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
Solution:
class Solution {
public int waysToSplitArray(int[] nums) {
if (nums ==
null || nums.length <= 1 ) return 0;
int sumSoFar = 0;
int total = 0;
int count = 0;
for (int i = 0; i < nums.length; i++) {
total += nums[i];
}
for (int i = 0; i < nums.length - 1; i++) {
sumSoFar +=
nums[i];
if (sumSoFar >=
total-sumSoFar) {
count += 1;
}
}
return count;
}
}
Test cases:
[0] => 0
[1] => 0
[0,1] => 0
[1,0] => 1
[0,0,1] => 0
[0,1,0] => 1
[0,1,1] => 1
[1,0,0] => 2
[1,0,1] => 2
[1,1,0] => 2
[1,1,1] => 1
No comments:
Post a Comment