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Shortest Interval in k-sorted list - Hard

Given k-sorted array, find the minimum interval such that there is at least one element from each array within the interval. Eg. [1,10,14],[2,5,10],[3,40,50]. Output : 1-3

To solve this problem, we perform a k-way merge as described here. At each point of 'popping' an element from an array. We keep track of the minimum and maximum head element (the first element) from all the k-lists. The minimum and maximum will obviously contain the rest of the header elements of the k-arrays. As we keep doing this, we find the smallest interval (max - min). That will be our solution. Here's a pictorial working of the algorithm.


And here's the Python code. Time Complexity : O(nlogk)


def longest_interval_lists(m):
heap = []
for i,lst in enumerate(m):
heapq.heappush(heap,(lst[0],i,0))
mn,mx = heap[0][0],max(heap)[0]
max_in_heap = mx
while heap:
x,index,i = heapq.heappop(heap)
if i + 1 == len(m[index]): break
heapq.heappush(heap, (m[index][i+1],index,i+1))
if max_in_heap < m[index][i+1]:
max_in_heap = m[index][i+1]
if max_in_heap - x < mx - mn:
mn,mx = x,max_in_heap
return mn,mx
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