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Merge pull request knaxus#6 from navneet15/branch1

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

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	`1`	`+//======================================Problem Statement=============================================`
	`2`	`+// --->> Print all possible path to reach the end of the GRID/MAZE (N x N) from starting point to ending point`
	`3`	`+// --->> One horizontal move will be represented by H and one vertical move will be represented by V`
	`4`	`+// --->> Complexity = Complexity will be exponential as there are many overlapping solutions`
	`5`	`+// --->> cr = current row`
	`6`	`+// --->> cc = current column`
	`7`	`+// --->> er = end row`
	`8`	`+// --->> ec = end column`
	`9`	`+`
	`10`	`+`
	`11`	`+`
	`12`	`+`
	`13`	`+let getMazePath = (cr, cc, er, ec) => {`
	`14`	`+ if(cr == er && cc == ec) { //============POSITIVE BASE CASE===========`
	`15`	`+ let br = [];`
	`16`	`+ br.push('');`
	`17`	`+ return br;`
	`18`	`+ }`
	`19`	`+`
	`20`	`+ if(cr > er \|\| cc > ec) { //============NEGATIVE BASE CASE===========`
	`21`	`+ let br = [];`
	`22`	`+ return br;`
	`23`	`+ }`
	`24`	`+`
	`25`	`+ let myResult = [];`
	`26`	`+`
	`27`	`+ let recResultH = getMazePath(cr, cc + 1, er, ec);`
	`28`	`+ recResultH.forEach((rrh) => {`
	`29`	`+ myResult.push("H" + rrh);`
	`30`	`+ });`
	`31`	`+`
	`32`	`+ let recResultV = getMazePath(cr + 1, cc, er, ec);`
	`33`	`+ recResultV.forEach((rrv) => {`
	`34`	`+ myResult.push("V" + rrv);`
	`35`	`+ });`
	`36`	`+`
	`37`	`+ return myResult;`
	`38`	`+}`
	`39`	`+`
	`40`	`+`
	`41`	`+let path = getMazePath(0, 0, 2, 2);`
	`42`	`+console.log(path);`

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	`1`	`+// GET PERMUTAION OF A GIVEN STRING`
	`2`	`+`
	`3`	`+let getPermutation = (str) => {`
	`4`	`+ if (str.length == 1) { // BASE CASE`
	`5`	`+ let array = [];`
	`6`	`+ array.push(str);`
	`7`	`+ return array;`
	`8`	`+ }`
	`9`	`+`
	`10`	`+ let currentCharacter = str.charAt(0);`
	`11`	`+ let restOfString = str.substring(1);`
	`12`	`+ let result = [];`
	`13`	`+ let returnResult = getPermutation(restOfString);`
	`14`	`+ for (j = 0; j < returnResult.length; j++) {`
	`15`	`+ for (i = 0; i <= returnResult[j].length; i++) {`
	`16`	`+ let value = returnResult[j].substring(0, i) + currentCharacter + returnResult[j].substring(i);`
	`17`	`+ result.push(value);`
	`18`	`+ }`
	`19`	`+ }`
	`20`	`+ return result;`
	`21`	`+}`
	`22`	`+`
	`23`	`+let permutation = getPermutation('abc');`
	`24`	`+console.log(permutation);`

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	`1`	`+// FIND SUBSEQUENCE OF A GIVEN SUBSTRING`
	`2`	`+// SUBSTRING OF 'abc' ---->>>> [ '', 'a', 'b', 'ab', 'c', 'ac', 'bc', 'abc' ]`
	`3`	`+// SUBSTRING OF 'bc' ---->>>> ['', 'b', 'c', 'bc']`
	`4`	`+// SUBSTRING OF 'c' ---->>>> ['', 'c']`
	`5`	`+// A pattern can be noticed in above three substrings. Technique followed is recursion.`
	`6`	`+// Time complexity : O(2^n) n is the length of the string provided.`
	`7`	`+`
	`8`	`+`
	`9`	`+let getSubesequence = (str) => {`
	`10`	`+ if (str.length == 0) {`
	`11`	`+ let array = [''];`
	`12`	`+ return array;`
	`13`	`+ }`
	`14`	`+`
	`15`	`+ let currentChar = str.charAt(0);`
	`16`	`+ let restOfString = str.substring(1);`
	`17`	`+`
	`18`	`+ let result = [];`
	`19`	`+ let returnResult = getSubesequence(restOfString);`
	`20`	`+ for (i = 0; i < returnResult.length; i++) {`
	`21`	`+ result.push(returnResult[i]);`
	`22`	`+ result.push(currentChar + returnResult[i]);`
	`23`	`+ }`
	`24`	`+ return result;`
	`25`	`+}`
	`26`	`+`
	`27`	`+`
	`28`	`+let subsequence = getSubesequence('abc');`
	`29`	`+console.log(subsequence);`
	`30`	`+`

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