Create 04 - Space Optimized | DP | Approach.cpp

Author: JawadSher

24 - Dynamic Programming Problems/40 - Wildcard Matching/04 - Space Optimized | DP | Approach.cpp

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+class Solution {
+public:
+    // Function to determine if the string matches the pattern using space-optimized dynamic programming
+    int solve(string &str, string &ptrn) {
+        int n = str.length(); // Length of the input string
+        int m = ptrn.length(); // Length of the pattern
+
+        // Two 1D vectors for storing the current and previous rows of the DP table
+        vector<int> prev(m + 1, 0); // Represents the DP state for the previous row
+        vector<int> curr(m + 1, 0); // Represents the DP state for the current row
+
+        // Base case: An empty string matches an empty pattern
+        prev[0] = true;
+
+        // Handle cases where the string is empty but the pattern has characters
+        for (int j = 1; j <= m; j++) {
+            bool flag = true; // Check if all characters in the pattern up to `j` are '*'
+            for (int k = 1; k <= j; k++) {
+                if (ptrn[k - 1] != '*') { // If a non-'*' character is found, stop
+                    flag = false;
+                    break;
+                }
+            }
+
+            // Set `prev[j]` to true if all characters in the pattern are '*', otherwise false
+            prev[j] = flag;
+        }
+
+        // Iterate over the characters in the string
+        for (int i = 1; i <= n; i++) {
+            // Iterate over the characters in the pattern
+            for (int j = 1; j <= m; j++) {
+                // Case 1: Characters match or the pattern has '?'
+                if (str[i - 1] == ptrn[j - 1] || ptrn[j - 1] == '?') {
+                    curr[j] = prev[j - 1]; // Match depends on the previous diagonal value
+                }
+                // Case 2: Pattern has '*'
+                // '*' can match the current character (`prev[j]`) or match zero characters (`curr[j-1]`)
+                else if (ptrn[j - 1] == '*') {
+                    curr[j] = prev[j] || curr[j - 1];
+                }
+                // Case 3: Characters do not match and there's no wildcard
+                else {
+                    curr[j] = false;
+                }
+            }
+
+            // After processing the current row, update `prev` to be the same as `curr`
+            prev = curr;
+        }
+
+        // The result for the full string and pattern is stored in `prev[m]`
+        return prev[m];
+    }
+
+    // Main function to check if the string matches the pattern
+    bool isMatch(string s, string p) {
+        return solve(s, p);
+    }
+};