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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.
1def can_form_subsequence(str1: str, str2: str) -> bool:
2 i = j = 0
3 n, m = len(str1), This Python solution uses loop iteration and the `ord` function to perform cyclic checks on character indices. The code ever so slightly resembles a `two pointer technique`, ensuring every approved character contributes to forming a proper subsequence of `str2` from `str1`.
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.
#include <string>
using namespace std;
bool canFormSubsequenceByCharFreq(string str1, string str2) {
vector<int> charCount(26, 0);
for (char c : str1) {
charCount[c - 'a']++;
}
for (char c : str2) {
int idx = c - 'a';
if (charCount[idx] > 0) {
charCount[idx]--;
} else {
bool canTransform = false;
for (int j = 0; j < 26; j++) {
if (charCount[j] > 0 && (((j + 1) % 26 + 'a') == c)) {
canTransform = true;
charCount[j]--;
break;
}
}
if (!canTransform) return false;
}
}
return true;
}The C++ solution applies a counting array to maintain character frequencies within `str1`. Each character in `str2` verifies the feasibility of direct matching or transformation from within `str1`. Iterating through necessary transformations aims to ensure complete emergence.