Create smallest-range.py

Author: kamyu104

Python/smallest-range.py

@@ -0,0 +1,47 @@
+# Time:  O(nlogk)
+# Space: O(k)
+
+# You have k lists of sorted integers in ascending order.
+# Find the smallest range that includes at least one number from each of the k lists.
+#
+# We define the range [a,b] is smaller than range [c,d] if b-a < d-c or a < c if b-a == d-c.
+#
+# Example 1:
+# Input:[[4,10,15,24,26], [0,9,12,20], [5,18,22,30]]
+# Output: [20,24]
+# Explanation: 
+# List 1: [4, 10, 15, 24,26], 24 is in range [20,24].
+# List 2: [0, 9, 12, 20], 20 is in range [20,24].
+# List 3: [5, 18, 22, 30], 22 is in range [20,24].
+# Note:
+# The given list may contain duplicates, so ascending order means >= here.
+# 1 <= k <= 3500
+# -105 <= value of elements <= 10^5.
+# For Java users, please note that the input type has been changed to List<List<Integer>>.
+# And after you reset the code template, you'll see this point.
+
+class Solution(object):
+    def smallestRange(self, nums):
+        """
+        :type nums: List[List[int]]
+        :rtype: List[int]
+        """
+        left, right = float("inf"), float("-inf")
+        min_heap = []
+        for row in nums:
+            left = min(left, row[0])
+            right = max(right, row[0])
+            it = iter(row)
+            heapq.heappush(min_heap, (next(it, None), it))
+        
+        result = (left, right)
+        while min_heap:
+            (val, it) = heapq.heappop(min_heap)    
+            val = next(it, None)
+            if val is None:
+                break
+            heapq.heappush(min_heap, (val, it))
+            left, right = min_heap[0][0], max(right, val);
+            if right - left < result[1] - result[0]:
+                result = (left, right)
+        return result