# Wildcard Matching | (DP-34)

Problem Statement: Wildcard Matching

We are given two strings ‘S1’ and ‘S2’. String S1 can have the following two special characters:

1. ‘?’ can be matched to a single character of S2.
2. ‘*’ can be matched to any sequence of characters of S2. (sequence can be of length zero or more).

We need to check whether strings S1 and S2 match or not.

```Example:  ```

Disclaimer: Don’t jump directly to the solution, try it out yourself first.

### Intuition:

For every index of string S1, we have different options to match that index with string S2. Therefore, we can think in terms of string matching path as we have done already in previous questions.

• Either the characters match already.
• Or, if there is a ‘?’, we can explicitly match a single character.
• For a ‘*’, the following figure explains the scenario. As there is no uniformity in data, there is no other way to find out than to try out all possible ways. To do so we will need to use recursion.

Steps to form the recursive solution:

We will first form the recursive solution by the three points mentioned in the Dynamic Programming Introduction

Step 1: Express the problem in terms of indexes.

We are given two strings. We can represent them with the help of two indexes i and j. Initially, i=n-1 and j=m-1, where n and m are lengths of strings S1 and S2. Initially, we will call f(n-1,m-1), which means whether string S1[0…n-1] matches with string S2[0…m-1].

We can generalize this as follows: Step 2: Try out all possible choices at a given index.

Now, i and j represent two characters from strings S1 and S2 respectively. There are only two options that make sense: either the characters represented by i and j match or they don’t.

(i) When the characters match

if(S1[i]==S2[j])

If this is true, the characters at i and j match, we can simply move to the next characters of both the strings. So we will just decrement both i and j by 1 and recursively find the answer for the remaining string portions. We return f(i-1,j-1). The following figure makes it clear. (ii) When the characters don’t match

If the characters don’t match, there are three possible scenarios:

1. S1[i] == ‘?’
2. S1[i] == ‘*’
3. S1[i] is some other character

Let us discuss them one by one:

(i) If S1[i] == ‘?’

In this case, we can explicitly match ‘?’ at index i of S1 with the corresponding character at index j of S2. And then recursively call f(i-1,j-1) to check for the remaining string.

(ii) If S1[i] == ‘*’

This is an interesting case as now ‘*’ can be replaced with any sequence of characters( of length 0 or more) from S2.

We will revisit this example: If any of these cases return true, we can say that the characters do match. The next question is how to try all possible ways?

We are using two pointers i and j to represent characters of strings S1 and S2. We can surely write a for loop to compare characters from 0 to j of S2 for the above scenario. Can we do it more smartly? Yes, we can. Please understand the approach explained below.

We are using a recursive function f(i,j). If we do only the following two recursive calls:

• Call f(i-1,j). i.e replace ‘*’ with nothing and act as if it was not present.
• Call f(i,j-1). i.e replace ‘*’ with a single character at index j and make the i pointer to still point at index i. In this, we matched it with a single character (one of the many options that need to be tried) and in the next recursive call, as i still point to ‘*’, we get the exact two recursive calls again.

The following recursive tree will help us to understand the recursion better. So we see how we can tackle all the required cases associated with ‘*’ by using recursion.

(iii) If S1[i] is neither ‘?’ nor ‘*’, then we can say as the characters at i and j don’t match then the strings don’t match, so we return false.

To summarise:

1. If S1[i] == ‘?’, return f(i-1,j)
2. Else if S1[i] == ‘*’, return f(i-1,j) || f(i,j-1)
3. Else return false

Step 3: Return logical OR (||) of all the choices

If any of the cases return true, we can say that strings do match. We can use OR operator (||) with the recursive calls.

Base Cases:

We are reducing i and j in our recursive relation, there can be two possibilities, either i becomes -1 or j becomes -1., i,e we exhaust either S1 or S2 respectively.

(i) When S1 is exhausted:

When S1 is exhausted (i<0), we know that in order for the strings to match, String S2 should also exhaust simultaneously. If it does, we return true, else we return false.

We can say:

• if(i<0 && j<0), return true.
• if(i<0 && j>=0), return false.

(ii) When S2 is exhausted:

When S2 is exhausted(j<0) and S1 has not, there is only one pattern that can account for true(matching of strings). It is if S1 is like this “*”,”****”,”***”, i.e: S1 contains only stars. Then we can replace every star with a sequence of length 0 and say that the string match.

If S1 is all-stars, we return true, else return false.

The final pseudocode after steps 1, 2, and 3: Steps to memoize a recursive solution:

If we draw the recursion tree, we will see that there are overlapping subproblems. In order to convert a recursive solution the following steps will be taken:

1. Create a dp array of size [n][m]. The size of S1 and S2 are n and m respectively, so the variable i will always lie between ‘0’ and ‘n-1’ and the variable j between ‘0’ and ‘m-1’.
2. We initialize the dp array to -1.
3. Whenever we want to find the answer to particular parameters (say f(i,j)), we first check whether the answer is already calculated using the dp array(i.e dp[i][j]!= -1 ). If yes, simply return the value from the dp array.
4. If not, then we are finding the answer for the given value for the first time, we will use the recursive relation as usual but before returning from the function, we will set dp[i][j] to the solution we get.

Code:

## C++ Code

