Find the Most Competitive Subsequence

This approach involves generating every possible subsequence of length k from the input array nums. After generating all such subsequences, we compare them lexicographically to find the 'most competitive' one, which is the lexicographically smallest subsequence.

• Initialize a list mostCompetitive of size k with a placeholder value (e.g., Integer.MAX_VALUE).
• Define a recursive function backtrack(start, current).
• Inside backtrack(start, current):
  • If current.size() == k:
    • Compare current with mostCompetitive.
    • If current is lexicographically smaller, update mostCompetitive with the elements of current.
    • Return.
  • Add a pruning step: if the number of elements picked (current.size()) plus the number of remaining elements in nums (nums.length - start) is less than k, we can't form a valid subsequence, so we return early.
  • Iterate i from start to nums.length - 1:
    • Add nums[i] to current.
    • Call backtrack(i + 1, current).
    • Remove the last element from current (this is the backtracking step).
• Start the process by calling backtrack(0, new ArrayList<>()).
• Return mostCompetitive.

We can use a recursive backtracking function to explore all combinations. The function, say findSubsequences(startIndex, currentSubsequence), would work as follows:

• The startIndex indicates the starting position in nums for the current recursive call to prevent duplicate combinations.
• currentSubsequence is the list of numbers we have picked so far.
• Base Case: If currentSubsequence.size() equals k, we have found a valid subsequence. We then compare it with the best subsequence found so far and update the best one if the current one is more competitive.
• Recursive Step: We iterate from startIndex to nums.length - 1. For each element nums[i], we add it to currentSubsequence and make a recursive call findSubsequences(i + 1, currentSubsequence). After the call returns, we backtrack by removing nums[i] from currentSubsequence to explore other possibilities.

To avoid storing all C(n, k) subsequences, which would consume a massive amount of memory, we can maintain a single 'best' subsequence found so far and update it whenever we find a more competitive one in the base case of our recursion.

class Solution {
    int[] mostCompetitive;
    int k;
    int[] nums;

    public int[] mostCompetitive(int[] nums, int k) {
        this.mostCompetitive = null;
        this.k = k;
        this.nums = nums;
        backtrack(0, new java.util.ArrayList<>());
        return mostCompetitive;
    }

    private void backtrack(int start, java.util.List<Integer> current) {
        // Pruning: if remaining elements are not enough to form a k-length subsequence
        if (current.size() + (nums.length - start) < k) {
            return;
        }

        if (current.size() == k) {
            int[] currentArr = current.stream().mapToInt(i -> i).toArray();
            if (mostCompetitive == null || isMoreCompetitive(currentArr, mostCompetitive)) {
                mostCompetitive = currentArr;
            }
            return;
        }

        for (int i = start; i < nums.length; i++) {
            current.add(nums[i]);
            backtrack(i + 1, current);
            current.remove(current.size() - 1); // backtrack
        }
    }

    private boolean isMoreCompetitive(int[] a, int[] b) {
        for (int i = 0; i < a.length; i++) {
            if (a[i] < b[i]) {
                return true;
            }
            if (a[i] > b[i]) {
                return false;
            }
        }
        return false; // they are equal
    }
}