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check-if-an-original-string-exists-given-two-encoded-strings.cpp
202 lines (194 loc) · 8.28 KB
/
check-if-an-original-string-exists-given-two-encoded-strings.cpp
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// Time: O(m * n * k), k is the max number of consecutive digits in s1 and s2
// Space: O(m * n * k)
// top-down dp (faster since accessing less states)
class Solution {
public:
bool possiblyEquals(string s1, string s2) {
vector<vector<unordered_map<int, bool>>> lookup(size(s1) + 1, vector<unordered_map<int, bool>>(size(s2) + 1));
return memoization(s1, s2, 0, 0, 0, &lookup);
}
private:
int memoization(const string& s1, const string& s2, int i, int j, int k, vector<vector<unordered_map<int, bool>>> *lookup) {
if (!(*lookup)[i][j].count(k)) {
if (i == size(s1) && j == size(s2)) {
(*lookup)[i][j][k] = (k == 0);
} else if (i != size(s1) && isdigit(s1[i])) {
(*lookup)[i][j][k] = false;
int ni = i + 1;
for (; ni < size(s1); ++ni) {
if (!isdigit(s1[ni])) {
break;
}
}
for (const auto& x : optimized_possible_numbers(s1.substr(i, ni - i))) {
if (memoization(s1, s2, ni, j, k + x, lookup)) {
(*lookup)[i][j][k] = true;
break;
}
}
} else if (j != size(s2) && isdigit(s2[j])) {
(*lookup)[i][j][k] = false;
int nj = j + 1;
for (; nj < size(s2); ++nj) {
if (!isdigit(s2[nj])) {
break;
}
}
for (const auto& x : optimized_possible_numbers(s2.substr(j, nj - j))) {
if (memoization(s1, s2, i, nj, k - x, lookup)) {
(*lookup)[i][j][k] = true;
break;
}
}
} else if (k < 0) {
(*lookup)[i][j][k] = (i != size(s1)) ? memoization(s1, s2, i + 1, j, k + 1, lookup) : false;
} else if (k > 0) {
(*lookup)[i][j][k] = (j != size(s2)) ? memoization(s1, s2, i, j + 1, k - 1, lookup) : false;
} else {
(*lookup)[i][j][k] = (i != size(s1) && j != size(s2) && s1[i] == s2[j]) ? memoization(s1, s2, i + 1, j + 1, k, lookup) : false;
}
}
return (*lookup)[i][j][k];
}
vector<int> optimized_possible_numbers(const string& s) {
assert(size(s) <= 3);
vector<int> result = {stoi(s)};
if (size(s) >= 2) {
if (s[1] != '0') {
result.emplace_back(stoi(s.substr(0, 1 - 0)) + stoi(s.substr(1)));
}
}
if (size(s) >= 3) {
if (s[2] != '0') {
result.emplace_back(stoi(s.substr(0, 2 - 0)) + stoi(s.substr(2)));
if (s[1] != '0') {
result.emplace_back(stoi(s.substr(0, 1 - 0)) + stoi(s.substr(1, 2 - 1)) + stoi(s.substr(2)));
}
}
}
return result; // vector is much faster than unordered_set even if vector has duplications in this case
}
unordered_set<int> general_possible_numbers(const string& s) { // Time: O(2^l), Space: O(2^l), l is the length of consecutive digits s, and l is at most 3
vector<unordered_set<int>> dp(size(s));
for (int i = 0; i < size(s); ++i) {
for (int j = i, curr = 0, basis = 1; j >= 0; --j, basis *= 10) {
curr += (s[j] - '0') * basis;
if (s[j] == '0') {
continue;
}
if (j == 0) {
dp[i].emplace(curr);
} else {
for (const auto& x : dp[j - 1]) {
dp[i].emplace(x + curr);
}
}
}
}
return dp.back();
}
};
