GET PERMUTATION AND SUBSEQUENCE OF A GIVEN STRING

Author: navneet

src/_Problems_/getStringPermutaion/getStringPermutaion.js

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+// 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_/getStringPermutation

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+// 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.
+
+
+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);