{ "data": { "question": { "questionId": "3947", "questionFrontendId": "3669", "categoryTitle": "Algorithms", "boundTopicId": 3767456, "title": "Balanced K-Factor Decomposition", "titleSlug": "balanced-k-factor-decomposition", "content": "
Given two integers n and k, split the number n into exactly k positive integers such that the product of these integers is equal to n.
Return any one split in which the maximum difference between any two numbers is minimized. You may return the result in any order.
\n\n\n
Example 1:
\n\nInput: n = 100, k = 2
\n\nOutput: [10,10]
\n\nExplanation:
\n\nThe split [10, 10] yields 10 * 10 = 100 and a max-min difference of 0, which is minimal.
Example 2:
\n\nInput: n = 44, k = 3
\n\nOutput: [2,2,11]
\n\nExplanation:
\n\n[1, 1, 44] yields a difference of 43[1, 2, 22] yields a difference of 21[1, 4, 11] yields a difference of 10[2, 2, 11] yields a difference of 9Therefore, [2, 2, 11] is the optimal split with the smallest difference 9.
\n
Constraints:
\n\n4 <= n <= 1052 <= k <= 5k is strictly less than the total number of positive divisors of n.给你两个整数 n 和 k,将数字 n 恰好分割成 k 个正整数,使得这些整数的 乘积 等于 n。
返回一个分割方案,使得这些数字中 最大值 和 最小值 之间的 差值 最小化。结果可以以 任意顺序 返回。
\n\n\n\n
示例 1:
\n\n输入:n = 100, k = 2
\n\n输出:[10,10]
\n\n解释:
\n\n分割方案 [10, 10] 的结果是 10 * 10 = 100,且最大值与最小值的差值为 0,这是最小可能值。
示例 2:
\n\n输入:n = 44, k = 3
\n\n输出:[2,2,11]
\n\n解释:
\n\n[1, 1, 44] 的差值为 43[1, 2, 22] 的差值为 21[1, 4, 11] 的差值为 10[2, 2, 11] 的差值为 9因此,[2, 2, 11] 是最优分割方案,其差值最小,为 9。
\n\n
提示:
\n\n4 <= n <= 1052 <= k <= 5k 严格小于 n 的正因数的总数。n and sort them into a list divs.",
"Use a recursive search dfs(start, picked, prod, path) that picks the next divisor from divs[start...], multiplies it into prod, and appends to path until you have k factors whose product is n.",
"During the search, keep track of the best path that minimizes max(path) – min(path) and return it."
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