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def can_partition(num): | |
s = sum(num) | |
# if 's' is a an odd number, we can't have two subsets with same total | |
if s % 2 != 0: | |
return False | |
# we are trying to find a subset of given numbers that has a total sum of 's/2'. | |
s = int(s / 2) | |
n = len(num) | |
dp = [[False for x in range(s+1)] for y in range(n)] | |
# populate the sum=0 column, as we can always have '0' sum without including | |
# any element | |
for i in range(0, n): | |
dp[i][0] = True | |
# with only one number, we can form a subset only when the required sum is | |
# equal to its value | |
for j in range(1, s+1): | |
dp[0][j] = num[0] == j | |
# process all subsets for all sums | |
for i in range(1, n): | |
for j in range(1, s+1): | |
# if we can get the sum 'j' without the number at index 'i' | |
if dp[i - 1][j]: | |
dp[i][j] = dp[i - 1][j] | |
elif j >= num[i]: # else if we can find a subset to get the remaining sum | |
dp[i][j] = dp[i - 1][j - num[i]] | |
# the bottom-right corner will have our answer. | |
return dp[n - 1][s] |
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