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LC_Combinational_Sum.py
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LC_Combinational_Sum.py
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"""
39. Combination Sum
Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
The same repeated number may be chosen from candidates unlimited number of times.
Note:
All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.
Example 1:
Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]
Example 2:
Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]
"""
"""
Approach:
=========
Do a depth based search, recursively.
-> Go down if target is positive.. If expected is zero save the Answer.
-> If expected is negative return there.
At ever point.. loop through all the candidates.. which are equal or greater than..
This is done so as to avoid duplicates.
"""
class Solution(object):
def combinationSum(self, candidates, target):
"""
:type candidates: List[int]
:type target: int
:rtype: List[List[int]]
"""
def rec_traverse (stack, expected, candidates, Ans):
# since all numbers are positive.. we cant make an negative expected number
if (expected < -1):
return
if (expected == 0):
Ans.append(stack)
for item in candidates:
# So as to avoid duplicate pairs
if (stack and stack[-1] < item):
continue;
# If there is still scope recurse again
if (expected - item >= 0):
rec_traverse(stack +[item] , expected-item, candidates, Ans)
Ans = []
rec_traverse ([], target, candidates, Ans)
return (Ans)