343. Integer Break LeetCode Solution

In this guide, you will get 343. Integer Break LeetCode Solution with the best time and space complexity. The solution to Integer Break problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Integer Break solution in C++
Integer Break solution in Java
Integer Break solution in Python
Additional Resources

343. Integer Break LeetCode Solution image

Problem Statement of Integer Break

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

Complexity Analysis

Time Complexity: O(n / 3)
Space Complexity: O(1)

343. Integer Break LeetCode Solution in C++

class Solution {
 public:
  int integerBreak(int n) {
    // If an optimal product contains a factor f >= 4, then we can replace it
    // with 2 and f - 2 without losing optimality. As 2(f - 2) = 2f - 4 >= f,
    // we never need a factor >= 4, meaning we only need factors 1, 2, and 3
    // (and 1 is wasteful).
    // Also, 3 * 3 is better than 2 * 2 * 2, so we never use 2 more than twice.
    if (n == 2)  // 1 * 1
      return 1;
    if (n == 3)  // 1 * 2
      return 2;

    int ans = 1;

    while (n > 4) {
      n -= 3;
      ans *= 3;
    }
    ans *= n;

    return ans;
  }
};
/* code provided by PROGIEZ */

343. Integer Break LeetCode Solution in Java

class Solution {
  public int integerBreak(int n) {
    // If an optimal product contains a factor f >= 4, then we can replace it
    // with 2 and f - 2 without losing optimality. As 2(f - 2) = 2f - 4 >= f,
    // we never need a factor >= 4, meaning we only need factors 1, 2, and 3
    // (and 1 is wasteful).
    // Also, 3 * 3 is better than 2 * 2 * 2, so we never use 2 more than twice.
    if (n == 2)
      return 1; // 1 * 1
    if (n == 3)
      return 2; // 1 * 2

    int ans = 1;

    while (n > 4) {
      n -= 3;
      ans *= 3;
    }
    ans *= n;

    return ans;
  }
}
// code provided by PROGIEZ

343. Integer Break LeetCode Solution in Python

class Solution:
  def integerBreak(self, n: int) -> int:
    # If an optimal product contains a factor f >= 4, then we can replace it
    # with 2 and f - 2 without losing optimality. As 2(f - 2) = 2f - 4 >= f,
    # we never need a factor >= 4, meaning we only need factors 1, 2, and 3
    # (and 1 is wasteful).
    # Also, 3 * 3 is better than 2 * 2 * 2, so we never use 2 more than twice.
    if n == 2:  # 1 * 1
      return 1
    if n == 3:  # 1 * 2
      return 2

    ans = 1

    while n > 4:
      n -= 3
      ans *= 3
    ans *= n

    return ans
 # code by PROGIEZ