Number of Ways to Split Array(Array,Prefix Sum)

https://leetcode.com/problems/number-of-ways-to-split-array/description/?envType=daily-question&envId=2025-01-03

You are given a 0-indexed integer array nums of length n.

nums contains a valid split at index i if the following are true:

• The sum of the first i + 1 elements is greater than or equal to the sum of the last n - i - 1 elements.
• There is at least one element to the right of i. That is, 0 <= i < n - 1.

Return the number of valid splits in nums.

 

Example 1:

Input: nums = [10,4,-8,7]
Output: 2
Explanation: 
There are three ways of splitting nums into two non-empty parts:
- Split nums at index 0. Then, the first part is [10], and its sum is 10. The second part is [4,-8,7], and its sum is 3. Since 10 >= 3, i = 0 is a valid split.
- Split nums at index 1. Then, the first part is [10,4], and its sum is 14. The second part is [-8,7], and its sum is -1. Since 14 >= -1, i = 1 is a valid split.
- Split nums at index 2. Then, the first part is [10,4,-8], and its sum is 6. The second part is [7], and its sum is 7. Since 6 < 7, i = 2 is not a valid split.
Thus, the number of valid splits in nums is 2.

Example 2:

Input: nums = [2,3,1,0]
Output: 2
Explanation: 
There are two valid splits in nums:
- Split nums at index 1. Then, the first part is [2,3], and its sum is 5. The second part is [1,0], and its sum is 1. Since 5 >= 1, i = 1 is a valid split. 
- Split nums at index 2. Then, the first part is [2,3,1], and its sum is 6. The second part is [0], and its sum is 0. Since 6 >= 0, i = 2 is a valid split.

 

Constraints:

• 2 <= nums.length <= 105
• -105 <= nums[i] <= 105

지난 번과 비슷한 prefix sum 문제이다.

 

문제 이해는 완료했고, 그리 어려워 보이진 않다만.. 역시 indexing을 어떻게 detail하게 잘 하냐가 관건일듯 싶다.

 

 

1. Python

다행이 나는 그래도 한 번 배우면 어느 정도까지는 풀이법을 기억할 줄 아는 Memory를 가지고 있었다..

최악은 아니라는 뜻 ㅎㅎ

 

쉽게 해결 完!

 

class Solution:
    def waysToSplitArray(self, nums: List[int]) -> int:
        n = len(nums)
        prefix = [0]*n
        prefix[0] = nums[0]
        # O(N)
        for i in range(1,n):
            prefix[i] = prefix[i-1] + nums[i]

        largest = prefix[-1]
        ans = 0
        # O(N)
        for j in range(n-1):
            # if prefix[j] >= largest - prefix[j] : ans += 1
            if prefix[j]*2 >= largest : ans += 1

        return ans

 

2. C / C++로도 함 해볼까나

 

int waysToSplitArray(int* nums, int numsSize) {
    long long prefix[numsSize];
    prefix[0] = nums[0];
    for(int i=1;i<numsSize;i++){
        prefix[i] = prefix[i-1] + nums[i];
    }
    long long largest = prefix[numsSize-1];
    int ans = 0;

    for(int j=0;j<numsSize-1;j++){
        if(prefix[j]*2 >= largest) ans++;
    }
    return ans;
}

 

무지막지한 것 -_-

 

C는 test case에서 overflow 발생하는 게 있어서 long long 변수를 사용했다;;

 

C++도 마찬가지..

class Solution {
public:
    int waysToSplitArray(vector<int>& nums) {
        int n = nums.size();
        vector<long long> prefix(n,0);
        prefix[0] = nums[0];

        for(int i=1;i<n;i++){
            prefix[i] = prefix[i-1] + nums[i];
        }
        int ans = 0;
        long long largest = prefix[n-1];
        for(int j=0;j<n-1;j++){
            if(prefix[j]*2 >= largest) ans++;
        }
        return ans;
    }
};