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343. Integer Break

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

Dynamic programming.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript
• Javascript
• C#

• 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;
      }
  };
  

• 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
  
  

• 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;
  }
  
  

• 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)
      }
  }
  
  

• /**
   * @param {number} n
   * @return {number}
   */
  var integerBreak = function (n) {
      const f = Array(n + 1).fill(1);
      for (let i = 2; i <= n; ++i) {
          for (let j = 1; j < i; ++j) {
              f[i] = Math.max(f[i], f[i - j] * j, (i - j) * j);
          }
      }
      return f[n];
  };
  
  

• public class Solution {
      public int IntegerBreak(int n) {
          int[] f = new int[n + 1];
          f[1] = 1;
          for (int i = 2; i <= n; ++i) {
              for (int j = 1; j < i; ++j) {
                  f[i] = Math.Max(Math.Max(f[i], f[i - j] * j), (i - j) * j);
              }
          }
          return f[n];
      }
  }
  

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