          // (ie. there is an overlap)
          if (ranges[i][0] < end) {
              // increase rem
              rem++;
              // Remove the interval
              // with the higher ending point
              end = min(ranges[i][1], end);
          }
          else
              end = ranges[i][1];
      }
    
      return rem;
  }
    
  // Driver code
  int main()
  {
      vector > input = { { 19, 25 }, 
                          { 10, 20 }, { 16, 20 } };
      cout << minRemovals(input) << endl;
  }

  Python3

  def minRemovels (ranges):
    
      size = len(ranges)
      rem = 0
    
      # Sort by minimum starting point
      ranges.sort()
    
      end = ranges[0][1]
      for i in range(1, size):
    
          # If the current starting point is less
          # than the previous interval's ending
          # point (ie. there is an overlap)
          if (ranges[i][0] < end):
    
              # Increase rem
              rem += 1
    
              # Remove the interval
              # with the higher ending point
              end = min(ranges[i][1], end)
                
          else:
              end = ranges[i][1]
    
      return rem
    
  # Driver Code
  if __name__ == '__main__':
        
      Input = [ [ 19, 25 ],
                [ 10, 20 ],
                [ 16, 20 ] ]
                  
      print(minRemovels(Input))
    
  # This code is contributed by himanshu77

  输出：

  2
  

  时间复杂度： O(n log n)

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