Merge pull request knaxus#6 from navneet15/branch1

Permutation & Subsequence of String and all paths of M x N grid

Author: ashokdey

src/_Problems_/get-mazePath/index.js

@@ -0,0 +1,42 @@
+//======================================Problem Statement=============================================
+// --->> Print all possible path to reach the end of the GRID/MAZE (N x N) from starting  point to ending point
+// --->> One horizontal move will be represented by H and one vertical move will be represented by V
+// --->> Complexity =  Complexity will be exponential as there are many overlapping solutions
+// --->> cr = current row
+// --->> cc = current column
+// --->> er = end row
+// --->> ec = end column
+
+
+
+
+let getMazePath = (cr, cc, er, ec) => {
+    if(cr == er && cc == ec) {              //============POSITIVE BASE CASE===========
+        let br = [];
+        br.push('');
+        return br;
+    }
+    
+    if(cr > er || cc > ec) {                //============NEGATIVE BASE CASE===========
+        let br = [];
+        return br;
+    }
+    
+    let myResult = [];
+    
+    let recResultH = getMazePath(cr, cc + 1, er, ec);
+    recResultH.forEach((rrh) => {
+        myResult.push("H" + rrh);
+    });
+    
+    let recResultV = getMazePath(cr + 1, cc, er, ec);
+    recResultV.forEach((rrv) => {
+        myResult.push("V" + rrv);
+    }); 
+    
+    return myResult;
+}
+
+
+let path = getMazePath(0, 0, 2, 2);
+console.log(path);

src/_Problems_/get_string_permutations/index.js

@@ -0,0 +1,24 @@
+// GET PERMUTAION OF A GIVEN STRING
+
+let getPermutation = (str) => {
+    if (str.length == 1) {                          // BASE CASE
+        let array = [];
+        array.push(str);
+        return array;
+    }
+
+    let currentCharacter = str.charAt(0);
+    let restOfString = str.substring(1);
+    let result = [];
+    let returnResult = getPermutation(restOfString);
+    for (j = 0; j < returnResult.length; j++) {
+        for (i = 0; i <= returnResult[j].length; i++) {
+            let value = returnResult[j].substring(0, i) + currentCharacter + returnResult[j].substring(i);
+            result.push(value);
+        }
+    }
+    return result;
+}
+
+let permutation = getPermutation('abc');
+console.log(permutation);

src/_Problems_/get_subsequence/index.js

@@ -0,0 +1,30 @@
+// FIND SUBSEQUENCE OF A GIVEN SUBSTRING
+// SUBSTRING OF 'abc' ---->>>>  [ '', 'a', 'b', 'ab', 'c', 'ac', 'bc', 'abc' ]
+// SUBSTRING OF 'bc'  ---->>>>  ['', 'b', 'c', 'bc']
+// SUBSTRING OF 'c'   ---->>>>  ['', 'c']
+// A pattern can be noticed in above three substrings. Technique followed is recursion.
+// Time complexity : O(2^n)  n is the length of the string provided.
+
+
+let getSubesequence = (str) => {
+    if (str.length == 0) {
+        let array = [''];
+        return array;
+    }
+
+    let currentChar = str.charAt(0);
+    let restOfString = str.substring(1);
+
+    let result = [];
+    let returnResult = getSubesequence(restOfString);
+    for (i = 0; i < returnResult.length; i++) {
+        result.push(returnResult[i]);
+        result.push(currentChar + returnResult[i]);
+    }
+    return result;
+}
+
+
+let subsequence = getSubesequence('abc');
+console.log(subsequence);
+