1712. Ways to Split Array Into Three Subarrays
Description
A split of an integer array is good if:
- The array is split into three non-empty contiguous subarrays - named
left
,mid
,right
respectively from left to right. - The sum of the elements in
left
is less than or equal to the sum of the elements inmid
, and the sum of the elements inmid
is less than or equal to the sum of the elements inright
.
Given nums
, an array of non-negative integers, return the number of good ways to split nums
. As the number may be too large, return it modulo 109 + 7
.
Example 1:
Input: nums = [1,1,1] Output: 1 Explanation: The only good way to split nums is [1] [1] [1].
Example 2:
Input: nums = [1,2,2,2,5,0] Output: 3 Explanation: There are three good ways of splitting nums: [1] [2] [2,2,5,0] [1] [2,2] [2,5,0] [1,2] [2,2] [5,0]
Example 3:
Input: nums = [3,2,1] Output: 0 Explanation: There is no good way to split nums.
Constraints:
3 <= nums.length <= 105
0 <= nums[i] <= 104
Solutions
Solution: Prefix Sum + Two Pointers
- Time complexity: O(n)
- Space complexity: O(n)
JavaScript
js
/**
* @param {number[]} nums
* @return {number}
*/
const waysToSplit = function (nums) {
const MODULO = 10 ** 9 + 7;
const prefixSum = [nums[0]];
const size = nums.length;
let mid = (right = result = 0);
for (let index = 1; index < size; index++) {
prefixSum[index] = prefixSum[index - 1] + nums[index];
}
for (let index = 0; index < size - 2; index++) {
const current = prefixSum[index];
mid = Math.max(index + 1, mid);
while (mid < size - 1 && prefixSum[mid] - current < current) mid += 1;
right = Math.max(mid, right);
while (right < size - 1 && prefixSum[right] - current <= prefixSum.at(-1) - prefixSum[right]) {
right += 1;
}
result += right - mid;
result %= MODULO;
}
return result;
};