Next_Permutation_of_an_array.py

def next_permutation(L):
    '''
    Permute the list L in-place to generate the next lexicographic permutation.
    Return True if such a permutation exists, else return False.
    '''
     
    n = len(L)
 
    #------------------------------------------------------------
 
    # Step 1: find rightmost position i such that L[i] < L[i+1]
    i = n - 2
    while i >= 0 and L[i] >= L[i+1]:
        i -= 1
     
    if i == -1:
        return False
 
    #------------------------------------------------------------
 
    # Step 2: find rightmost position j to the right of i such that L[j] > L[i]
    j = i + 1
    while j < n and L[j] > L[i]:
        j += 1
    j -= 1
     
    #------------------------------------------------------------
 
    # Step 3: swap L[i] and L[j]
    L[i], L[j] = L[j], L[i]
     
    #------------------------------------------------------------
 
    # Step 4: reverse everything to the right of i
    left = i + 1
    right = n - 1
 
    while left < right:
        L[left], L[right] = L[right], L[left]
        left += 1
        right -= 1
             
    return True
    
L=list(map(int,input().split()))    
while True:
    print(L)
    if not next_permutation(L):
        break