Split a String into two Substring such that the sum of unique characters is maximum

Given a string str, the task is to partition the string into two substrings such that the sum of unique characters of both substrings is the maximum possible.

Examples:

Input: str = "abcabcd” 
Output: 7
Explanation: Partition the given string into "abc" and "abcd", the sum of unique characters is 3 + 4 = 7 which is maximum possible.

Input: str = "aaaaa”
Output: 2
Explanation: Partition the given string into "aa" and "aaa", the sum of unique characters is 1 + 1 = 2 which is maximum possible. Given string can be partitioned into many other ways but this partition gives the maximum sum of unique characters.

Approach: This problem can easily be solved using prefix and suffix arrays. 

We can create a prefix array which will store the count of unique characters starting from index 0 to last index of the string. Similarly, We can create a suffix array which will store the count  of unique characters starting from the last index to 0^th index. Here, set can be used to check if the current character appeared before in the iterated string or not. We can calculate the maximum sum by doing prefix[i-1] + suffix[i] for each 1 <= i < n. Take prefix[i-1] + suffix[i] so that intersecting point never counts.

Follow the steps mentioned below to implement the idea:

• Declare two arrays of prefix and suffix
• Declare two sets that store the visited characters so far.
• Store prefix[i] = prefix[i-1] if the current character has already appeared, else prefix[i] = prefix[i-1] + 1
• Store suffix[i] = suffix[i] if the current character has already appeared, else suffix[i] = suffix[i+1] + 1.
• Declare a variable maxi = -1 which stores the maximum sum
• Iterate through the array and compare the sum of prefix[i-1] + suffix[i] with maxi in each iteration.
• Return maxi.

Below is the implementation of the above approach.

Maximum value is 7

Time Complexity: O(N)
Auxiliary Space: O(N) 

Efficient Approach: To solve this problem, we will use two Hash Maps and do the below steps:

• Firstly, make a frequency map of characters and store in Hash Map characters.
• Now, we will traverse from the start of the given string str, also we will maintain another frequency map using Hash Map freq.
• While traversing, we will keep track of the maximum sum of the sizes of characters and freq, whenever our maximum value changes we will change our maximum pointer.
• At last, we will return the maximum possible answer.

Implementation of the approach:

Output:

Maximum sum is 7

Time Complexity: O(N)

Auxiliary Space: Constant space is used as there can be only at most 26 characters stored in hash map.