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This approach involves maintaining a running maximum for the left subarray and a suffix minimum for the right subarray. By iterating through the array and comparing these values, we can determine the appropriate partition point where all conditions are satisfied.
Time Complexity: O(n) - Linear scan of the array.
Space Complexity: O(1) - Only variables used for tracking, no extra storage required.
1function partitionDisjoint(nums) {
2 let maxLeft = nums[0];
3 let maxSoFar = nums[0];
4 let partitionIdx = 0;
5
6 for (let i = 1; i < nums.length; i++) {
7 if (nums[i] < maxLeft) {
8 maxLeft = maxSoFar;
9 partitionIdx = i;
10 } else {
11 maxSoFar = Math.max(maxSoFar, nums[i]);
12 }
13 }
14 return partitionIdx + 1;
15}
16
17console.log(partitionDisjoint([5, 0, 3, 8, 6])); // Output: 3
18In JavaScript, the solution leverages running maximum values and selectively updates the partition index upon finding an element less than maxLeft, ensuring partition criteria are followed.
This approach involves using two additional arrays: one to track the maximum values until any index from the left and another to track minimum values from the right. These auxiliary arrays help determine where a valid partition can be made in the original array.
Time Complexity: O(n) - Needs three linear passes through the array.
Space Complexity: O(n) - Additional space for two auxiliary arrays.
Using auxiliary arrays in JavaScript allows this solution to efficiently compute potential partition points where maximums and minimums from respective ends meet the conditions required for a valid array partitioning.