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  2. Algorithms
  3. Strings
  4. Highest Value Palindrome
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Highest Value Palindrome

  • aagam29
     + 2 comments

    Yes, I thought of something similar and here you have done the same! Python

    def highestValuePalindrome(s: str, n, k):
    left = list(s[:n//2])
    right = list(s[n//2+1:][::-1]) if n % 2 == 1 else list(s[n//2:][::-1])
    remainingCount = k
    difference = 0
    length = len(left)
    for i in range(length):
        if(left[i] != right[i]):
            difference += 1

    for i in range(length):
        x = left[i]
        y = right[i]
        if(x == y):
            if(x == '9'):
                continue
            elif(remainingCount-2 >= (difference)):
                left[i] = '9'
                right[i] = '9'
                remainingCount -= 2
        elif(x != y):
            if(x == '9'):
                if(remainingCount-1 >= (difference-1)):
                    right[i] = '9'
                    difference -= 1
                    remainingCount -= 1
                else:
                    return -1
            elif(y == '9'):
                if(remainingCount-1 >= (difference-1)):
                    left[i] = '9'
                    difference -= 1
                    remainingCount -= 1
                else:
                    return -1
            else:
                if(remainingCount-2 >= (difference-1)):
                    left[i] = '9'
                    right[i] = '9'
                    difference -= 1
                    remainingCount -= 2
                elif(remainingCount-1 >= (difference-1)):
                    if(x > y):
                        right[i] = left[i]
                    else:
                        left[i] = right[i]
                    difference -= 1
                    remainingCount -= 1
                else:
                    return -1
    if (n % 2 == 0):
        ans = "".join(left + (right[::-1]))
    if(n % 2 == 1):
        mid = ['9'] if(remainingCount > 0) else [s[n//2]]
        ans = "".join(left + mid + (right[::-1]))
    return(ans)