Formatted question description: https://leetcode.ca/all/915.html

915. Partition Array into Disjoint Intervals

Medium

Description

Given an array A, 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. It is guaranteed that such a partitioning exists.

Example 1:

Input: [5,0,3,8,6]

Output: 3

Explanation: left = [5,0,3], right = [8,6]

Example 2:

Input: [1,1,1,0,6,12]

Output: 4

Explanation: left = [1,1,1,0], right = [6,12]

Note:

1. 2 <= A.length <= 30000
2. 0 <= A[i] <= 10^6
3. It is guaranteed there is at least one way to partition A as described.

Solution

Create an array minRight that has the same length as A, where minRight[i] represents the minimum element in the subarray A[i..A.length-1]. Then loop over A from left to right and maintain the maximum element met so far, which is maxLeft. Once an index such that maxLeft <= minRight[index + 1] is met, return index + 1. Since it is guaranteed that such a partitioning exists, the maximum possible return value is A.length - 1.

• class Solution {
public int partitionDisjoint(int[] A) {
int length = A.length;
int[] minRight = new int[length];
minRight[length - 1] = A[length - 1];
for (int i = length - 2; i >= 0; i--)
minRight[i] = Math.min(minRight[i + 1], A[i]);
int maxLeft = Integer.MIN_VALUE;
for (int i = 0; i < length - 1; i++) {
maxLeft = Math.max(maxLeft, A[i]);
if (maxLeft <= minRight[i + 1])
return i + 1;
}
return length - 1;
}
}

• // OJ: https://leetcode.com/problems/partition-array-into-disjoint-intervals/
// Time: O(N)
// Space: O(N)
class Solution {
public:
int partitionDisjoint(vector<int>& A) {
deque<int> q; // index
int N = A.size(), mx = 0;
for (int i = 0; i < N; ++i) {
while (q.size() && A[q.back()] >= A[i]) q.pop_back();
q.push_back(i);
}
for (int i = 0; i < N; ++i) {
if (q.front() == i) q.pop_front();
mx = max(mx, A[i]);
if (mx <= A[q.front()]) return i + 1;
}
return -1;
}
};

• 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

############

class Solution(object):
def partitionDisjoint(self, A):
"""
:type A: List[int]
:rtype: int
"""
disjoint = 0
v = A[0]
max_so_far = v
for i in range(len(A)):
max_so_far = max(max_so_far, A[i])
if A[i] < v:
v = max_so_far
disjoint = i
return disjoint + 1

• 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
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

func min(a, b int) int {
if a < b {
return a
}
return b
}