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Intervals

35. 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.

Examples

Example 1

Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlap, merge them into [1,6].

Example 2

Input: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

Constraints

  • 1 <= intervals.length <= 10^4
  • intervals[i].length == 2
  • 0 <= starti <= endi <= 10^4

Dado un arreglo de intervals donde intervals[i] = [starti, endi], fusiona todos los intervalos superpuestos y devuelve un arreglo de intervalos no superpuestos que cubran todos los intervalos de la entrada.

Ejemplos

Ejemplo 1

Entrada: intervals = [[1,3],[2,6],[8,10],[15,18]]
Salida: [[1,6],[8,10],[15,18]]
Explicacion: Como los intervalos [1,3] y [2,6] se superponen, se fusionan en [1,6].

Ejemplo 2

Entrada: intervals = [[1,4],[4,5]]
Salida: [[1,5]]
Explicacion: Los intervalos [1,4] y [4,5] se consideran superpuestos.

Restricciones

  • 1 <= intervals.length <= 10^4
  • intervals[i].length == 2
  • 0 <= starti <= endi <= 10^4
merge_intervals.py 
intervals = [[1, 3], [2, 6], [8, 10], [15, 18]]

# Sort intervals by start time
intervals.sort()

# Start with the first interval
result = [intervals[0]]

for start, end in intervals[1:]:
    # If current interval overlaps with the last merged one
    if result[-1][1] >= start:
        # Extend the end to the max of both
        result[-1][1] = max(result[-1][1], end)
    else:
        # No overlap, add as new interval
        result.append([start, end])

print(result)
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