// Time: O(m * n * k), k is the max number of consecutive digits in s1 and s2
// Space: O(m * n * k)
// top-down dp (faster since accessing less states)
class Solution2 {
public:
bool possiblyEquals(string s1, string s2) {
vector<vector<unordered_map<int, bool>>> lookup(size(s1) + 1, vector<unordered_map<int, bool>>(size(s2) + 1));
return memoization(s1, s2, 0, 0, 0, &lookup);
}
private:
int memoization(const string& s1, const string& s2, int i, int j, int k, vector<vector<unordered_map<int, bool>>> *lookup) {
if (!(*lookup)[i][j].count(k)) {
if (i == size(s1) && j == size(s2)) {
(*lookup)[i][j][k] = (k == 0);
} else if (i != size(s1) && isdigit(s1[i])) {
(*lookup)[i][j][k] = false;
for (int ni = i + 1; ni <= size(s1); ++ni) {
if ((ni == size(s1) || s1[ni] != '0') && memoization(s1, s2, ni, j, k + stoi(s1.substr(i, ni - i)), lookup)) {
(*lookup)[i][j][k] = true;
break;
}
if (ni == size(s1) || !isdigit(s1[ni])) {
break;
}
}
} else if (j != size(s2) && isdigit(s2[j])) {
(*lookup)[i][j][k] = false;
for (int nj = j + 1; nj <= size(s2); ++nj) {
if ((nj == size(s2) || s2[nj] != '0') && memoization(s1, s2, i, nj, k - stoi(s2.substr(j, nj - j)), lookup)) {
(*lookup)[i][j][k] = true;
break;
}
if (nj == size(s2) || !isdigit(s2[nj])) {
break ;
}
}
} else if (k < 0) {
(*lookup)[i][j][k] = (i != size(s1)) ? memoization(s1, s2, i + 1, j, k + 1, lookup) : false;
} else if (k > 0) {
(*lookup)[i][j][k] = (j != size(s2)) ? memoization(s1, s2, i, j + 1, k - 1, lookup) : false;
} else {
(*lookup)[i][j][k] = (i != size(s1) && j != size(s2) && s1[i] == s2[j]) ? memoization(s1, s2, i + 1, j + 1, k, lookup) : false;
}
}
return (*lookup)[i][j][k];
}
};
// Time: O(m * n * k), k is the max number of consecutive digits in s1 and s2
// Space: O(min(m, n) * k)
// bottom-up dp
class Solution3 {
public:
bool possiblyEquals(string s1, string s2) {
static const int MAX_DIGIT_LEN = 3;
static const int w = 1 + MAX_DIGIT_LEN;
if (size(s1) < size(s2)) {
swap(s1, s2);
}
vector<vector<unordered_set<int>>> dp(w, vector<unordered_set<int>>(size(s2) + 1));
dp[0][0].emplace(0);
for (int i = 0; i <= size(s1); ++i) {
if (i >= 1) {
dp[(i - 1) % w] = vector<unordered_set<int>>(size(s2) + 1);
}
if (i != size(s1) && s1[i] == '0') {
continue;
}
for (int j = 0; j <= size(s2); ++j) {
for (const auto& k : dp[i % w][j]) {
if (i != size(s1) && j != size(s2) && s1[i] == s2[j] && k == 0) {
dp[(i + 1) % w][j + 1].emplace(k);
}
if (k <= 0 && i != size(s1)) {
if (!isdigit(s1[i])) {
if (k) {
dp[(i + 1) % w][j].emplace(k + 1);
}
} else if (s1[i] != '0') {
int curr = 0;
for (int ni = i; ni < size(s1) && isdigit(s1[ni]); ++ni) {
curr = curr * 10 + (s1[ni] - '0');
dp[(ni + 1) % w][j].emplace(k + curr);
}
}
}
if (k >= 0 && j != size(s2)) {
if (!isdigit(s2[j])) {
if (k) {
dp[i % w][j + 1].emplace(k - 1);
}
} else if (s2[j] != '0') {
int curr = 0;
for (int nj = j; nj < size(s2) && isdigit(s2[nj]); ++nj) {
curr = curr * 10 + (s2[nj] - '0');
dp[i % w][nj + 1].emplace(k - curr);
}
}
}
}
}
}
return dp[size(s1) % w][size(s2)].count(0);
}
};