2560. House Robber IV

, 
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class Solution {
    public int minCapability(int[] nums, int k) {
        int left = 0, right = 0;
        for (int x : nums) {
            right = Math.max(right, x);
        }

        // 二分搜索小偷的最小窃取能力
        while (left + 1 < right) { // 开区间写法
            int mid = (left + right) >>> 1;
            if (check(nums, k, mid)) { // 此处尽量满足条件才能收缩区间，便于寻找最小窃取能力
                right = mid;
            } else {
                left = mid;
            }
        }
        return right;
    }

    /**
     * 检查窃取能力为 maxMoney 时是否至少偷窃 k 个房屋
     */
    private boolean check(int[] nums, int k, int maxMoney) {
        if (nums.length == 1) {
            return (nums[0] <= maxMoney ? 1 : 0) >= k;
        }
        if (nums.length == 2) {
            if (nums[0] <= maxMoney || nums[1] <= maxMoney) {
                return 1 >= k;
            } else {
                return 0 >= k;
            }
        }

        // 定义 dp[i] 为窃取能力为 money 时最多能偷窃的房屋数
        int[] dp = new int[nums.length];

        dp[0] = nums[0] <= maxMoney ? 1 : 0; // 小于等于 money 才能偷，不然就超了
        dp[1] = (nums[0] <= maxMoney || nums[1] <= maxMoney) ? 1 : 0;
        for (int i = 2; i < dp.length; i++) {
            if (nums[i] <= maxMoney) {
                dp[i] = Math.max(dp[i - 1], dp[i - 2] + 1);
            } else {
                dp[i] = dp[i - 1];
            }
        }

        return dp[dp.length - 1] >= k;
    }
}

References

2560. House Robber IV