Solution for max product

content/4_Gold/Digit_DP.problems.json

@@ -38,8 +38,8 @@
       "isStarred": false,
       "tags": ["DP"],
       "solutionMetadata": {
-        "kind": "autogen-label-from-site",
-        "site": "CF"
+        "kind": "internal",
+        "hasHints": true
       }
     },
     {

solutions/gold/cf-100886G.mdx

@@ -0,0 +1,69 @@
+---
+id: cf-100886G
+source: CF
+title: Maximum Product
+author: Daniel Zhu
+---
+
+<Spoiler title="Hint">
+</Spoiler>
+## Explanation
+### $l = 1$
+If $l = 1$, what will the best solution always be?
+Let's start with an example: $l = 1$ and $r = 875$. Observe that we really need to consider three numbers as viable answers: $875$, $799$, and $869$. Try to think about why, and how this generalizes!
+### $l \neq 1$
+We can apply a very similar approach when $l \neq 1$. In fact, it's almost identical, with one minor difference. Consider the case where $l = 855$ and $r = 875$, just as before. The only difference is we can no longer consider $799$ as a viable answer, as it is less than $l$. Again, try to think about how to generalize this!
+
+## Implementation
+
+**Time Complexity:** $\mathcal{O}(D^2)$
+
+_Note:_ $D$ is the max number of digits (19).
+
+<LanguageSection>
+
+<CPPSection>
+
+```cpp
+#include <bits/stdc++.h>
+#define sz(v) (int)v.size()
+using ll = long long;
+using namespace std;
+// gets product of digits of given string
+// and returns -1 for empty string
+ll prod(string s) {
+	if (!s.length()) return -1;
+	ll res = 1;
+	for (char c : s) res *= c - '0';
+	return res;
+}
+int main() {
+	string l, r;
+	cin >> l >> r;
+	// pad beginning of l with zeros
+	// l = 12, r = 132 -> l = 012, r = 132
+	while (l.size() < r.size()) l.insert(l.begin(), '0');
+	string ans = "";
+	// i -> length of LCP (longest common prefix) of str and r
+	bool eq = true;  // whether l[0, i) = r[0, i)
+	for (int i = 0; i <= sz(r); i++) {
+		string cur = r.substr(0, i);
+		eq = eq && l[i] == r[i];
+		// in this case, str[i] must equal r[i]
+		// and now the shared prefix has length i + 1
+		// which contradicts our assumption that the LCP has length i
+		if (i < sz(r) && eq) continue;
+		if (i < sz(r)) cur += char(r[i] - 1);
+		// fill remaining digits with 9's
+		cur += string(sz(r) - cur.size(), '9');
+		if (prod(cur) > prod(ans)) ans = cur;
+	}
+	// remove any leading zeros from ans
+	while (ans[0] == '0') ans.erase(ans.begin());
+	cout << ans << endl;
+}
+```
+
+</CPPSection>
+
+</LanguageSection>