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All suggeted changes within additional time limit tests (#1815)
* With all suggested changes ✅ possibly covered all the recommended guidelines * Updated with both slow and faster algorithms possibally covered all the recomendations * removed the time comparision part! * Update data_structures/stacks/next_greater_element.py Co-Authored-By: Christian Clauss <cclauss@me.com> * Update data_structures/stacks/next_greater_element.py Co-Authored-By: Christian Clauss <cclauss@me.com> * Update data_structures/stacks/next_greater_element.py Co-Authored-By: Christian Clauss <cclauss@me.com> * Update data_structures/stacks/next_greater_element.py Co-Authored-By: Christian Clauss <cclauss@me.com> * Add benchmark using timeit https://docs.python.org/3/library/timeit.html The performance delta between these two implementation is quite small... ``` next_greatest_element_slow(): 1.843442126 next_greatest_element(): 1.828941414 ``` * Optimize slow() to create fast() - Three algorithms in the race Three algorithms in the race * Use a bigger test array with floats, negatives, zero * Setup import next_greatest_element_fast Co-authored-by: Christian Clauss <cclauss@me.com>
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def printNGE(arr):
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arr = [-10, -5, 0, 5, 5.1, 11, 13, 21, 3, 4, -21, -10, -5, -1, 0]
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"""
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Function to print element and Next Greatest Element (NGE) pair for all elements of list
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NGE - Maximum element present afterwards the current one which is also greater than current one
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>>> printNGE([11,13,21,3])
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11 -- 13
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13 -- 21
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21 -- -1
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3 -- -1
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"""
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for i in range(0, len(arr), 1):
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def next_greatest_element_slow(arr):
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"""
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Function to get Next Greatest Element (NGE) pair for all elements of list
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Maximum element present afterwards the current one which is also greater than current one
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>>> next_greatest_element_slow(arr)
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[-5, 0, 5, 5.1, 11, 13, 21, -1, 4, -1, -10, -5, -1, 0, -1]
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"""
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result = []
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for i in range(0, len(arr), 1):
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next = -1
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next = -1
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for j in range(i + 1, len(arr), 1):
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for j in range(i + 1, len(arr), 1):
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if arr[i] < arr[j]:
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if arr[i] < arr[j]:
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next = arr[j]
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next = arr[j]
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break
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break
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result.append(next)
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print(str(arr[i]) + " -- " + str(next))
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return result
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# Driver program to test above function
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def next_greatest_element_fast(arr):
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arr = [11, 13, 21, 3]
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"""
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printNGE(arr)
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Like next_greatest_element_slow() but changes the loops to use
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enumerate() instead of range(len()) for the outer loop and
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for in a slice of arr for the inner loop.
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>>> next_greatest_element_fast(arr)
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[-5, 0, 5, 5.1, 11, 13, 21, -1, 4, -1, -10, -5, -1, 0, -1]
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"""
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result = []
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for i, outer in enumerate(arr):
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next = -1
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for inner in arr[i + 1:]:
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if outer < inner:
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next = inner
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break
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result.append(next)
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return result
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def next_greatest_element(arr):
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"""
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Function to get Next Greatest Element (NGE) pair for all elements of list
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Maximum element present afterwards the current one which is also greater than current one
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Naive way to solve this is to take two loops and check for the next bigger number but that will make the
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time complexity as O(n^2). The better way to solve this would be to use a stack to keep track of maximum
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number givig a linear time complex solution.
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>>> next_greatest_element(arr)
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[-5, 0, 5, 5.1, 11, 13, 21, -1, 4, -1, -10, -5, -1, 0, -1]
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"""
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stack = []
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result = [-1]*len(arr)
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for index in reversed(range(len(arr))):
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if len(stack):
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while stack[-1] <= arr[index]:
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stack.pop()
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if len(stack) == 0:
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break
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if len(stack) != 0:
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result[index] = stack[-1]
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stack.append(arr[index])
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return result
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if __name__ == "__main__":
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from doctest import testmod
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from timeit import timeit
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testmod()
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print(next_greatest_element_slow(arr))
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print(next_greatest_element_fast(arr))
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print(next_greatest_element(arr))
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setup = ("from __main__ import arr, next_greatest_element_slow, "
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"next_greatest_element_fast, next_greatest_element")
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print("next_greatest_element_slow():",
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timeit("next_greatest_element_slow(arr)", setup=setup))
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print("next_greatest_element_fast():",
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timeit("next_greatest_element_fast(arr)", setup=setup))
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print(" next_greatest_element():",
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timeit("next_greatest_element(arr)", setup=setup))
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