Optimized the code

Dynamic-Programming/LongestCommonSubsequence.js

@@ -1,31 +1,36 @@
 /*
-    * Given two sequences, find the length of longest subsequence present in both of them. 
+    * Given two sequences, find the length of longest subsequence present in both of them.
-    * A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous. 
+    * A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous.
     * For example, “abc”, “abg”, “bdf”, “aeg”, ‘”acefg”, .. etc are subsequences of “abcdefg”
 */
 
-function longestCommonSubsequence(x, y, str1, str2, dp) {
+function longestCommonSubsequence (x, y, str1, str2, dp) {
-  if (x == -1 || y == -1) return 0;
+  if (x === -1 || y === -1) {
+    return 0
+  }
   else {
-    if (dp[x][y] != 0) return dp[x][y];
+    if (dp[x][y] !== 0){
+      return dp[x][y]
+    } 
     else {
-      if (str1[x] == str2[y]) {
+      if (str1[x] === str2[y]) {
-        return dp[x][y] = 1 + longestCommonSubsequence(x - 1, y - 1, str1, str2, dp);
+        dp[x][y] = 1 + longestCommonSubsequence(x - 1, y - 1, str1, str2, dp);
+        return dp[x][y]
       }
       else {
-        return dp[x][y] = Math.max(longestCommonSubsequence(x - 1, y, str1, str2, dp), longestCommonSubsequence(x, y - 1, str1, str2, dp))
+        dp[x][y] = Math.max(longestCommonSubsequence(x - 1, y, str1, str2, dp), longestCommonSubsequence(x, y - 1, str1, str2, dp))
+        return dp[x][y]
       }
     }
   }
-
 }
 
-function main() {
+function main () {
-  const str1 = "ABCDGH"
+  const str1 = 'ABCDGH'
-  const str2 = "AEDFHR"
+  const str2 = 'AEDFHR'
   const dp = new Array(str1.length + 1).fill(0).map(x => new Array(str2.length + 1).fill(0))
   const res = longestCommonSubsequence(str1.length - 1, str2.length - 1, str1, str2, dp)
-  console.log(res);
+  console.log(res)
 }
 
 main()