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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 <stdbool.h>
2#include <string.h>
3
4bool canFormSubsequenceByCyclicIncrement(char *str1, char *str2) {
5 int len1
This C solution uses two indices, `i` and `j`, to iterate through `str1` and `str2`. It checks if a character in `str1` directly matches with or is a cyclic increment of a character in `str2`. If a match is found, we move to the next character in `str2`. If the end of `str2` is reached, it means it's possible to form `str2` as a subsequence.
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.
The C solution leverages character counting: a frequency array tracks occurrences of each `str1` character. Each `str2` character checks whether direct presence or cyclic traversal can fulfill its emergence in `str1`. Absence discontinues success, embracing incremental dispatches.