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Program to count number of minimum swaps required to make it palindrome in Python
Suppose we have a string s, we have to find the minimum number of adjacent swaps needed to make it into a palindrome. If there is no such way to solve, then return -1.
So, if the input is like s = "xxyy", then the output will be 2, as we can swap the middle "x" and "y" so string is "xyxy" and then swap the first two "x" and "y" to get "yxxy", and this is palindrome.
To solve this, we will follow these steps −
Define a function util() . This will take s
seen := a new map
for each i in s, do
seen[i] := 1 + (seen[i] if exists otherwise 0)
odd_count := 0
for each key value pair of seen, do
if value is odd, then
odd_count := odd_count + 1
if odd_count is same as 2, then
return False
return True
From the main method do the following −
swaps := 0
if util(s) is true, then
left := 0
right := size of s - 1
s := a new list of characters by taking from s
while left < right, do
if s[left] is not same as s[right], then
k := right
while k > left and s[k] is not same as s[left], do
k := k - 1
if k is same as left, then
swaps := swaps + 1
s[left], s[left + 1] := s[left + 1], s[left]
otherwise,
while k < right, do
swap s[k], s[k + 1]
k := k + 1
swaps := swaps + 1
left := left + 1
right := right - 1
otherwise,
left := left + 1
right := right - 1
return swaps
return -1
Example(Python)
Let us see the following implementation to get better understanding −
class Solution:
def solve(self, s):
def util(s):
seen = {}
for i in s:
seen[i] = seen.get(i, 0) + 1
odd_count = 0
for k, val in seen.items():
if val & 1 == 1:
odd_count += 1
if odd_count == 2:
return False
return True
swaps = 0
if util(s):
left = 0
right = len(s) - 1
s = list(s)
while left < right:
if s[left] != s[right]:
k = right
while k > left and s[k] != s[left]:
k -= 1
if k == left:
swaps += 1
s[left], s[left + 1] = s[left + 1], s[left]
else:
while k < right:
s[k], s[k + 1] = s[k + 1], s[k]
k += 1
swaps += 1
left += 1
right -= 1
else:
left += 1
right -= 1
return swaps
return -1
ob = Solution()
s = "xxyy"
print(ob.solve(s))Input
"xxyy"
Output
2
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