Wildcard Matching
Problem Statement :
Given an input string (s) and a pattern (p), implement wildcard pattern matching with support for '?' and '*' where: '?' Matches any single character. '*' Matches any sequence of characters (including the empty sequence). The matching should cover the entire input string (not partial). Example 1: Input: s = "aa", p = "a" Output: false Explanation: "a" does not match the entire string "aa". Example 2: Input: s = "aa", p = "*" Output: true Explanation: '*' matches any sequence. Example 3: Input: s = "cb", p = "?a" Output: false Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'. Constraints: 0 <= s.length, p.length <= 2000 s contains only lowercase English letters. p contains only lowercase English letters, '?' or '*'.
Solution :
Solution in C :
bool isMatch(char * s, char * p){
int m = strlen(s), n = strlen(p);
int i = 0, j = 0;
int star_idx = -1, s_idx = -1;
while (i < m) {
if (j < n && (s[i] == p[j] || p[j] == '?')) {
i++;
j++;
} else if (j < n && p[j] == '*') {
star_idx = j;
s_idx = i;
j++;
} else if (star_idx != -1) {
j = star_idx + 1;
i = s_idx + 1;
s_idx++;
} else {
return false;
}
}
while (j < n && p[j] == '*') {
j++;
}
return j == n;
}
Solution in C++ :
class Solution {
public:
bool isMatch(string s, string p) {
int n = s.size();
int m = p.size();
vector<vector<bool>> dp(n+1, vector<bool>(m+1, false));
dp[0][0] = true;
if(p[0] == '*') dp[0][1] = true;
for(int j=2; j<=m; j++){
if(p[j-1] == '*') dp[0][j] = dp[0][j-1];
}
for(int i=1; i<=n; i++){
for(int j=1; j<=m; j++){
if(s[i-1] == p[j-1] || p[j-1] == '?'){
dp[i][j] = dp[i-1][j-1];
}
else{
if(p[j-1] == '*'){
dp[i][j] = dp[i-1][j] | dp[i][j-1] ;
}
}
}
}
return dp[n][m];
}
};
Solution in Java :
// TC -> O(|s|*|p|)
// SC -> O(|s|*|p|)
class Solution {
public boolean isMatch(String s, String p) {
return solve(s,p,0,0,new Boolean[s.length()][p.length()]);
}
private boolean solve(String s,String t,int i,int j,Boolean[][] dp){
if(i==s.length()){
return hasNoLCLeft(t,j);
}
if(j==t.length()){
return i==s.length();
}
if(dp[i][j]!=null){
return dp[i][j];
}
if(Character.isLowerCase(t.charAt(j)) && s.charAt(i)!=t.charAt(j)){
return dp[i][j] = false;
}
if(t.charAt(j)=='*'){
if(solve(s,t,i+1,j,dp)){
return true;
}
if(solve(s,t,i,j+1,dp)){
return true;
}
}
return dp[i][j] = solve(s,t,i+1,j+1,dp)? true: false;
}
private boolean hasNoLCLeft(String t,int j){
for(int i=j;i<t.length();i++){
if(t.charAt(i)!='*'){
return false;
}
}
return true;
}
}
Solution in Python :
# DP
def isMatch(self, s: str, p: str) -> bool:
dp = [[False]*(len(p)+1) for i in range(len(s)+1)]
dp[0][0] = True
for j in range(1,len(p)+1):
if p[j-1] == '*':
dp[0][j] = dp[0][j-1]
for i in range(1,len(s)+1):
for j in range(1,len(p)+1):
dp[i][j] = (p[j-1] in [s[i-1],'?','*'] and dp[i-1][j-1]) or (p[j-1] == '*' and (dp[i][j-1] or dp[i-1][j]))
return dp[len(s)][len(p)]
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