upload Lesson 15

Author: Mickey0521

15-CaterpillarMethod_cckao.pdf

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+# you can write to stdout for debugging purposes, e.g.
+# print("this is a debug message")
+
+def solution(A):
+    # write your code in Python 3.6
+    
+    my_dictionary = {}
+    
+    for item in A:
+        temp = abs(item)
+        if temp not in my_dictionary:
+            my_dictionary[temp] = True 
+    
+    # print(my_dictionary)
+    #print(len(my_dictionary))
+    return len(my_dictionary)

AbsDistinct.py

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+# you can write to stdout for debugging purposes, e.g.
+# print("this is a debug message")
+
+def solution(M, A):
+    # write your code in Python 3.6
+    
+	# try to remember which value was counted
+    current_dictionary = {} 
+    
+    len_A = len(A)
+    count_slice = 0
+    begin = 0
+    end = 0
+    
+    # cbombine the two loops (using one while loop)
+    while (begin < len_A) and (end < len_A):
+        # distinct value (end +=1)
+        if A[end] not in current_dictionary:
+            current_dictionary[ A[end] ] = 1
+            # super important: not just plus 1 (the key to be faster)
+            count_slice += (end - begin +1)
+            if count_slice > 1_000_000_000:
+                return 1_000_000_000
+            end +=1
+        # smae value
+        else:
+			# the same value happens (begin +=1)
+            # important: remove the value of begin
+            current_dictionary.pop(A[begin])
+            begin+=1
+
+    return count_slice

CountDistinctSlices_high_performance.py

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+# you can write to stdout for debugging purposes, e.g.
+# print("this is a debug message")
+
+def solution(M, A):
+    # write your code in Python 3.6
+    
+	# try to remember which value was counted
+    current_dictionary = {} 
+    
+    len_A = len(A)
+    count_slice = 0
+    begin = 0
+    end = 0
+    
+    for begin in range(len_A):
+        end = begin
+        # print(begin)
+        while end < len_A:
+            # print(end)
+            if A[end] not in current_dictionary:
+                current_dictionary[ A[end] ] = 1
+                count_slice += 1
+                # print('count plus 1')
+                if count_slice > 1_000_000_000:
+                    return 1_000_000_000
+                end +=1
+            else:
+				# the same value happens (begin +=1)
+                current_dictionary = {}
+                break
+        
+		# important:
+		# begin will plus 1, and so we reset 'current_dictionary'
+        current_dictionary = {}
+
+    return count_slice

CountDistinctSlices_low_performance.py

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+# you can write to stdout for debugging purposes, e.g.
+# print("this is a debug message")
+
+def solution(A):
+    # write your code in Python 3.6
+    
+    my_list = A
+    my_list.sort()
+    # print(my_list)
+    
+    if len(A) < 3:
+        return 0
+    
+    count_triangular = 0 
+
+    for index_one in range( len(A) -2):
+        index_two = index_one + 1
+        index_three = index_one + 2
+        while (index_two< len(A)-1):
+            if ( index_three<len(A) ) and (my_list[index_one] + my_list[index_two] > my_list[index_three]) :
+                index_three +=1
+            else: 
+                count_triangular += (index_three - index_two - 1)
+                index_two += 1
+
+    return count_triangular