Question

343. Integer Break

中文文档

Description

Given an integer n, break it into the sum of k positive integers, where k >= 2, and maximize the product of those integers.

Return _the maximum product you can get_.

 

Example 1:

Input: n = 2
Output: 1
Explanation: 2 = 1 + 1, 1 × 1 = 1.

Example 2:

Input: n = 10
Output: 36
Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36.

 

Constraints:

• 2 <= n <= 58

Solutions

Solution 1

class Solution:
  def integerBreak(self, n: int) -> int:
    dp = [1] * (n + 1)
    for i in range(2, n + 1):
      for j in range(1, i):
        dp[i] = max(dp[i], dp[i - j] * j, (i - j) * j)
    return dp[n]

class Solution {
  public int integerBreak(int n) {
    int[] dp = new int[n + 1];
    dp[1] = 1;
    for (int i = 2; i <= n; ++i) {
      for (int j = 1; j < i; ++j) {
        dp[i] = Math.max(Math.max(dp[i], dp[i - j] * j), (i - j) * j);
      }
    }
    return dp[n];
  }
}

class Solution {
public:
  int integerBreak(int n) {
    vector<int> dp(n + 1);
    dp[1] = 1;
    for (int i = 2; i <= n; ++i) {
      for (int j = 1; j < i; ++j) {
        dp[i] = max(max(dp[i], dp[i - j] * j), (i - j) * j);
      }
    }
    return dp[n];
  }
};

func integerBreak(n int) int {
  dp := make([]int, n+1)
  dp[1] = 1
  for i := 2; i <= n; i++ {
    for j := 1; j < i; j++ {
      dp[i] = max(max(dp[i], dp[i-j]*j), (i-j)*j)
    }
  }
  return dp[n]
}

function integerBreak(n: number): number {
  let dp = new Array(n + 1).fill(1);
  for (let i = 3; i <= n; i++) {
    for (let j = 1; j < i; j++) {
      dp[i] = Math.max(dp[i], j * (i - j), j * dp[i - j]);
    }
  }
  return dp.pop();
}

impl Solution {
  pub fn integer_break(n: i32) -> i32 {
    if n < 4 {
      return n - 1;
    }
    let count = (n - 2) / 3;
    (3i32).pow(count as u32) * (n - count * 3)
  }
}

int integerBreak(int n) {
  if (n < 4) {
    return n - 1;
  }
  int count = (n - 2) / 3;
  return pow(3, count) * (n - count * 3);
}

Solution 2

class Solution:
  def integerBreak(self, n: int) -> int:
    if n < 4:
      return n - 1
    if n % 3 == 0:
      return pow(3, n // 3)
    if n % 3 == 1:
      return pow(3, n // 3 - 1) * 4
    return pow(3, n // 3) * 2

class Solution {
  public int integerBreak(int n) {
    if (n < 4) {
      return n - 1;
    }
    if (n % 3 == 0) {
      return (int) Math.pow(3, n / 3);
    }
    if (n % 3 == 1) {
      return (int) Math.pow(3, n / 3 - 1) * 4;
    }
    return (int) Math.pow(3, n / 3) * 2;
  }
}

class Solution {
public:
  int integerBreak(int n) {
    if (n < 4) {
      return n - 1;
    }
    if (n % 3 == 0) {
      return pow(3, n / 3);
    }
    if (n % 3 == 1) {
      return pow(3, n / 3 - 1) * 4;
    }
    return pow(3, n / 3) * 2;
  }
};

func integerBreak(n int) int {
  if n < 4 {
    return n - 1
  }
  if n%3 == 0 {
    return int(math.Pow(3, float64(n/3)))
  }
  if n%3 == 1 {
    return int(math.Pow(3, float64(n/3-1))) * 4
  }
  return int(math.Pow(3, float64(n/3))) * 2
}

function integerBreak(n: number): number {
  if (n < 4) {
    return n - 1;
  }
  const m = Math.floor(n / 3);
  if (n % 3 == 0) {
    return 3 ** m;
  }
  if (n % 3 == 1) {
    return 3 ** (m - 1) * 4;
  }
  return 3 ** m * 2;
}