0410. Split Array Largest Sum

Hard | Binary Search | 32 ms (92.40%), 14.3 MB (65.39%)

❓ Problem Statement

Source: LeetCode - Aplit Array Largest Sum GitHub: Solution / Performance

Given an array nums which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays.

Write an algorithm to minimize the largest sum among these m subarrays.

✍🏻 Constraints & Example

Constraints:

• 1 <= nums.length <= 1000

• 0 <= nums[i] <= 10^6

• 1 <= m <= min(50, nums.length)

Input: nums = [7,2,5,10,8], m = 2
Output: 18
Explanation:
  There are four ways to split nums into two subarrays.
  The best way is to split it into [7,2,5] and [10,8],
  where the largest sum among the two subarrays is only 18.

Input: nums = [1,2,3,4,5], m = 2
Output: 9

Input: nums = [1,4,4], m = 3
Output: 4

💡 Ideas

Same as LeetCode 1011

Binary Search Problem Find a minimum largest sum among these m subarrays.

Boundary / Search Space

Left (Minimum) = MAX number (cannot seperate a single number) Right (Maximum) = TOTAL amount

Condition

While Loop = Left < Right

Return Value

Left

🤖 Python3

class Solution:
    def splitArray(self, nums: List[int], m: int) -> int:
        
        # ==================================================
        #  Binary Search                                   =
        # ==================================================
        # time  : O(nlog(m)), m is the search space
        # space : O(1)
        
        l, r = max(nums), sum(nums)
        
        while l < r:
            mid = (l + r) // 2
            
            count, groups = 0, 1
            for num in nums:
                count += num
                if count > mid:
                    count = num
                    groups += 1
                    
                if groups > m: break
                    
            if groups <= m: r = mid
            else: l = mid + 1
            
        return l

🤖 Java

class Solution {
    /**
     * @time  : O(nlog(m)), m is the search space
     * @space : O(1)
     */
    
    public int splitArray(int[] nums, int m) {
        int l = 0, r = 0;
        for (int n: nums) {
            l = Math.max(l, n);
            r += n;
        }
        
        while(l < r) {
            int mid = (l + r) / 2;
            
            int count = 0, groups = 1;
            for(int n: nums) {
                count += n;
                if(count > mid) {
                    count = n;
                    groups += 1;
                }
                
                if(groups > m) break;
            }
            
            if(groups <= m) r = mid;
            else l = mid + 1;
        }
        
        return l;
    }
}