String Merging #JC2018

CP/January Challenge 2018/string.cpp

@@ -0,0 +1,85 @@
+#include<bits/stdc++.h>
+using namespace std;
+typedef long long ll;
+/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
+ll lcs( string X, string Y, ll m, ll n )
+{
+   int L[m+1][n+1];
+   int i, j;
+
+   /* Following steps build L[m+1][n+1] in bottom up fashion. Note
+      that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */
+   for (i=0; i<=m; i++)
+   {
+     for (j=0; j<=n; j++)
+     {
+       if (i == 0 || j == 0)
+         L[i][j] = 0;
+
+       else if (X[i-1] == Y[j-1])
+         L[i][j] = L[i-1][j-1] + 1;
+
+       else
+         L[i][j] = max(L[i-1][j], L[i][j-1]);
+     }
+   }
+
+   /* L[m][n] contains length of LCS for X[0..n-1] and Y[0..m-1] */
+   return L[m][n];
+}
+
+char * removeDuplicates(char S[]){
+
+   int n = strlen(S);
+
+   // We don't need to do anything for
+   // empty or single character string.
+   if (n < 2)
+     return S;
+
+   // j is used to store index is result
+   // string (or index of current distinct
+   // character)
+   int j = 0;
+
+   // Traversing string
+   for (int i=1; i<n; i++)
+   {
+       // If current character S[i]
+       // is different from S[j]
+       if (S[j] != S[i])
+       {
+           j++;
+           S[j] = S[i];
+       }
+   }
+
+   // Putting string termination
+   // character.
+   j++;
+   S[j] = '\0';
+   return S;
+}
+
+int main()
+{
+	ll t,n,m,i,x,y;
+	cin >> t;
+	while(t--)
+	{
+		cin >> n >> m;
+		char a[n],b[m];
+    string a1,b1;
+		scanf("%s",a);
+		scanf("%s",b);
+    a1 = removeDuplicates(a); //cout<<removeDuplicates(a)<<endl;
+    b1 = removeDuplicates(b); //cout<<removeDuplicates(b)<<endl;
+    x = a1.length();
+		y = b1.length();
+    //cout << x << " "<< y;
+		ll l = lcs(a1,b1,x,y);
+		//cout<< l;
+		printf("%lld\n",(x+y-l));
+	}
+	return 0;
+}

CP/January Challenge 2018/string.py

@@ -1,19 +1,54 @@
-def longest_common_subsequence(sequence1, sequence2):
-    cols = len(sequence1) + 1   # Add 1 to represent 0 valued column for DP
-    rows = len(sequence2) + 1   # Add 1 to represent 0 valued row for DP
+def lcs(X , Y):
+    # find the length of the strings
+    m = len(X)
+    n = len(Y)
 
-    T = [[0 for _ in range(cols)] for _ in range(rows)]
+    # declaring the array for storing the dp values
+    L = [[None]*(n+1) for i in range(m+1)]
 
-    max_length = 0
-
-    for i in range(1, rows):
-        for j in range(1, cols):
-            if sequence2[i - 1] == sequence1[j - 1]:
-                T[i][j] = 1 + T[i - 1][j - 1]
+    """Following steps build L[m+1][n+1] in bottom up fashion
+    Note: L[i][j] contains length of LCS of X[0..i-1]
+    and Y[0..j-1]"""
+    for i in range(m+1):
+        for j in range(n+1):
+            if i == 0 or j == 0 :
+                L[i][j] = 0
+            elif X[i-1] == Y[j-1]:
+                L[i][j] = L[i-1][j-1]+1
             else:
-                T[i][j] = max(T[i - 1][j], T[i][j - 1])
-
-            max_length = max(max_length, T[i][j])
+                L[i][j] = max(L[i-1][j] , L[i][j-1])
 
-    return max_length
+    # L[m][n] contains the length of LCS of X[0..n-1] & Y[0..m-1]
+    return L[m][n]
 
+for _ in range(int(input())):
+	n,m = input().split()
+	n,m = int(n),int(m)
+	a = input()
+	b = input()
+	i=1
+	pre=a[0]
+	c=""
+	c+=pre
+	while i<n:
+		if a[i]!=pre:
+			 pre = a[i]
+			 c += pre
+		i+=1
+	a=c
+	i=1
+	pre=b[0]
+	c=""
+	c=pre
+	while i<m:
+		if b[i]!=pre:
+			 pre = b[i]
+			 c += pre
+		i+=1
+	b=c
+	n=len(a)
+	m=len(b)
+	lc = lcs(a,b)
+	f1 = n-lc
+	f1+= (m-lc)+lc
+	print(f1)

dynamic/longest_common_substring.py

@@ -5,10 +5,6 @@
 Given two sequences A = [A1, A2, A3,..., An] and B = [B1, B2, B3,..., Bm], find the length of the longest common
 substring.
 
-Video
------
-
-* https://youtu.be/BysNXJHzCEs
 
 Complexity
 ----------
@@ -77,4 +73,3 @@ def longest_common_substring(str1, str2):
     str2 = "zcdemf"
     expected = 3
     assert expected == longest_common_substring(str1, str2)
-