Find all possible Palindromic Partitions of the given String in Python

October 14, 2022

Palindromic Partitions of the given String in Python

Here on this page, We will learn How to Find All Possible Palindromic Partitions of the given String in Python Programming Language. We will discuss both approach and code on this page.

All Possible Palindromic Partitions of the given String in Python

Algorithm

  • We go through every substring starting from the first character
  • Check if it is a palindrome
  • If yes, then add the substring to the solution
  • And recur for the remaining part
All Possible Palindromic Partitions of the given String

Python Code

Run 
def isPalindrome(str, low, high):
    while low < high:
        if str[low] != str[high]:
            return False
        low += 1
        high -= 1
    return True


def allPalPartUtil(allPart, currPart, start, n, str):

    if start >= n:
        x = currPart.copy()
        allPart.append(x)
        return

    for i in range(start, n):
        if isPalindrome(str, start, i):
            currPart.append(str[start:i + 1])
            allPalPartUtil(allPart, currPart, i + 1, n, str)
            currPart.pop()


def allPalPartitions(str):

    n = len(str)
    allPart = []
    currPart = []
    allPalPartUtil(allPart, currPart, 0, n, str)

    for i in range(len(allPart)):
        for j in range(len(allPart[i])):
            print(allPart[i][j], end=" ")
        print()


string = "nitin"
allPalPartitions(string)

Output

n i t i n 
n iti n 
nitin

Hexadecimal to Decimal conversion

Decimal to Binary conversion

Decimal to Octal Conversion

Decimal to Hexadecimal Conversion

Binary to Octal conversion

Octal to Binary conversion

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2 comments on “Find all possible Palindromic Partitions of the given String in Python”


  • Yash

    def palindrome(st):
    index=0
    l=len(st)-1
    while index<l:
    if st[index]!=st[l]:
    return False
    index+=1
    return True

    def partiotions(st,result,temp):
    if len(st)==0:
    print(temp,'temp')
    result.append(temp[:])
    for i in range(0,len(st)):
    left=st[0:i+1]
    if palindrome(left):
    temp.append(left)
    partiotions(st[i+1:],result,temp)
    temp.pop()
    return(result)

    a=(partiotions('aacbb',[],[]))
    output=0
    for i in range(0,len(a)-1):
    if len(a[i])<=len(a[i+1]):
    output=len(a[i])-1
    else:
    output=len(a[i+1])-1
    print(output)

    0