Check if a string has all characters with same frequency with one variation allowed in Python

Suppose we have a lowercase string s, we have to check whether we can convert s into a valid string by deleting at most 1 character. Here a valid string means a string str such that for all unique characters in str each character’s frequency is same.

So, if the input is like s = "xyyzx", then the output will be True as we can delete z then string will be "xyyx" where occurrences of x and y are same.

To solve this, we will follow these steps −

• size := 26
• occurrence := an array of size 26. This is storing frequencies of each character in s
• occr1 := 0
• occr1_cnt := 0
• for i in range 0 to size - 1, do
  • if occurrence[i] is not 0, then
    • occr1 := occurrence[i]
    • occr1_cnt := 1
    • come out from loop
• occr2 := 0
• occr2_cnt := 0
• for j in range i+1 to size - 1, do
  • if occurrence[j] is not 0, then
    • if occurrence[j] is same as occr1, then
      • occr1_cnt := occr1_cnt + 1
    • otherwise,
      • occr2_cnt := 1
      • occr := occurrence[j]
      • come out from loop
• for k in range j+1 to size - 1, do
  • if occurrence[k] is not 0, then
    • if occurrence[k] is same as occr1, then
      • occr1_cnt := occr1_cnt + 1
    • if occurrence[k] is same as occr2, then
      • occr2_cnt := occr2_cnt + 1
    • otherwise,
      • return False
  • if occr1_cnt > 1 and occr2_cnt > 1, then
    • return False
• return True

Let us see the following implementation to get better understanding −

Example

 Live Demo

size = 26
def solve(str):
   occurrence = [0]*size
   for i in range(len(str)): occurrence[ord(str[i])-ord('a')] += 1
      occr1 = 0
      occr1_cnt = 0
      for i in range(size):
         if (occurrence[i] != 0):
            occr1 = occurrence[i]
            occr1_cnt = 1
            break
   occr2 = 0
   occr2_cnt = 0
   for j in range(i+1,size):
      if (occurrence[j] != 0):
         if (occurrence[j] == occr1):
            occr1_cnt += 1
         else:
            occr2_cnt = 1
            occr = occurrence[j]
            break
   for k in range(j+1,size):
      if occurrence[k] != 0:
         if (occurrence[k] == occr1):
            occr1_cnt += 1
      if (occurrence[k] == occr2):
         occr2_cnt += 1
      else:
         return False
      if occr1_cnt > 1 and occr2_cnt > 1:
         return False
   return True
s = "xyyzx"
print(solve(s))

Input

"xyyzx"

Output

True