minimum insertions to form a palindrome

— Written by — 3 min read
#python  #algorithm 

Brute-force approach

Here I present a few approaches to deduce “minimum insertions” required to convert a string into a palindrome.

The basic brute force approach is quite simple, given a string with length L, start comparing, the first character from left and the last character while scanning inwards.

Here is a basic test for a palindrome.

   L = len(s) 
   for i in range(L):
       if s[i] != s[L - i - 1]:
           return False,i,L-i -1
   return True,0,0

The above code returns True if the string is a palindrome or returns False with mismatching indices.

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print two-dimensional array in spiral order

— Written by — 1 min read
#algorithm  #python 

So I saw this problem in a book today about printing a 2d matrix in spiral order

Here are two solutions to it

Solution one. 

def printSpiralTL(m,x1,y1,x2,y2):
    for i in range(x1,x2):
        print m[y1][i]
    for j in range(y1+1,y2+1):
        print m[j][x2-1]

    if x2-x1 > 0:
        printSpiralBL(m, x1, y1 + 1, x2-1, y2)
    

def printSpiralBL(m,x1,y1,x2,y2):
    for i in range(x2-1,x1-1,-1):
        print m[y2][i]
    for j in range(y2-1,y1-1,-1):
        print m[j][x1]
    if x2-x1 > 0:
        printSpiralTL(m, x1+1, y1, x2, y2-1)
    
m = [
    
    [1, 2, 3, 4], 
    [5, 6, 7, 8],
    [9, 0, 1, 2],   
    [3, 4, 5, 6], 
    [7, 8, 9, 1]
        
    ]

Output:

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