# 198.house-robber

## Statement

• Difficulty: Medium
• Tag: `数组` `动态规划`

``````输入：[1,2,3,1]

偷窃到的最高金额 = 1 + 3 = 4 。``````

``````输入：[2,7,9,3,1]

偷窃到的最高金额 = 2 + 9 + 1 = 12 。
``````

• `1 <= nums.length <= 100`
• `0 <= nums[i] <= 400`

• Difficulty: Medium
• Tag: `Array` `Dynamic Programming`

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array `nums` representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Example 1:

``````Input: nums = [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.
``````

Example 2:

``````Input: nums = [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.
``````

Constraints:

• `1 <= nums.length <= 100`
• `0 <= nums[i] <= 400`

## Solution

``````#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif

class Solution {
public:
int rob(vector<int> &nums) {
int n = nums.size();
auto f = vector<int>(n + 5, 0);
f = nums;
for (int i = 2; i <= n; i++) {
for (int j = 0; j < i - 1; j++) {
f[i] = max(f[i], nums[i - 1] + f[j]);
}
}

return *max_element(all(f));
}
};

#ifdef LOCAL

int main() {
return 0;
}

#endif
``````