Update StringInterleaving.java

Author: 17tanya

DynamicProgramming/StringInterleaving.java

@@ -8,6 +8,9 @@
 
 */
 
+//For only recursion --> Worst Case Time Complexity - O(2^(n+m)) --> When both strings are the same
+//n and m are lengths of the two strings
+
 //MEMOIZED RECUSRION
 public boolean isInterleaving(String X, String Y, String S, HashMap<String, Boolean> map){
     if(X.length() == 0 && Y.length() == 0 && S.length() == 0) return true;
@@ -28,24 +31,43 @@ public boolean isInterleaving(String X, String Y, String S){
 
     if(X.length()+Y.length() != S.length()) return false;
 
+    //T[i][j] indicates if S( 0....(i+j-1) ) is an interleaving of X(0...i) and Y(0...j)
     int T[][] = new int[X.length()+1][Y.length()+1];
 
     for(int i=0 ; i < X.length() ; i++){
         for(int j=0 ; j < Y.length() ; j++){
+            
+            //total length of S to be checked for interleaving when first i charachters from X and first j characters from Y are considered
+            int l = i + j - 1;
+            
+            //is interleaving of two empty strings also an empty string? ==> True
             if(i==0 && j==0) T[i][j] = true;
 
+            //when X is an empty string ==> Only check Y and S
             else if(i==0){
-                if(Y.charAt(j-1) == S.charAt(j-1)) T[i][j] = T[i][j-1];
+                //Check is the current characters match? ==> If True, check the subproblem ==>
+                //==> does Y(0...j-1) form an interleaving in S(0...j-1)
+                //solution to current problem is true only when - 
+                //  1. current characters match
+                //  2. the subproblem also forms an interleaving ie. the subproblem returns true
+                if(Y.charAt(j-1) == S.charAt(l)) T[i][j] = T[i][j-1];
             }
 
             else if(j==0){
-                if(X.charAt(i-1) == S.charAt(i-1)) T[i][j] = T[i-1][j];
+                if(X.charAt(i-1) == S.charAt(l)) T[i][j] = T[i-1][j];
             }
+            //checks if either of the current characters X[i-1] or Y[j-1] matches the current character from interleaved string S[l]
             else{
-                T[i][j] = (X.charAt(i-1) == S.charAt(i-1) ? T[i-1][j] : false) || (Y.charAt(j-1) == S.charAt(j-1) ? T[i][j-1] : false);
+                //If there is a match of current characters, solve the subproblem ie. T[i-1][j] or T[i][j-1]
+                T[i][j] = (X.charAt(i-1) == S.charAt(l) ? T[i-1][j] : false) || (Y.charAt(j-1) == S.charAt(l) ? T[i][j-1] : false);
             }
         }
     }
 
     return T[X.length()][Y.length()];
 }
+
+/*
+Time Complexity and Space Complexity - O(X.length()*Y.length())
+
+*/