``````#include <bits/stdc++.h>

using namespace std;

bool isAllStars(string & S1, int i) {
for (int j = 0; j <= i; j++) {
if (S1[j] != '*')
return false;
}
return true;
}

bool wildcardMatchingUtil(string & S1, string & S2, int i, int j, vector < vector < bool >> & dp) {

//Base Conditions
if (i < 0 && j < 0)
return true;
if (i < 0 && j >= 0)
return false;
if (j < 0 && i >= 0)
return isAllStars(S1, i);

if (dp[i][j] != -1) return dp[i][j];

if (S1[i] == S2[j] || S1[i] == '?')
return dp[i][j] = wildcardMatchingUtil(S1, S2, i - 1, j - 1, dp);

else {
if (S1[i] == '*')
return wildcardMatchingUtil(S1, S2, i - 1, j, dp) || wildcardMatchingUtil(S1, S2, i, j - 1, dp);
else return false;
}
}

bool wildcardMatching(string & S1, string & S2) {

int n = S1.size();
int m = S2.size();

vector < vector < bool >> dp(n, vector < bool > (m, -1));
return wildcardMatchingUtil(S1, S2, n - 1, m - 1, dp);

}

int main() {

string S1 = "ab*cd";
string S2 = "abdefcd";

if (wildcardMatching(S1, S2))
cout << "String S1 and S2 do match";
else cout << "String S1 and S2 do not match";
}``````

Output:

String S1 and S2 do match

Time Complexity: O(N*M)

Reason: There are N*M states therefore at max ‘N*M’ new problems will be solved.

Space Complexity: O(N*M) + O(N+M)

Reason: We are using a recursion stack space(O(N+M)) and a 2D array ( O(N*M)).

