Recursive DFS accepted - try unique nums only

  • 0

    Count the frequency of nums, Try to make sum(nums)//2 (if even). For each unique num, subtract it from target and decrement its count in the frequency map then recurse. Increment frequency map if no solution found. This avoids duplication e.g. in the test case where there are many 1s.
    Might well be faster if dictionary was ordered from largest key first.

    from collections import Counter
    class Solution(object):
        def canPartition(self, nums):
            nums_sum = sum(nums)
            if nums_sum % 2 == 1:
                return False
            freq = Counter(nums)
            return self.partition(freq, nums_sum // 2)
        def partition(self, freq, target):
            if target == 0:
                return True
            if target < 0:
                return False
            for num in freq:
                if freq[num] == 0:
                freq[num] -= 1
                if self.partition(freq, target - num):
                    return True
                freq[num] += 1
            return False

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