README_EN.md

comments	difficulty	edit_url	tags
true	Hard	https://github.com/doocs/leetcode/edit/main/solution/0800-0899/0805.Split%20Array%20With%20Same%20Average/README_EN.md	Bit Manipulation Array Math Dynamic Programming Bitmask

805. Split Array With Same Average

中文文档

Description

You are given an integer array nums.

You should move each element of nums into one of the two arrays A and B such that A and B are non-empty, and average(A) == average(B).

Return true if it is possible to achieve that and false otherwise.

Note that for an array arr, average(arr) is the sum of all the elements of arr over the length of arr.

 

Example 1:

Input: nums = [1,2,3,4,5,6,7,8]
Output: true
Explanation: We can split the array into [1,4,5,8] and [2,3,6,7], and both of them have an average of 4.5.

Example 2:

Input: nums = [3,1]
Output: false

 

Constraints:

• 1 <= nums.length <= 30
• 0 <= nums[i] <= 104

Solutions

Solution 1

Python3

class Solution:
    def splitArraySameAverage(self, nums: List[int]) -> bool:
        n = len(nums)
        if n == 1:
            return False
        s = sum(nums)
        for i, v in enumerate(nums):
            nums[i] = v * n - s
        m = n >> 1
        vis = set()
        for i in range(1, 1 << m):
            t = sum(v for j, v in enumerate(nums[:m]) if i >> j & 1)
            if t == 0:
                return True
            vis.add(t)
        for i in range(1, 1 << (n - m)):
            t = sum(v for j, v in enumerate(nums[m:]) if i >> j & 1)
            if t == 0 or (i != (1 << (n - m)) - 1 and -t in vis):
                return True
        return False

Java

class Solution {
    public boolean splitArraySameAverage(int[] nums) {
        int n = nums.length;
        if (n == 1) {
            return false;
        }
        int s = Arrays.stream(nums).sum();
        for (int i = 0; i < n; ++i) {
            nums[i] = nums[i] * n - s;
        }
        int m = n >> 1;
        Set<Integer> vis = new HashSet<>();
        for (int i = 1; i < 1 << m; ++i) {
            int t = 0;
            for (int j = 0; j < m; ++j) {
                if (((i >> j) & 1) == 1) {
                    t += nums[j];
                }
            }
            if (t == 0) {
                return true;
            }
            vis.add(t);
        }
        for (int i = 1; i < 1 << (n - m); ++i) {
            int t = 0;
            for (int j = 0; j < (n - m); ++j) {
                if (((i >> j) & 1) == 1) {
                    t += nums[m + j];
                }
            }
            if (t == 0 || (i != (1 << (n - m)) - 1) && vis.contains(-t)) {
                return true;
            }
        }
        return false;
    }
}

C++

class Solution {
public:
    bool splitArraySameAverage(vector<int>& nums) {
        int n = nums.size();
        if (n == 1) return false;
        int s = accumulate(nums.begin(), nums.end(), 0);
        for (int& v : nums) v = v * n - s;
        int m = n >> 1;
        unordered_set<int> vis;
        for (int i = 1; i < 1 << m; ++i) {
            int t = 0;
            for (int j = 0; j < m; ++j)
                if (i >> j & 1) t += nums[j];
            if (t == 0) return true;
            vis.insert(t);
        }
        for (int i = 1; i < 1 << (n - m); ++i) {
            int t = 0;
            for (int j = 0; j < (n - m); ++j)
                if (i >> j & 1) t += nums[m + j];
            if (t == 0 || (i != (1 << (n - m)) - 1 && vis.count(-t))) return true;
        }
        return false;
    }
};

Go

func splitArraySameAverage(nums []int) bool {
	n := len(nums)
	if n == 1 {
		return false
	}
	s := 0
	for _, v := range nums {
		s += v
	}
	for i, v := range nums {
		nums[i] = v*n - s
	}
	m := n >> 1
	vis := map[int]bool{}
	for i := 1; i < 1<<m; i++ {
		t := 0
		for j, v := range nums[:m] {
			if (i >> j & 1) == 1 {
				t += v
			}
		}
		if t == 0 {
			return true
		}
		vis[t] = true
	}
	for i := 1; i < 1<<(n-m); i++ {
		t := 0
		for j, v := range nums[m:] {
			if (i >> j & 1) == 1 {
				t += v
			}
		}
		if t == 0 || (i != (1<<(n-m))-1 && vis[-t]) {
			return true
		}
	}
	return false
}