What is the longest increasing subsequence (LIS)?

The Longest Increasing Subsequence (LIS) problem finds the length of the longest strictly increasing subsequence in an array. Example: in [10, 9, 2, 5, 3, 7, 101, 18], LIS = [2, 3, 7, 101], length 4. DP solution (O(n²)): dp[i] = LIS ending at index i. For each i, dp[i] = max(dp[j] + 1) for all j < i where arr[j] < arr[i]. Base: dp[i] = 1. Answer: max(dp[i]). Optimal O(n log n) solution using binary search: maintain a sorted array tails where tails[i] is the smallest possible tail element of all increasing subsequences of length i+1. For each element x: binary search (lower_bound) for the first element in tails ≥ x; if found, replace it with x; otherwise append x. The length of tails is the LIS length. This gives the LIS length but not the actual sequence (requires additional tracking). Applications: (1) Patience sorting (O(n log n) deck of cards sort related to LIS); (2) Dilworth's theorem — minimum number of non-increasing subsequences to partition the array = LIS length; (3) Edit distance variant; (4) Stock trading problems. Often combined with other DP problems (LIS in 2D — the Russian doll envelopes problem).