Standard dynamic programming problems

0-1 Knapsack.cpp

@@ -0,0 +1,21 @@
+int dp[1001][1001];
+int knap(vector<int> &A, vector<int> &B, int n,int C)
+{
+    if(n==0)
+        return 0;
+    if(dp[n][C]!=-1)
+        return dp[n][C];
+    if(B[n-1]<=C)
+    {
+        return dp[n][C]=max(A[n-1]+knap(A,B,n-1,C-B[n-1]),knap(A,B,n-1,C));
+    }
+    else
+        return dp[n][C]=knap(A,B,n-1,C);
+}
+int Solution::solve(vector<int> &A, vector<int> &B, int C) {
+    int n=A.size();
+    for(int i=0;i<=n;i++)
+        for(int j=0;j<=C;j++)
+            dp[i][j]=-1;
+    return knap(A,B,n,C);
+}

Edit distance.cpp

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+int Solution::minDistance(string A, string B) {
+    vector<vector<int>>t(B.size()+1,vector<int>(A.size()+1,-1));
+    int n=B.size(),m=A.size();
+    for(int i=0;i<n+1;i++)
+    for(int j=0;j<m+1;j++)
+    {
+        if(i==0)
+            t[i][j]=j;
+        else if(j==0)
+            t[i][j]=i;
+        else if(A[j-1]==B[i-1])
+        {
+            t[i][j]=t[i-1][j-1];
+        }
+        else
+        {
+            t[i][j]=1+min(t[i-1][j],min(t[i][j-1],t[i-1][j-1]));
+        }
+    }
+    return t[n][m];
+}

Longest increasing subsequence.cpp

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+int Solution::lis(const vector<int> &A) {
+    int n=A.size();
+    int ma=1;
+    if(n==1||n==0)
+        return n;
+    vector<int>inc(n,1);
+    for(int k=0;k<n;k++)
+    {
+        for(int i=0;i<k;i++)
+        {
+            if(A[i]<A[k])
+            {
+                inc[k]=max(inc[k],inc[i]+1);
+                ma=max(ma,inc[k]);
+            }
+        }
+    }
+    return ma;
+}

Longest palindromic subsequence.cpp

@@ -0,0 +1,21 @@
+int dp[1006][1006];
+int solve1(string A,int n,int m)
+{
+    if(n==m)
+        return 1;
+    if(n>m)
+        return 0;
+    if(dp[n][m]!=-1)
+        return dp[n][m];
+    if(A[n-1]==A[m-1])
+        return dp[n][m]=2+solve1(A,n+1,m-1);
+    else
+        return dp[n][m]=max(solve1(A,n+1,m),solve1(A,n,m-1));
+}
+int Solution::solve(string A) {
+    int n=A.length(),m=A.length();
+    for(int i=0;i<=n;i++)
+        for(int j=0;j<=m;j++)
+            dp[i][j]=-1;
+    return solve1(A,1,m);
+}