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

915. Partition Array into Disjoint Intervals

Level

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.

• Java
• C++
• Python
• Go

• 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
  }
  

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