Solved "Reverse Shuffle Merge"

Author: jonasraoni

reverseShuffleMerge.js

@@ -0,0 +1,38 @@
+//+ Jonas Raoni Soares Silva
+//@ http://raoni.org
+
+function reverseShuffleMerge(s) {
+	const required = new Map();
+	const used = new Map();
+	const available = new Map();
+	for (const c of s) {
+		// As the string "S" is formed by the reverse of S + scrambled S, it means each character should appear twice...
+		// So we know which characters should be part of the original string and how many times they should appear
+		const amount = (required.get(c) || 0) + .5;
+		// Keep the required amount (static reference) of characters to form the string
+		required.set(c, amount);
+		// As we collect chars, this will be used to monitor how many still exists ahead...
+		available.set(c, amount * 2);
+	}
+	const origin = [];
+	// Reverse the string by making a loop from the end to the beginning
+	for (let i = s.length; i--; ) {
+		const c = s[i];
+		const count = used.get(c) || 0;
+		// If the current character is useful
+		if (count < required.get(c)) {
+			// Check if we can replace the last character we stored by a lexicographically smaller one...
+			for (let last; origin.length && (last = origin[origin.length - 1]) > c; origin.pop()) {
+				if (used.get(last) + available.get(last) <= required.get(last))
+					break;
+				used.set(last, used.get(last) - 1);
+			}
+			// Use character
+			used.set(c, count + 1);
+			origin.push(c);
+		}
+		// Decrease the availability
+		available.set(c, available.get(c) - 1);
+	}
+	return origin.join('');
+}