Select K Disjoint Special Substrings

Let dp[i] be the maximum number of disjoint special substrings in the prefix s[0...i-1]. Precompute the first and last occurrence index for each character in s. Initialize dp array of size n+1 to all zeros. Iterate i from 1 to n to compute dp[i]: Initialize dp[i] = dp[i-1] (option to not end a substring at i-1). Iterate j from i-1 down to 0 (potential start of a substring s[j...i-1]): Keep track of min_first and max_last for characters in s[j...i-1]. If min_first >= j and max_last < i, then s[j...i-1] is a special substring. If it is special (and not the whole string), update dp[i] = max(dp[i], (j > 0 ? dp[j] : 0) + 1). After the loops, dp[n] holds the maximum number of disjoint special substrings. Return dp[n] >= k.

Let dp[i] be the maximum number of disjoint special substrings that can be formed entirely within the prefix s[0...i-1].

Our goal is to compute dp[n] and check if it's greater than or equal to k.

The base case is dp[0] = 0, as no substrings can be formed from an empty prefix.

For the transition, to compute dp[i], we have two choices:

1. We don't form a special substring ending at index i-1. In this case, the maximum number of substrings is the same as for the prefix s[0...i-2]. So, dp[i] = dp[i-1].
2. We form a special substring ending at index i-1. Let this substring be s[j...i-1] for some 0 <= j < i. If s[j...i-1] is a valid special substring, we can potentially form one more substring than the maximum we could form in the prefix s[0...j-1]. This would give dp[j] + 1 substrings. We should try this for all possible start indices j.

Combining these, the recurrence relation is: dp[i] = max(dp[i-1], max(dp[j] + 1)) for all j such that s[j...i-1] is a special substring.

To implement this, we first need an efficient way to check if a substring s[j...i-1] is special. A substring is special if every character within it does not appear outside of it. This means for every character c in s[j...i-1], its first and last occurrences in the original string s must be within the bounds [j, i-1].

We can precompute the first and last occurrences of all characters in s in O(n) time. Let's call them first[c] and last[c].

The check for s[j...i-1] being special becomes:

• min(first[s[p]]) >= j for p from j to i-1.
• max(last[s[p]]) <= i-1 for p from j to i-1.
• The substring is not the entire string s.

The overall algorithm is:

1. Precompute first and last arrays for all 26 lowercase letters. O(n).
2. Initialize a dp array of size n+1 with zeros.
3. Iterate i from 1 to n: a. Set dp[i] = dp[i-1]. b. Iterate j from i-1 down to 0: i. Check if s[j...i-1] is a special substring. This check involves iterating from j to i-1 to find the minimum of first occurrences and maximum of last occurrences for the characters in the substring. This can be optimized by updating these min/max values as j decreases. ii. If s[j...i-1] is special (and not the whole string), update dp[i] = max(dp[i], dp[j] + 1).
4. Finally, return dp[n] >= k.

1class Solution {2    public boolean canSelect(String s, int k) {3        if (k == 0) {4            return true;5        }6        int n = s.length();7        int[] first = new int[26];8        int[] last = new int[26];9        java.util.Arrays.fill(first, -1);10        java.util.Arrays.fill(last, -1);11 12        for (int i = 0; i < n; i++) {13            int charIndex = s.charAt(i) - 'a';14            if (first[charIndex] == -1) {15                first[charIndex] = i;16            }17            last[charIndex] = i;18        }19 20        int[] dp = new int[n + 1];21 22        for (int i = 1; i <= n; i++) {23            dp[i] = dp[i - 1];24            int minFirst = n;25            int maxLast = -1;26            for (int j = i - 1; j >= 0; j--) {27                int charIndex = s.charAt(j) - 'a';28                minFirst = Math.min(minFirst, first[charIndex]);29                maxLast = Math.max(maxLast, last[charIndex]);30 31                if (minFirst >= j && maxLast < i) {32                    if (j == 0 && i == n) continue; // Not the entire string33                    dp[i] = Math.max(dp[i], (j > 0 ? dp[j] : 0) + 1);34                }35            }36        }37 38        return dp[n] >= k;39    }40}

This approach is a recursive (or memoized) version of the DP. Let solve(i) be the max number of disjoint special substrings in s[i...n-1]. The base case is solve(n) = 0. To compute solve(i): One option is to not start a special substring at i. The result is solve(i+1). Another option is to find a special substring s[i...j]. If one is found, the result is 1 + solve(j+1). We iterate j from i to n-1 to find all possible special substrings s[i...j]. A substring s[i...j] is special if min_first(i,j) >= i and max_last(i,j) <= j. We take the maximum over all these possibilities. solve(i) = max(solve(i+1), max_{j | s[i..j] is special} (1 + solve(j+1))). Memoization is used to store the results of solve(i) to avoid recomputing, effectively making it a DP. The final answer for the whole string is solve(0).

