Description#
Given an integer array nums, partition it into two (contiguous) subarrays left and right so that:
- Every element in
left is less than or equal to every element in right. left and right are non-empty.left has the smallest possible size.
Return the length of left after such a partitioning.
Test cases are generated such that partitioning exists.
Example 1:
Input: nums = [5,0,3,8,6]
Output: 3
Explanation: left = [5,0,3], right = [8,6]
Example 2:
Input: nums = [1,1,1,0,6,12]
Output: 4
Explanation: left = [1,1,1,0], right = [6,12]
Constraints:
2 <= nums.length <= 1050 <= nums[i] <= 106- There is at least one valid answer for the given input.
Solutions#
Solution 1#
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| class Solution:
def partitionDisjoint(self, nums: List[int]) -> int:
n = len(nums)
mi = [inf] * (n + 1)
for i in range(n - 1, -1, -1):
mi[i] = min(nums[i], mi[i + 1])
mx = 0
for i, v in enumerate(nums, 1):
mx = max(mx, v)
if mx <= mi[i]:
return i
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| class Solution {
public int partitionDisjoint(int[] nums) {
int n = nums.length;
int[] mi = new int[n + 1];
mi[n] = nums[n - 1];
for (int i = n - 1; i >= 0; --i) {
mi[i] = Math.min(nums[i], mi[i + 1]);
}
int mx = 0;
for (int i = 1; i <= n; ++i) {
int v = nums[i - 1];
mx = Math.max(mx, v);
if (mx <= mi[i]) {
return i;
}
}
return 0;
}
}
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| class Solution {
public:
int partitionDisjoint(vector<int>& nums) {
int n = nums.size();
vector<int> mi(n + 1, INT_MAX);
for (int i = n - 1; ~i; --i) mi[i] = min(nums[i], mi[i + 1]);
int mx = 0;
for (int i = 1; i <= n; ++i) {
int v = nums[i - 1];
mx = max(mx, v);
if (mx <= mi[i]) return i;
}
return 0;
}
};
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| func partitionDisjoint(nums []int) int {
n := len(nums)
mi := make([]int, n+1)
mi[n] = nums[n-1]
for i := n - 1; i >= 0; i-- {
mi[i] = min(nums[i], mi[i+1])
}
mx := 0
for i := 1; i <= n; i++ {
v := nums[i-1]
mx = max(mx, v)
if mx <= mi[i] {
return i
}
}
return 0
}
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