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915. Partition Array into Disjoint Intervals

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 <= 105
  • 0 <= nums[i] <= 106
  • There is at least one valid answer for the given input.

Solutions

  • 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;
    }
}

  • 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;
    }
};

  • 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


  • 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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