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class Solution {
    public int minimumCardPickup(int[] cards) {
        Map<Integer, Integer> lastOccur = new HashMap<>();
        int ans = Integer.MAX_VALUE;
        for (int i = 0; i < cards.length; i++) {
            if (lastOccur.containsKey(cards[i])) {
                ans = Math.min(ans, i - lastOccur.get(cards[i]) + 1);
            }
            lastOccur.put(cards[i], i);
        }
        return ans == Integer.MAX_VALUE ? -1 : ans;
    }
}

用hashmap记录每个值最后一次出现的index是多少.

时间复杂度: O(n)
空间复杂度: O(n)