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This approach uses two pointers to simultaneously traverse both strings. The goal is to see if `str2` can be formed as a subsequence of `str1` with at most one cyclic increment operation. During the traversal, if `str1[i]` can be incremented to match `str2[j]` by a cyclic shift, move both pointers forward. If `str1[i]` cannot be shifted to match `str2[j]`, and only one operation is allowed, then return false. Otherwise, continue checking until either `str2` is fully traversed (return true) or `str1` is exhausted (return false).
Time Complexity: O(n + m), where n is the length of `str1` and m is the length of `str2` since we process each character once.
Space Complexity: O(1), since no additional space proportional to input size is used.
1#include <string>
2using namespace std;
3
4bool canFormSubsequenceByCyclicIncrement(string str1, string str2) {
5 int i = 0, j = 0, n = str1.size(), m = str2.size();
6 while (i < n && j < m) {
7 if (str1[i] == str2[j] || (((str1[i] - 'a' + 1) % 26 + 'a') == str2[j])) {
8 j++;
9 }
10 i++;
11 }
12 return j == m;
13}This C++ solution employs string manipulation similar to the C approach. We use two pointers to traverse and match `str1` and `str2`, where matches consider cyclic increments. The cyclic increment is checked via modulo arithmetic to simulate a circular alphabet. The function returns true if all characters in `str2` can be matched.
This method determines the feasibility of forming `str2` as a subsequence by transforming `str1` using character frequency transitions, considering cyclic alphabet rules. Using a frequency map for `str1`, iterate through `str2`, checking if corresponding increments can map `str1` characters to those in `str2`. For each character needed in `str2`, see if it's obtainable via allowable operations on `str1` characters, optimizing through potential full rotations using the alphabet cycle property.
Time Complexity: O(n + m + 26) to cover all characters and frequency evaluations.
Space Complexity: O(1), preserving essential space outside of constant size arrays.
Java's solution involves preparing an array to track `str1` character frequencies, iteratively checking the possibility of forming each `str2` character through available data post-cyclic application. Mismatches halting transformations obstruct outcomes.