## Java Code

``````import java.util.*;

class TUF {
static boolean isAllStars(String S1, int i) {
for (int j = 0; j <= i; j++) {
if (S1.charAt(j) != '*')
return false;
}
return true;
}

static int wildcardMatchingUtil(String S1, String S2, int i, int j, int[][] dp) {

//Base Conditions
if (i < 0 && j < 0)
return 1;
if (i < 0 && j >= 0)
return 0;
if (j < 0 && i >= 0)
return isAllStars(S1, i) ? 1 : 0;

if (dp[i][j] != -1) return dp[i][j];

if (S1.charAt(i) == S2.charAt(j) || S1.charAt(i) == '?')
return dp[i][j] = wildcardMatchingUtil(S1, S2, i - 1, j - 1, dp);

else {
if (S1.charAt(i) == '*')
return (wildcardMatchingUtil(S1, S2, i - 1, j, dp) == 1 || wildcardMatchingUtil(S1, S2, i, j - 1, dp) == 1) ? 1 : 0;
else return 0;
}
}

static int wildcardMatching(String S1, String S2) {

int n = S1.length();
int m = S2.length();

int dp[][] = new int[n][m];
for (int row[]: dp)
Arrays.fill(row, -1);
return wildcardMatchingUtil(S1, S2, n - 1, m - 1, dp);

}

public static void main(String args[]) {

String S1 = "ab*cd";
String S2 = "abdefcd";

if (wildcardMatching(S1, S2) == 1)
System.out.println("String S1 and S2 do match");
else System.out.println("String S1 and S2 do not match");
}
}``````

Output:

Output:

String S1 and S2 do match

Time Complexity: O(N*M)

Reason: There are N*M states therefore at max ‘N*M’ new problems will be solved.

Space Complexity: O(N*M) + O(N+M)

Reason: We are using a recursion stack space(O(N+M)) and a 2D array ( O(N*M)).

Steps to convert Recursive Solution to Tabulation one.

In the recursive logic, we set the base case too if(i<0 ) and if(j<0) but we can’t set the dp array’s index to -1. Therefore a hack for this issue is to shift every index by 1 towards the right. • First we initialise the dp array of size [n+1][m+1] as zero.
• Next, we set the base condition (keep in mind 1-based indexing), we set the top-left cell as ‘true’, then we set the first column’s value as ‘false’; and for the first row, we will run isAllStars() for every cell value.
• Similarly, we will implement the recursive code by keeping in mind the shifting of indexes, therefore S1[i] will be converted to S1[i-1]. Same for S2.
• At last we will print dp[n][m] as our answer.

Code:

## C++ Code

``````#include <bits/stdc++.h>

using namespace std;

bool isAllStars(string & S1, int i) {

// S1 is taken in 1-based indexing
for (int j = 1; j <= i; j++) {
if (S1[j - 1] != '*')
return false;
}
return true;
}

bool wildcardMatching(string & S1, string & S2) {

int n = S1.size();
int m = S2.size();

vector < vector < bool >> dp(n + 1, vector < bool > (m, false));

dp = true;

for (int j = 1; j <= m; j++) {
dp[j] = false;
}
for (int i = 1; i <= n; i++) {
dp[i] = isAllStars(S1, i);
}

for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {

if (S1[i - 1] == S2[j - 1] || S1[i - 1] == '?')
dp[i][j] = dp[i - 1][j - 1];

else {
if (S1[i - 1] == '*')
dp[i][j] = dp[i - 1][j] || dp[i][j - 1];

else dp[i][j] = false;
}
}
}

return dp[n][m];

}

int main() {

string S1 = "ab*cd";
string S2 = "abdefcd";

if (wildcardMatching(S1, S2))
cout << "String S1 and S2 do match";
else cout << "String S1 and S2 do not match";
}``````

Output:

String S1 and S2 do match

Time Complexity: O(N*M)

Reason: There are two nested loops

Space Complexity: O(N*M)

Reason: We are using an external array of size ‘N*M’. Stack Space is eliminated.

