With the help of sympy.combinatorics.partitions.IntegerPartition().prev_lex() method, we can get the previous partition of integer n in lexical order by using sympy.combinatorics.partitions.IntegerPartition().prev_lex() method.
Syntax :
sympy.combinatorics.partitions.IntegerPartition().prev_lex()
Return : Return the lexical value of previous partition of integer n.
Example #1 :
In this example we can see that by using sympy.combinatorics.partitions.IntegerPartition().prev_lex() method, we are able to get the lexical value of previous partition of integer n.
# import sympy and IntegerPartition from sympy.combinatorics.partitions import IntegerPartition from sympy import *Â Â Â Â # Using sympy.combinatorics.partitions.IntegerPartition().prev_lex() method gfg = IntegerPartition([1, 2, 3]) Â Â Â Â print(gfg.prev_lex()) |
Output :
[3, 1, 1, 1]
Example #2 :
# import sympy and IntegerPartition from sympy.combinatorics.partitions import IntegerPartition from sympy import *Â Â Â Â # Using sympy.combinatorics.partitions.IntegerPartition().prev_lex() method gfg = IntegerPartition([1, 2, 3, 4, 3, 2, 1]) Â Â Â Â print(gfg.prev_lex()) |
Output :
[4, 3, 3, 2, 1, 1, 1, 1]
