fix: linter fixes, revision in eslint config

Author: ashokdey

.eslintrc.json

@@ -3,5 +3,6 @@
     "node": true,
     "jest": true
   },
-  "extends": "airbnb-base"
+  "extends": ["prettier", "airbnb-base"],
+  "plugins": ["prettier"]
 }

package-lock.json

@@ -18,7 +18,10 @@
   "devDependencies": {
     "eslint": "^5.11.1",
     "eslint-config-airbnb-base": "^13.1.0",
+    "eslint-config-prettier": "^6.5.0",
     "eslint-plugin-import": "^2.14.0",
-    "jest": "^25.0.0"
+    "eslint-plugin-prettier": "^3.1.1",
+    "jest": "^25.0.0",
+    "prettier": "^1.19.1"
   }
 }

package.json

@@ -22,7 +22,7 @@ describe('Compose Largest Number', () => {
     });
   });
 
-  describe('Testing `compare()`', () => {
+  describe('Testing custom `compare()` for `sort()`', () => {
     it('Should return [60, 548] instead of [548, 60]', () => {
       expect([60, 548].sort(compare)).toEqual([60, 548]);
     });

src/_Problems_/compose-largest-number/compose-largest.test.js

@@ -0,0 +1,28 @@
+// 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.
+
+function getSubesequence(str) {
+  if (str.length === 0) {
+    const array = [''];
+    return array;
+  }
+
+  const currentChar = str.charAt(0);
+  const restOfString = str.substring(1);
+
+  const result = [];
+  const returnResult = getSubesequence(restOfString);
+  for (let i = 0; i < returnResult.length; i += 1) {
+    result.push(returnResult[i]);
+    result.push(currentChar + returnResult[i]);
+  }
+  return result;
+}
+
+module.exports = {
+  getSubesequence,
+};

src/_Problems_/factorial/index.test.js src/_Problems_/factorial/factorial.test.js

@@ -2,7 +2,7 @@ const { getSubesequence } = require('.');
 
 describe('GetSubesequence', () => {
   it('Sequence of abc', () => {
-    expect(getSubesequence('abc').sort()).toEqual([ '', 'a', 'ab', 'abc', 'ac', 'b', 'bc', 'c' ]);
+    expect(getSubesequence('abc').sort()).toEqual(['', 'a', 'ab', 'abc', 'ac', 'b', 'bc', 'c']);
   });
 
   it('Sequence of bc', () => {