 # LUCKPAL - Editorial

EASY

### EXPLANATION

The problem is solvable by Greedy Algorithm. For each i >= 0 and j >=0, i < j such that string[i….j] = “lucky”, find the lexicographically smallest palindrome that can be constructed in a greedy manner using minimum number of replacement operations. So the problem can be divided into two tasks:

1. Place the substring “lucky” at all possible locations in the string.
2. For each such location of substring “lucky”, find the lexicographically smallest palindrome using minimum number of replacement operations while maintaining the substring “lucky”.

Let’s define each string obtained in task 1 as: “xxxxxluckyxxxxx” with str[a…b] = “lucky”. Now, second task can be achieved by iterating over the string starting with i = 0 and j = n-1 (n = string length), until i <= j. At each iteration, we have the following possibilities:

1. str[i] = str[j] , requires no replacement operation
2. str[i] != str[j] , either requires 1 replacement or no solution is possible
a. i < a and j > b, requires 1 replacement
i. if str[i] < str[j], str[j] = str[i]
ii. str[i] > str[j], str[i] = str[j]
b. a >= i and i <= b and j > b, requires 1 replacement
i. str[j] = str[i], cannot change str[i] as it is part of “lucky”
c. i < a and a >= j and j <= b, requires 1 replacement
i. str[i] = str[j], cannot change str[j] as it is part of “lucky”
d. a >= i,j <= b, cannot be converted to palindrome containing “lucky”
Final answer is the string obtained after task 2 with minimum number of replacement operations and lexicographically smallest one. If no such string exists, then “unlucky”.

### SETTER’S SOLUTION

Can be found here.

### TESTER’S SOLUTION

Can be found here.

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