Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,4,4,5,6,7] might become:
- [4,5,6,7,0,1,4] if it was rotated 4 times.
- [0,1,4,4,5,6,7] if it was rotated 7 times.
Notice that rotating an array [a[0], a[1], a[2], …, a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], …, a[n-2]].
Given the sorted rotated array nums that may contain duplicates, return the minimum element of this array.
You must decrease the overall operation steps as much as possible.
Example 1:
Input: nums = [1,3,5]
Output: 1
Example 2:
Input: nums = [2,2,2,0,1]
Output: 0
We use a `modified` binary search to find the minimum element:-
- At every point, compare (‘i’, ‘mid’) with (‘mid+1’, ‘j’). Whichever section has the lower value, let’s move that direction. Because of duplicates, we might be in a situation where there’s no one clear winner. In that case, move in both directions.
if ( (a < c && a < d) || (b < c && b < d) ) // go left else if ((c > a && c > b) || (d > a && d > b)) // go right else // go both directions.
- Return the last number remaining in the recursion (when i === j)
/** * @param {number[]} nums * @return {number} */ var findMin = function(nums) { return findMinRecursive(nums, 0, nums.length-1); }; var findMinRecursive = function(nums, i, j) { if (i === j) return nums[i]; let mid = i + Math.floor((j-i)/2); const a = nums[i]; const b = nums[mid]; const c = nums[mid + 1]; const d = nums[j]; let left = Number.MAX_VALUE; let right = Number.MAX_VALUE; if ((a < c && a < d) || (b < c && b < d)) { // go left left = findMinRecursive(nums, i, mid); } else if ((c > a && c > b) || (d > a && d > b)) { right = findMinRecursive(nums, mid+1, j); } else { // go both left = findMinRecursive(nums, i, mid); right = findMinRecursive(nums, mid+1, j); } return Math.min(left, right); }; |
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