Here’s the LeetCode-ready Java implementation for #154 Find Minimum in Rotated Sorted Array II. Core IdeaThis is binary search, but duplicates break the usual “discard half” logic. We compare“nums[mid]” with“nums[right]”:Condition Meaning Action“nums[mid] nums[right]” Min is in left half (incl. mid)“right mid”“nums[mid] nums[right]” Min is in right half“left mid 1”“nums[mid] nums[right]” Can’t decide (duplicates)“right–” (shrink safely)Why“right–” is safe: If“nums[mid] nums[right]”, even if“nums[right]” were the min, there’s an identical value at“mid”, so we never lose the true minimum.✅ Java Solution (LeetCode format)class Solution {public int findMin(int[] nums) {int left 0;int right nums.length - 1;while (left right) { int mid left (right - left) / 2; if (nums[mid] nums[right]) { // Minimum is in the left half (mid could be the min) right mid; } else if (nums[mid] nums[right]) { // Minimum is in the right half left mid 1; } else { // nums[mid] nums[right], ambiguous due to duplicates right--; } } return nums[left]; }} ComplexityTime:“O(log n)” average / best case → degrades to“O(n)” worst case (e.g.,“[1,1,1,1,1]”)Space:“O(1)” Quick Dry Runnums [2,2,2,0,1]left0, right4, mid2 → nums[2]2 nums[4]1 → left3left3, right4, mid3 → nums[3]0 nums[4]1 → right3leftright → return nums[3] 0 ✅ vs. LeetCode 153 (no duplicates)153: strict comparison, always“O(log n)”,“right mid” /“left mid 1” only.154: adds the“nums[mid] nums[right]” branch →“right–”, which is why worst case can be“O(n)”.Want me to add a recursive version, test cases, or a visual diagram of the search range?
