Find the Most Competitive Subsequence

Build the lexicographically smallest subsequence of length k with a monotonic increasing stack. At index i, a larger previous value should be removed if the current value is smaller and there are enough elements left to still fill k positions after removing it. A convenient condition is stack length minus one plus (n-i) at least k. Pop while that condition and top greater than current both hold. Then push current if the stack has fewer than k elements; otherwise skip it. The greedy exchange argument is that replacing an earlier larger value with a smaller available value improves the earliest differing position, provided feasibility of final length is preserved. Test k zero if permitted, k equal n, equal values, decreasing input, and repeated candidates. Each element is pushed once and popped at most once, so time is O(n) and stack space is O(k) or O(n) during processing depending on implementation.