replace dsa code

Author: trungIsOnGithhub

java_datastructures_algorithms/dynamic_programming/code2_longest_increasing_subsequence.java

@@ -1,64 +1,25 @@
-/* A Naive Java Program for LIS Implementation */
-class LIS
-{
-static int max_ref; // stores the LIS
-
-/* To make use of recursive calls, this function must return
-two things:
-1) Length of LIS ending with element arr[n-1]. We use
-	max_ending_here for this purpose
-2) Overall maximum as the LIS may end with an element
-	before arr[n-1] max_ref is used this purpose.
-The value of LIS of full array of size n is stored in
-*max_ref which is our final result */
-static int _lis(int arr[], int n)
-{
-	// base case
-	if (n == 1)
-		return 1;
-
-	// 'max_ending_here' is length of LIS ending with arr[n-1]
-	int res, max_ending_here = 1;
-
-		/* Recursively get all LIS ending with arr[0], arr[1] ...
-		arr[n-2]. If arr[i-1] is smaller than arr[n-1], and
-		max ending with arr[n-1] needs to be updated, then
-		update it */
-		for (int i = 1; i < n; i++)
-		{
-			res = _lis(arr, i);
-			if (arr[i-1] < arr[n-1] && res + 1 > max_ending_here)
-				max_ending_here = res + 1;
-		}
-
-		// Compare max_ending_here with the overall max. And
-		// update the overall max if needed
-		if (max_ref < max_ending_here)
-		max_ref = max_ending_here;
-
-		// Return length of LIS ending with arr[n-1]
-		return max_ending_here;
-}
-
-	// The wrapper function for _lis()
-	static int lis(int arr[], int n)
-	{
-		// The max variable holds the result
-		max_ref = 1;
-
-		// The function _lis() stores its result in max
-		_lis( arr, n);
-
-		// returns max
-		return max_ref;
-	}
-
-	// driver program to test above functions
-	public static void main(String args[])
-	{
-		int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
-		int n = arr.length;
-		System.out.println("Length of LIS is "
-						+ lis(arr, n) + "n");
-	}
+class Solution {
+	// problem statement and judging: https://leetcode.com/problems/longest-increasing-subsequence
+    public int lengthOfLIS(int[] nums) {
+        int len = nums.length;
+
+        int[] dp = new int[len];
+        
+        for(int i=0; i<len; ++i) { dp[i] = 1; }
+        
+        for(int i=0; i<len; ++i) {
+            for(int j=0; j<i; ++j) {
+                if( nums[j] < nums[i] ) {
+                    dp[i] = Math.max(dp[i], dp[j]+1);
+                }
+            }
+        }
+        
+        int max_dp = 1;
+        for(int i=1; i<len; ++i) {
+            max_dp = Math.max(dp[i], max_dp);
+        }
+        
+        return max_dp;
+    }
 }

java_datastructures_algorithms/dynamic_programming/code4_edit_distance.java

@@ -1,45 +1,27 @@
-// A Naive recursive Java solution for Edit Distance
-// to find minimum number operations to convert str1 to str2
-class EDIST
-{
-	static int min(int x,int y,int z)
-	{
-		if (x<=y && x<=z) return x;
-		if (y<=x && y<=z) return y;
-		else return z;
-	}
+class Solution {
+	// problem statement and judging: https://leetcode.com/problems/edit-distance
+    public int minDistance(String word1, String word2) {
+        int n1 = word1.length(), n2 = word2.length();
 
-	static int editDist(String str1 , String str2 , int m ,int n)
-	{
-		// If first string is empty, the only option is to
-	// insert all characters of second string into first
-	if (m == 0) return n;
-	
-	// If second string is empty, the only option is to
-	// remove all characters of first string
-	if (n == 0) return m;
-	
-	// If last characters of two strings are same, nothing
-	// much to do. Ignore last characters and get count for
-	// remaining strings.
-	if (str1.charAt(m-1) == str2.charAt(n-1))
-		return editDist(str1, str2, m-1, n-1);
-	
-	// If last characters are not same, consider all three
-	// operations on last character of first string, recursively
-	// compute minimum cost for all three operations and take
-	// minimum of three values.
-	return 1 + min ( editDist(str1, str2, m, n-1), // Insert
-					editDist(str1, str2, m-1, n), // Remove
-					editDist(str1, str2, m-1, n-1) // Replace					 
-				);
-	}
+        if(n1==0)
+            return n2;
+        if(n2==0)
+            return n1;
 
-	public static void main(String args[])
-	{
-		String str1 = "sunday";
-		String str2 = "saturday";
+        int dp[][]=new int[n1+1][n2+1];
 
-		System.out.println( editDist( str1 , str2 , str1.length(), str2.length()) );
-	}
+        for(int i=0;i<=n1;i++){
+            for(int j=0;j<=n2;j++){
+                if( i==0 )
+					dp[i][j] = j;
+                else if( j==0 )
+					dp[i][j]=i;
+                else if( word1.charAt(i-1) == word2.charAt(j-1) )
+					dp[i][j] = dp[i-1][j-1];
+                else
+					dp[i][j] = Math.min( dp[i-1][j-1], Math.min( dp[i-1][j],dp[i][j-1] ) ) + 1;
+            }
+        }
+        return dp[n1][n2];
+    }
 }