A key observation is how special substrings are constrained by the last occurrences of their characters. Let's define a "suffix-closed" block. A substring s[i...j] is suffix-closed if for any character c in s[i...j], its last occurrence in the whole string s is at most j. This is equivalent to max(last[s[p]]) for p from i to j being equal to j.

The string s can be uniquely partitioned into a sequence of disjoint suffix-closed blocks. For example, if s = "abacaca", last['a']=6, last['b']=2, last['c']=5. The partitions are s[0..2] (aba) and s[3..6] (caca). Any special substring must be entirely contained within one of these blocks.

This suggests a divide-and-conquer approach:

1. Partition the string s into its suffix-closed blocks.
2. The total number of special substrings is the sum of the maximum number of special substrings found in each block.
3. For each block, we can recursively apply the same logic. A block is solved by partitioning it into its own suffix-closed sub-blocks (based on first and last occurrences re-calculated for the block's substring).

When a block cannot be partitioned further, it's an "irreducible" block. For such an irreducible block, we must resort to the O(m^2) DP approach, where m is the length of the block.

The overall algorithm:

1. Precompute first and last arrays for the original string s.
2. Implement a function, say solve(start, end), that computes the max number of disjoint special substrings within s[start...end].
3. solve(start, end): a. Partition s[start...end] into suffix-closed blocks. This is done by iterating from start to end, keeping track of the maximum last occurrence seen so far. A block [block_start, i] is found when this maximum equals i. b. If the partition results in multiple blocks, the total is the sum of solve() called on each block. c. If s[start...end] is itself an irreducible block, we must check if this block itself is a special substring with respect to the original string s. This is true if min(first[s[p]]) >= start for p from start to end. If it is, we can either take this whole block as 1 special substring and add the result of solve() on its interior (by breaking its irreducibility, a complex step), or more simply, we can just run the O(m^2) DP on this block to find the max number of special substrings strictly inside it. A simpler implementation is to always run the DP on irreducible blocks.

This approach is faster on average because for many inputs, the string decomposes into small blocks, and sum(m_i^2) is much smaller than n^2. The worst-case is still O(n^2) if the string is one large irreducible block.

1class Solution {2    int[] first;3    int[] last;4    String s;5    int n;6    int[] memo;7 8    public boolean canSelect(String s, int k) {9        if (k == 0) {10            return true;11        }12        this.s = s;13        this.n = s.length();14        this.first = new int[26];15        this.last = new int[26];16        java.util.Arrays.fill(first, -1);17        java.util.Arrays.fill(last, -1);18 19        for (int i = 0; i < n; i++) {20            int charIndex = s.charAt(i) - 'a';21            if (first[charIndex] == -1) {22                first[charIndex] = i;23            }24            last[charIndex] = i;25        }26        27        this.memo = new int[n + 1];28        java.util.Arrays.fill(memo, -1);29 30        return solve(0) >= k;31    }32 33    private int solve(int start) {34        if (start >= n) {35            return 0;36        }37        if (memo[start] != -1) {38            return memo[start];39        }40 41        // Option 1: Skip s.charAt(start) and solve for the rest42        int res = solve(start + 1);43 44        // Option 2: Try to form a special substring starting at `start`45        int minFirst = n;46        int maxLast = -1;47        for (int end = start; end < n; end++) {48            int charIndex = s.charAt(end) - 'a';49            minFirst = Math.min(minFirst, first[charIndex]);50            maxLast = Math.max(maxLast, last[charIndex]);51 52            if (minFirst >= start && maxLast == end) {53                if (start == 0 && end == n - 1) {54                    // This is the whole string, not allowed as a single special substring.55                    // We must find substrings inside it.56                    // A full DP on the inside is complex. A simpler interpretation is to just not count it.57                } else {58                    res = Math.max(res, 1 + solve(end + 1));59                }60            }61        }62        63        return memo[start] = res;64    }65}

The provided code snippet is a memoized recursion which is equivalent to the DP but explores the state space differently. It tries to find a special substring starting at start and if it finds one s[start...end], it adds 1 and recurses on solve(end+1). This is another way to frame the DP, which can be more intuitive. Its complexity remains O(n^2) in the worst case but performs better if special substrings are found early, pruning the search space.