In [1]:
# The following algorithm generates the next permutation lexicographically after a given permutation.
# It changes the given permutation in-place.
#
# 1. Find the largest index k such that a[k] < a[k + 1].
# If no such index exists, the permutation is the last permutation.
# 2. Find the largest index l greater than k such that a[k] < a[l].
# 3. Swap the value of a[k] with that of a[l].
# 4. Reverse the sequence from a[k + 1] up to and including the final element a[n].
def next_lex(a):
k = len(a)-2
while (k >= 0 and a[k] >= a[k+1]):
k -= 1
if k < 0:
return False # Biggest
l = len(a)-1
while (a[l] <= a[k]):
l -= 1
a[k], a[l] = a[l], a[k]
a[k+1:] = a[k+1:][::-1]
return True
In [2]:
# hardcoded test data
nums = [123,279134399742,987]
for n in nums:
print(n, end=" -> ")
sn=str(n)
l1=list(sn)
if (next_lex(l1)):
x="".join(x for x in l1)
print(x)
else:
print("BIGGEST")
