Implement integer-break

name: integer-break
url: https://leetcode.com/problems/integer-break
difficulty: 2

time: 0 ms
time-rank: 100.0 %
time-complexity: O(N)

space: 6.5 MB
space-rank: 35.36 %
space-complexity: O(N)

Fixes #313

Signed-off-by: Christopher Friedt <[email protected]>

Author: cfriedt

integer-break-test.cpp

@@ -0,0 +1,19 @@
+/*
+ * Copyright (c) 2021, Christopher Friedt
+ *
+ * SPDX-License-Identifier: MIT
+ */
+
+#include <gtest/gtest.h>
+
+#include "integer-break.cpp"
+
+using namespace std;
+
+TEST(IntegerBreak, 1) { EXPECT_EQ(1, Solution().integerBreak(1)); }
+
+TEST(IntegerBreak, 2) { EXPECT_EQ(1, Solution().integerBreak(2)); }
+
+TEST(IntegerBreak, 3) { EXPECT_EQ(2, Solution().integerBreak(3)); }
+
+TEST(IntegerBreak, 10) { EXPECT_EQ(36, Solution().integerBreak(10)); }

integer-break.cpp

@@ -0,0 +1,66 @@
+/*
+ * Copyright (c) 2021, Christopher Friedt
+ *
+ * SPDX-License-Identifier: MIT
+ */
+
+// clang-format off
+// name: integer-break
+// url: https://leetcode.com/problems/integer-break
+// difficulty: 2
+// clang-format on
+
+#include <climits>
+#include <unordered_map>
+#include <vector>
+
+using namespace std;
+
+class Solution {
+public:
+  using dp_t = unordered_map<int, int>;
+  int integerBreak(int n) {
+    /*
+    Let's say we're given the number 3. How many different ways can we generate
+    the number 3 from the sum of at least two positive integers?
+    1 + 1 + 1, product is 1
+    2 + 1, product is 2
+    1 + 2, product is 2
+    So MP(3) is 2.
+
+             MP(3)
+          -1 /  \ -2
+        MP(2)    MP(1)
+
+    We know that MP(2) is 1 and we can extrapolate that MP(1) is also 1. Then we
+    can figure out that
+
+    MP(n) = max(i * MP(n - i), i * (n - i)), for i in [1, n)
+          = 1, if n <= 2
+
+    It becomes evident, that for larger numbers, there is very obvious
+    overlapping of subproblems.
+    */
+    dp_t dp;
+    dp[1] = 1;
+    dp[2] = 1;
+
+    return MP(dp, n);
+  }
+
+  int MP(dp_t &dp, int n) {
+    auto it = dp.find(n);
+    if (dp.end() != it) {
+      return it->second;
+    }
+
+    int max_product = INT_MIN;
+    for (int i = 1; i < n; ++i) {
+      max_product = max(max_product, i * MP(dp, n - i));
+      max_product = max(max_product, i * (n - i));
+    }
+
+    dp[n] = max_product;
+    return max_product;
+  }
+};