🚀 31-Jul-2020

Author: sureshmangs

competitive programming/leetcode/2020-July-Challenge/Day-30-Word Break II.cpp

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+Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.
+
+Note:
+
+The same word in the dictionary may be reused multiple times in the segmentation.
+You may assume the dictionary does not contain duplicate words.
+Example 1:
+
+Input:
+s = "catsanddog"
+wordDict = ["cat", "cats", "and", "sand", "dog"]
+Output:
+[
+  "cats and dog",
+  "cat sand dog"
+]
+Example 2:
+
+Input:
+s = "pineapplepenapple"
+wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
+Output:
+[
+  "pine apple pen apple",
+  "pineapple pen apple",
+  "pine applepen apple"
+]
+Explanation: Note that you are allowed to reuse a dictionary word.
+Example 3:
+
+Input:
+s = "catsandog"
+wordDict = ["cats", "dog", "sand", "and", "cat"]
+Output:
+[]
+
+
+
+
+
+
+
+
+
+string s;int n;
+unordered_map<string, int> mp;
+
+bool cut(int start, int end, string a,vector<vector<char> >& dp)
+{
+    if(start > end)
+        return 1;
+
+    if(dp[start][end] != '?')
+    {
+        if(dp[start][end] == '1')
+            return 1;
+        return 0;
+    }
+    string ans;
+
+    for(int i = start; i <= end; i++)
+    {
+        ans += a[i];
+        if(mp.find(ans) != mp.end())
+            if(cut(i+1, end, a, dp))
+            {
+                dp[start][end] = '1';
+                return 1;
+            }
+    }
+    dp[start][end] = '0';
+    return 0;
+}
+
+void backtrack(string s, vector<string> &ans, string yet, int idx)
+{
+	if(idx == n)
+	{
+		yet.pop_back();
+		ans.push_back(yet);
+		return;
+	}
+	string temp = "";
+	for(int i = idx; i < n; i++)
+	{
+		temp += s[i];
+		if(mp.find(temp) != mp.end())
+		{
+			string ss = yet;
+			ss += temp;
+			ss += " ";
+			backtrack(s, ans, ss, i+1);
+		}
+	}
+}
+
+class Solution {
+public:
+    vector<string> wordBreak(string A, vector<string>& B) {
+        vector<string> ans;
+        mp.clear();
+        s = A;
+        n = A.size();
+        for(int i = 0; i < B.size(); i++)
+            mp[B[i]] = 1;
+        vector<vector<char> > dp(n, vector<char>(n, '?'));
+        int x = cut(0, n-1, A, dp);
+        if(x == 0)
+        	return ans;
+        backtrack(A, ans, "", 0);
+        return ans;
+    }
+};