LeetCode5897. 将数组分成两个数组并最小化数组和的差

思路#

此题做法与LC 1755. 最接近目标值的子序列和类似 详情见这篇Blog
还是把数组分为两部分,分别用状态压缩进行枚举,然后对第二部分所凑成的值在第一部分中二分查找即可

AC_CODE#

#define fi first    
#define se second    
#define pb push_back
#define mk make_pair
#define len(a) (int)a.size()
#define mod(x) ((x) % MOD)
#define lowbit(x) (x&-x)
#define LL long long
#define fix(a) fixed << setprecision(a) 
#define debug(x) cout<<#x" ----> "<<x<<endl
#define rep(i, b, s) for(register int i = (b); i <= (s); ++i)
#define pre(i, b, s) for(register int i = (b); i >= (s); --i)
#define all(v) (v).begin(),(v).end()
#define leng(v) ((int)v.size())
const int INF = INT_MAX;
const LL INFF = INT64_MAX;
const int MOD = 1e9 + 7; //// 998244353
class Solution {
public:
    // map<int, vector<int> > mp;
    vector<int> v[20];
    int minimumDifference(vector<int>& a) {
        for(int i = 0; i < 20; i ++ ) v[i].clear();
        int n = len(a), px = n / 2;
        vector<int> l, r, res;
        int anss = 0;
        int p;
        for(int i = 0; i < px; i ++ ) {
            l.pb(a[i]);
            r.pb(a[i + px]);
            anss += a[i] + a[i + px];
        }
        p = anss;
        anss /= 2;
        for(int i = 0; i < 1 << px; i ++ ) {
            int cnt = 0, sum = 0;
            for(int j = 0; j < px; j ++ ) 
                if(i >> j & 1)
                    cnt ++, sum += l[j];
            v[cnt].pb(sum);
        }
        int ans = INF;
        for(int i = 1; i <= px; i ++ ) sort(all(v[i]));
        for(int i = 0; i < 1 << px; i ++ ) {
            int cnt = 0, sum = 0;
            for(int j = 0; j < px; j ++ ) {
                if(i >> j & 1) 
                    cnt ++, sum += r[j];
            }
            int pp = sum;
            sum = anss - sum;
            auto p = lower_bound(all(v[px - cnt]), sum);

            if(p != v[px - cnt].end()) ans = min(ans, abs( *p + pp - anss) * 2);
            if(p != v[px - cnt].begin()) ans = min(ans, abs(*prev(p) + pp - anss) * 2 );
        }
        if(p & 1) ans ++;
        return ans;     
    }
};