2560. House Robber IV
Explanation:
To solve this problem, we can use dynamic programming. We can define a dp array where dp[i] represents the minimum capability of the robber if we rob at least i houses. We iterate over all possible house combinations and update the dp array accordingly. Finally, we return dp[k] as the result.
- Initialize a dp array of size n+1 where n is the length of the input array nums.
- Initialize dp[0] = 0 and dp[1] = nums[0].
- Iterate over i from 2 to n and for each i, calculate dp[i] as the minimum of maximum of the last i houses robbed.
- Return dp[k].
Time Complexity: O(n^2), where n is the length of the input array nums.
Space Complexity: O(n)
:
class Solution {
public int minCapability(int[] nums, int k) {
int n = nums.length;
int[] dp = new int[n + 1];
dp[0] = 0;
dp[1] = nums[0];
for (int i = 2; i <= n; i++) {
dp[i] = Integer.MAX_VALUE;
for (int j = 1; j <= i; j++) {
dp[i] = Math.min(dp[i], Math.max(dp[i - j], nums[i - j]));
}
}
return dp[k];
}
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