详细介绍：LeetCode //C - 915. Partition Array into Disjoint Intervals

915. Partition Array into Disjoint Intervals

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.
  Returnthe 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:

• $2<=nums.length<=10^5$
• $0<=nums[i]<=10^6$
• There is at least one valid answer for the given input.

From: LeetCode
Link: 915. Partition Array into Disjoint Intervals

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Solution:

Ideas:

• Time: O(n) single pass

• Space: O(1)

• Why it works: Whenever an element in the right candidate is < leftMax, the partition is invalid; we extend left to include it and update leftMax to the best max seen so far (curMax), restoring the invariant that all in left ≤ all in future right.

Code:

int partitionDisjoint(int* nums, int numsSize) {
// Invariants:
// left = [0 .. idx], right = [idx+1 .. end)
// leftMax = max(nums[0..idx])
// curMax  = max(nums[0..i]) while scanning
int idx = 0;
int leftMax = nums[0];
int curMax  = nums[0];
for (int i = 1; i < numsSize; ++i) {
if (nums[i] < leftMax) {
// Need to extend left to include i
idx = i;
leftMax = curMax;   // new leftMax becomes the best we've seen so far
} else if (nums[i] > curMax) {
curMax = nums[i];
}
}
return idx + 1; // length of left
}