🔥Easy explanation | Why 3s why not 4s?| 🚀Dynamic Programming Solution with Best Approach ✅✅✅✅✅
\n## Upvote if you like the  eexplanation and solution \uD83D\uDD25\n\n---\n\n\n# Intuition\n Describe your first thoughts on how to solve this problem. \n We w
\n## Upvote if you like the  eexplanation and solution \uD83D\uDD25\n\n---\n\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n We want to break the integer n into the sum of positive integers to maximize the product. By observation, it\'s more beneficial to use 3s instead of 4s because 3 * 3 > 4 * 4.\n\n\n---\n\n\n# Using 3s instead of 4s maximizes the product because:\n\n- 3 * 3 is greater than 4 * 4 (9 > 16).\n- When using 4s, we can always replace two 4s with three 3s, resulting in a larger product (4 * 4 = 16, but 3 * 3 * 3 > 4 * 4). This is why 3s are preferred.\n\n---\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe use dynamic programming to find the maximum product for each integer from 2 to n. We initialize the base cases for 2 and 3. Then, for each integer from 4 to n, we calculate the maximum product based on the previous results.\n\n---\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n The time complexity is O(n) because we iterate through all integers from 4 to n once.\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity is O(n) due to the dynamic programming array dp.\n\n---\n\n# Code\n```\nclass Solution {\npublic:\n    int integerBreak(int n) {\n        if (n <= 3) return n - 1;\n        \n        vector<int> dp(n + 1, 0);\n        dp[2] = 2;\n        dp[3] = 3;\n        \n        for (int i = 4; i <= n; ++i) {\n            dp[i] = max(dp[i - 2] * 2, dp[i - 3] * 3);\n        }\n        \n        return dp[n];\n    }\n};\n\n```
['Array', 'Hash Table', 'Math', 'Dynamic Programming', 'C', 'Heap (Priority Queue)', 'Python', 'C++', 'Java', 'Python3']