You are given a string s.
A string t is called good if all characters of t occur the same number of times.
You can perform the following operations any number of times:
\n\ns.s.s to its next letter in the alphabet.Note that you cannot change 'z' to 'a' using the third operation.
Return the minimum number of operations required to make s good.
\n
Example 1:
\n\nInput: s = "acab"
\n\nOutput: 1
\n\nExplanation:
\n\nWe can make s good by deleting one occurrence of character 'a'.
Example 2:
\n\nInput: s = "wddw"
\n\nOutput: 0
\n\nExplanation:
\n\nWe do not need to perform any operations since s is initially good.
Example 3:
\n\nInput: s = "aaabc"
\n\nOutput: 2
\n\nExplanation:
\n\nWe can make s good by applying these operations:
'a' to 'b''c' into s\n
Constraints:
\n\n3 <= s.length <= 2 * 104s contains only lowercase English letters.occ where occ[x] is the number of occurrences of the x character of the alphabet. How do the described operations change occ?",
"We have three types of operations: increase any occ[x] by 1, decrease any occ[x] by 1, or decrease any occ[x] by 1 and simultaneously increase occ[x + 1] by 1 at the same time. To make s good, we need to make occ good. occ is good if and only if every occ[x] equals either 0 or some constant c.",
"If you know the value of c, how can you calculate the minimum operations required to make occ good?",
"Observation 1: It is never optimal to apply the third type of operation (simultaneous decrease and increase) on two continuous elements occ[x] and occ[x + 1]. Instead, we can decrease occ[x] by 1 then increase occ[x + 2] by 1 to achieve the same effect.",
"Observation 2: It is never optimal to increase an element of occ then decrease it, or vice versa.",
"Use dynamic programming where dp[i] is the minimum number of operations required to make occ[0..i] good. You will need to use the above observations to come up with the transitions."
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