LeetCode-python/problem_070.py at main · saitejamanchi/LeetCode-python · GitHub

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"""
Problem: https://leetcode.com/problems/climbing-stairs/

Solution: Number of ways to reach a step n would be sum of number of ways to reach n-1 and n-2 steps.

Time Complexity: O(n) as we're iterating n times.

Space Complexity: O(1) as we're using a constant space data structure to keep tarck of steps.
"""
classSolution:
defclimbStairs(self, n: int) ->int:
ifn<=2:
returnn
step_tracker= [1, 2]
foriinrange(n-2):
step_tracker.append(step_tracker[0] +step_tracker[1])
delstep_tracker[0]

returnstep_tracker[-1]
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"""
Problem: https://leetcode.com/problems/climbing-stairs/

Solution: Number of ways to reach a step n would be sum of number of ways to reach n-1 and n-2 steps.

Time Complexity: O(n) as we're iterating n times.

Space Complexity: O(1) as we're using a constant space data structure to keep tarck of steps.
"""
classSolution:
defclimbStairs(self, n: int) ->int:
ifn<=2:
returnn
step_tracker= [1, 2]
foriinrange(n-2):
step_tracker.append(step_tracker[0] +step_tracker[1])
delstep_tracker[0]

returnstep_tracker[-1]
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"""
Problem: https://leetcode.com/problems/climbing-stairs/

Solution: Number of ways to reach a step n would be sum of number of ways to reach n-1 and n-2 steps.

Time Complexity: O(n) as we're iterating n times.

Space Complexity: O(1) as we're using a constant space data structure to keep tarck of steps.
"""
classSolution:
defclimbStairs(self, n: int) ->int:
ifn<=2:
returnn
step_tracker= [1, 2]
foriinrange(n-2):
step_tracker.append(step_tracker[0] +step_tracker[1])
delstep_tracker[0]

returnstep_tracker[-1]
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"""
Problem: https://leetcode.com/problems/climbing-stairs/

Solution: Number of ways to reach a step n would be sum of number of ways to reach n-1 and n-2 steps.

Time Complexity: O(n) as we're iterating n times.

Space Complexity: O(1) as we're using a constant space data structure to keep tarck of steps.
"""
classSolution:
defclimbStairs(self, n: int) ->int:
ifn<=2:
returnn
step_tracker= [1, 2]
foriinrange(n-2):
step_tracker.append(step_tracker[0] +step_tracker[1])
delstep_tracker[0]

returnstep_tracker[-1]
1
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"""
Problem: https://leetcode.com/problems/climbing-stairs/

Solution: Number of ways to reach a step n would be sum of number of ways to reach n-1 and n-2 steps.

Time Complexity: O(n) as we're iterating n times.

Space Complexity: O(1) as we're using a constant space data structure to keep tarck of steps.
"""
classSolution:
defclimbStairs(self, n: int) ->int:
ifn<=2:
returnn
step_tracker= [1, 2]
foriinrange(n-2):
step_tracker.append(step_tracker[0] +step_tracker[1])
delstep_tracker[0]

returnstep_tracker[-1]
1
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"""
Problem: https://leetcode.com/problems/climbing-stairs/

Solution: Number of ways to reach a step n would be sum of number of ways to reach n-1 and n-2 steps.

Time Complexity: O(n) as we're iterating n times.

Space Complexity: O(1) as we're using a constant space data structure to keep tarck of steps.
"""
classSolution:
defclimbStairs(self, n: int) ->int:
ifn<=2:
returnn
step_tracker= [1, 2]
foriinrange(n-2):
step_tracker.append(step_tracker[0] +step_tracker[1])
delstep_tracker[0]

returnstep_tracker[-1]
1
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"""
Problem: https://leetcode.com/problems/climbing-stairs/

Solution: Number of ways to reach a step n would be sum of number of ways to reach n-1 and n-2 steps.

Time Complexity: O(n) as we're iterating n times.

Space Complexity: O(1) as we're using a constant space data structure to keep tarck of steps.
"""
classSolution:
defclimbStairs(self, n: int) ->int:
ifn<=2:
returnn
step_tracker= [1, 2]
foriinrange(n-2):
step_tracker.append(step_tracker[0] +step_tracker[1])
delstep_tracker[0]

returnstep_tracker[-1]
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18
19
"""
Problem: https://leetcode.com/problems/climbing-stairs/

Solution: Number of ways to reach a step n would be sum of number of ways to reach n-1 and n-2 steps.

Time Complexity: O(n) as we're iterating n times.

Space Complexity: O(1) as we're using a constant space data structure to keep tarck of steps.
"""
classSolution:
defclimbStairs(self, n: int) ->int:
ifn<=2:
returnn
step_tracker= [1, 2]
foriinrange(n-2):
step_tracker.append(step_tracker[0] +step_tracker[1])
delstep_tracker[0]

returnstep_tracker[-1]