## Java Code

``````import java.util.*;

class TUF {
static boolean isAllStars(String S1, int i) {

// S1 is taken in 1-based indexing
for (int j = 1; j <= i; j++) {
if (S1.charAt(j - 1) != '*')
return false;
}
return true;
}

static boolean wildcardMatching(String S1, String S2) {

int n = S1.length();
int m = S2.length();

boolean dp[][] = new boolean[n + 1][m + 1];
dp = true;

for (int j = 1; j <= m; j++) {
dp[j] = false;
}
for (int i = 1; i <= n; i++) {
dp[i] = isAllStars(S1, i);
}

for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {

if (S1.charAt(i - 1) == S2.charAt(j - 1) || S1.charAt(i - 1) == '?')
dp[i][j] = dp[i - 1][j - 1];

else {
if (S1.charAt(i - 1) == '*')
dp[i][j] = dp[i - 1][j] || dp[i][j - 1];

else dp[i][j] = false;
}
}
}

return dp[n][m];

}

public static void main(String args[]) {

String S1 = "ab*cd";
String S2 = "abdefcd";

if (wildcardMatching(S1, S2))
System.out.println("String S1 and S2 do match");
else System.out.println("String S1 and S2 do not match");
}
}``````

Output:

String S1 and S2 do match

Time Complexity: O(N*M)

Reason: There are two nested loops

Space Complexity: O(N*M)

Reason: We are using an external array of size ‘N*M’. Stack Space is eliminated.

Part 3: Space Optimization

If we closely look the relation,

dp[i][j] =  dp[i-1][j-1], dp[i][j] = dp[i-1][j] ||dp[i][j-1]

We see that to calculate a value of a cell of the dp array, we need only the previous row values (say prev) and current row’s previous columns values. So, we don’t need to store an entire array. Hence we can space optimise it.

Approach:

We will space optimize in the following way:

• We take two rows ‘prev’ and ‘cur’.
• We initialize it to the base condition. We first initialize the prev row. Its first value needs to be true. Rest all the values of the prev row meeds to be false.
• Moreover, the cur variable whenever declared should have its first cell’s value given by isAllStarts() function.
• Next, we implement the memoization logic. We replace dp[i-1] with prev and dp[i] by cur.
• After every inner loop execution, we set prev=cur, for the next iteration.
• At last, we return prev[m] as our answer.

Code:

## C++ Code

``````#include <bits/stdc++.h>

using namespace std;

bool isAllStars(string & S1, int i) {

// S1 is taken in 1-based indexing
for (int j = 1; j <= i; j++) {
if (S1[j - 1] != '*')
return false;
}
return true;
}

bool wildcardMatching(string & S1, string & S2) {

int n = S1.size();
int m = S2.size();

vector < bool > prev(m + 1, false);
vector < bool > cur(m + 1, false);

prev = true;

for (int i = 1; i <= n; i++) {
cur = isAllStars(S1, i);
for (int j = 1; j <= m; j++) {

if (S1[i - 1] == S2[j - 1] || S1[i - 1] == '?')
cur[j] = prev[j - 1];

else {
if (S1[i - 1] == '*')
cur[j] = prev[j] || cur[j - 1];

else cur[j] = false;
}
}
prev = cur;
}

return prev[m];

}

int main() {

string S1 = "ab*cd";
string S2 = "abdefcd";

if (wildcardMatching(S1, S2))
cout << "String S1 and S2 do match";
else cout << "String S1 and S2 do not match";
}``````

Output:

String S1 and S2 do match

Time Complexity: O(N*M)

Reason: There are two nested loops.

Space Complexity: O(M)

Reason: We are using an external array of size ‘M+1’ to store two rows.

## Java Code

``````import java.util.*;

class TUF {
static boolean isAllStars(String S1, int i) {

// S1 is taken in 1-based indexing
for (int j = 1; j <= i; j++) {
if (S1.charAt(j - 1) != '*')
return false;
}
return true;
}

static boolean wildcardMatching(String S1, String S2) {

int n = S1.length();
int m = S2.length();

boolean prev[] = new boolean[m + 1];
boolean cur[] = new boolean[m + 1];

prev = true;

for (int i = 1; i <= n; i++) {
cur = isAllStars(S1, i);
for (int j = 1; j <= m; j++) {

if (S1.charAt(i - 1) == S2.charAt(j - 1) || S1.charAt(i - 1) == '?')
cur[j] = prev[j - 1];

else {
if (S1.charAt(i - 1) == '*')
cur[j] = prev[j] || cur[j - 1];

else cur[j] = false;
}
}
prev = (boolean[]) cur.clone();
}

return prev[m];

}

public static void main(String args[]) {

String S1 = "ab*cd";
String S2 = "abdefcd";

if (wildcardMatching(S1, S2))
System.out.println("String S1 and S2 do match");
else System.out.println("String S1 and S2 do not match");
}
}``````

Output:

String S1 and S2 do match

Time Complexity: O(N*M)

Reason: There are two nested loops.

Space Complexity: O(M)

Reason: We are using an external array of size ‘M+1’ to store two rows.