Day 26: q1

Day-26/q1: Palindrome Partitioning II/Tech-neophyte--cp.md

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+## Approach 1: Using recursion
+## cpp code
+```
+class Recursion {
+private:
+    int n;
+    bool isPalindrome(string& s, int l, int r)
+    {
+        while(l < r)
+        {
+            if(s[l++] != s[r--])
+            {
+                return false;
+            }
+        }
+        return true;
+    }
+    int solve(string& s, int idx)
+    {
+        if(idx >= n)
+        {
+            return 0;
+        }
+        int minSteps = INT_MAX;
+        for(int k=idx; k<n; k++)
+        {
+            if(isPalindrome(s, idx, k))
+            {
+                int steps = 1 + solve(s, k+1);
+                minSteps = min(minSteps, steps);
+            }
+        }
+        return minSteps;
+    }
+public:
+    int minCut(string s) {
+        n = s.size();
+        return solve(s, 0) - 1;
+    }
+};
+```
+## Approach 2: Using dp and memorisatiom
+<br /> Step 1: Take all substrings and check that it is palindrome or not and store in check matrix.
+<br /> Step 2: For each i, store all indexes such that s[i]....s[j] is a palindrome.
+<br /> Step 3: Now apply dp with memoisation to find minimum partitions.
+## CPP code
+```
+class Solution {
+public:
+    vector<int> adj[2005];
+    int dp[2005];
+    int help(int i,int n){
+        if(i==n){
+            return 0;
+        }
+        if(dp[i]!=-1){
+            return dp[i];
+        }
+        int ans=1e9;
+        for(auto x:adj[i]){
+            int tmp=1+help(x+1,n);
+            ans=min(ans,tmp);
+        }
+        return dp[i]=ans;
+    }
+    int minCut(string s) {
+        int n=s.size();
+        int check[n][n];      
+        for(int g=0;g<n;g++){
+            for(int i=0,j=g;j<n;i++,j++){
+                if(g==0){
+                    check[i][j]=1;
+                }
+                else if(g==1){
+                    check[i][j]=(s[i]==s[j]);
+                }
+                else{
+                    check[i][j]=(check[i+1][j-1]==1 ? (s[i]==s[j] ? 1 : 0) : 0);
+                }
+            }
+        }
+        for(int i=0;i<n;i++){
+            for(int j=i;j<n;j++){
+                if(check[i][j]){
+                    adj[i].push_back(j);
+                }
+            }
+        }
+
+        memset(dp,-1,sizeof(dp));
+        return help(0,n)-1;       
+    }  
+};
+```
+## Python code:
+```
+# python code
+
+class Solution(object):
+    def minCut(self, s):
+        """
+        s="abcgcbafj"
+        :type s: str
+        :rtype: int
+        """
+
+        if s == s[::-1]:
+            return 0
+
+        for i in range(len(s)):
+            if s[:i] == s[:i][::-1] and s[i:] == s[i:][::-1]:
+                return 1
+
+        l=len(s)
+
+        #d=[[0 for i  in range(l)] for j in range(l)]
+        x=[[0 for i  in range(l)] for j in range(l)]
+        for i in range(l):
+          for j in range(i,l):
+            st=s[i:j+1]
+            #d[i][j]=st
+            x[i][j]= (st==st[::-1])
+
+
+        p=[0 for c in range(l)]
+        for i in range(1,l):
+          if x[0][i]:
+            p[i]=0
+          else:
+            m=float("inf")
+            for j in range(i,0,-1):
+              if x[j][i]: 
+                m=min(m,p[j-1])
+            p[i]=m+1
+        return p[-1]
+
+```