96 lines
2.6 KiB
C++
96 lines
2.6 KiB
C++
#include <string>
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/**
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* 1616. Split Two Strings to Make Palindrome
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*
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* You are given two strings a and b of the same length. Choose an index and split both strings at the same index, splitting a into two strings: aprefix and asuffix where a = aprefix + asuffix, and splitting b into two strings: bprefix and bsuffix where b = bprefix + bsuffix. Check if aprefix + bsuffix or bprefix + asuffix forms a palindrome.
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* When you split a string s into sprefix and ssuffix, either ssuffix or sprefix is allowed to be empty. For example, if s = "abc", then "" + "abc", "a" + "bc", "ab" + "c" , and "abc" + "" are valid splits.
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* Return true if it is possible to form a palindrome string, otherwise return false.
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* Notice that x + y denotes the concatenation of strings x and y.
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*/
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class Solution {
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public:
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static bool checkPalindromeFormation(const std::string& a, const std::string& b) {
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const int m = a.length(), n = b.length();
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if (m == 1 || n == 1)
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return true;
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// prefix[shorter](b) + suffix[longer](a)
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// prefix b, base a
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if (n >= ((m >> 1) + (m & 1))) {
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bool OK = true;
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// prefix (i) operates on a first and optionally switches to b
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// suffix (j) always operates on a
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for (int i = (m >> 1) - !(m & 1), j = m >> 1, switched = 0; i >= 0; --i, ++j) {
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if ((switched ? b : a)[i] == a[j])
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continue;
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switched = 1;
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if (b[i] == a[j])
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continue;
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OK = false;
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break;
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}
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if (OK) return true;
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}
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// prefix(a) + suffix(b)
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// prefix a, base b
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if (m >= ((n >> 1) + (n & 1))) {
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bool OK = true;
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for (int i = (n >> 1) - !(n & 1), j = m >> 1, switched = 0; i >= 0; --i, ++j) {
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if ((switched ? a : b)[i] == b[j])
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continue;
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switched = 1;
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if (a[i] == b[j])
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continue;
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OK = false;
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break;
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}
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if (OK) return true;
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}
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// prefix a, base a
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if (n >= m) {
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bool OK = true;
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for (int i = (m >> 1) - !(m & 1), j = m >> 1, switched = 0; i >= 0; --i, ++j) {
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if (((switched |= (j >= m)) ? b : a)[j] == a[i])
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continue;
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switched = 1;
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if (a[i] == b[j])
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continue;
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OK = false;
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break;
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}
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if (OK) return true;
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}
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// prefix b, base b
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if (m >= n) {
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bool OK = true;
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for (int i = (n >> 1) - !(n & 1), j = m >> 1, switched = 0; i >= 0; --i, ++j) {
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if (((switched |= (j >= n)) ? a : b)[j] == b[i])
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continue;
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switched = 1;
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if (b[i] == a[j])
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continue;
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OK = false;
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break;
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}
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if (OK) return true;
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}
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// can NEVER reach here
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return false;
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}
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};
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int main() {
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bool ans = Solution::checkPalindromeFormation(
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"aejbaalflrmkswrydwdkdwdyrwskmrlfqizjezd",
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"uvebspqckawkhbrtlqwblfwzfptanhiglaabjea"
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);
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std::printf(ans ? "true" : "false");
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return 0;
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}
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