README_EN.md

410. Split Array Largest Sum

中文文档

Description

Given an integer array nums and an integer k, split nums into k non-empty subarrays such that the largest sum of any subarray is minimized.

Return the minimized largest sum of the split.

A subarray is a contiguous part of the array.

 

Example 1:

Input: nums = [7,2,5,10,8], k = 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.

Example 2:

Input: nums = [1,2,3,4,5], k = 2
Output: 9
Explanation: There are four ways to split nums into two subarrays.
The best way is to split it into [1,2,3] and [4,5], where the largest sum among the two subarrays is only 9.

 

Constraints:

• 1 <= nums.length <= 1000
• 0 <= nums[i] <= 106
• 1 <= k <= min(50, nums.length)

Solutions

Binary search.

Python3

class Solution:
    def splitArray(self, nums: List[int], k: int) -> int:
        def check(mx):
            s, cnt = inf, 0
            for x in nums:
                s += x
                if s > mx:
                    s = x
                    cnt += 1
            return cnt <= k

        left, right = max(nums), sum(nums)
        return left + bisect_left(range(left, right + 1), True, key=check)

Java

class Solution {
    public int splitArray(int[] nums, int k) {
        int left = 0, right = 0;
        for (int x : nums) {
            left = Math.max(left, x);
            right += x;
        }
        while (left < right) {
            int mid = (left + right) >> 1;
            if (check(nums, mid, k)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

    private boolean check(int[] nums, int mx, int k) {
        int s = 1 << 30, cnt = 0;
        for (int x : nums) {
            s += x;
            if (s > mx) {
                ++cnt;
                s = x;
            }
        }
        return cnt <= k;
    }
}

C++

class Solution {
public:
    int splitArray(vector<int>& nums, int k) {
        int left = 0, right = 0;
        for (int& x : nums) {
            left = max(left, x);
            right += x;
        }
        auto check = [&](int mx) {
            int s = 1 << 30, cnt = 0;
            for (int& x : nums) {
                s += x;
                if (s > mx) {
                    s = x;
                    ++cnt;
                }
            }
            return cnt <= k;
        };
        while (left < right) {
            int mid = (left + right) >> 1;
            if (check(mid)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
};

Go

func splitArray(nums []int, k int) int {
	left, right := 0, 0
	for _, x := range nums {
		left = max(left, x)
		right += x
	}
	return left + sort.Search(right-left, func(mx int) bool {
		mx += left
		s, cnt := 1<<30, 0
		for _, x := range nums {
			s += x
			if s > mx {
				s = x
				cnt++
			}
		}
		return cnt <= k
	})
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

TypeScript

function splitArray(nums: number[], k: number): number {
    let left = 0;
    let right = 0;
    for (const x of nums) {
        left = Math.max(left, x);
        right += x;
    }
    const check = (mx: number) => {
        let s = 1 << 30;
        let cnt = 0;
        for (const x of nums) {
            s += x;
            if (s > mx) {
                s = x;
                ++cnt;
            }
        }
        return cnt <= k;
    };
    while (left < right) {
        const mid = (left + right) >> 1;
        if (check(mid)) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

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