Binary-Search

LeetCode 611 note: sort, fix the longest side k, and use two pointers for the shorter sides; when the sum is large enough every i in between works, adding j - i at once

LeetCode 540 note: before the single element every pair starts at an even index, so binary search whether nums[mid] equals nums[mid + 1] at an even mid

LeetCode 275 note: with sorted citations, find the first i where citations[i] >= n - i; the answer is n - i, in O(log n) with binary search

LeetCode 1337 note: soldiers come first in each row, so binary search the first 0 for the count; sort by (count, row index) and take k

LeetCode 222 note: in a complete tree, equal leftmost depths on both sides mean one side is perfect, counted as 2^h - 1, for O(log² n)

LeetCode 240 note: start at the top-right corner, moving left when larger and down when smaller, eliminating a row or column each step

LeetCode 162 note: nums[mid] < nums[mid + 1] guarantees a peak to the right (rising until the edge at worst), otherwise one at or left of mid

LeetCode 287 note: treat i -> nums[i] as a linked list; the duplicate is the cycle entry, found with Floyd’s algorithm in O(1) space without modifying the array

LeetCode 981 note: store a (timestamp, value) list per key, with timestamps increasing; get binary searches the last timestamp <= the query

LeetCode 875 note: binary search the speed for the smallest k with total hours <= h, continuing left even when the hours equal h

LeetCode 153 note: compare mid with the right end: nums[mid] > nums[r] puts the minimum to the right, otherwise mid may be it, so r = mid

LeetCode 167 note: sorted, so move two pointers inward, shifting right when the sum is too large and left when too small; indices are 1-based