HackerEarth Minimum Flip problem solution

1. Select a non-empty subarray of the array and a number j. Now flip the j-th bit of all the elements of this subarray.
2. Output the minimum sum of the elements of the given array that can be obtained after doing the above operation at most K times.

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define f first
#define s second
#define sc(n) scanf("%d",&n)
#define scl(n) scanf("%lld",&n)
#define pr(n) printf("%d",n)
#define prl(n) printf("%lld",n)
#define nl printf("n")
#define fr(i,n) for(i=0;i<n;i++)
#define rep(i,st,en) for(i=st;i<=en;i++)
#define repv(i,en,st) for(i=en;i>=st;i--)
#define fout cout<<fixed<<setprecision(7)
#define bi(n) __builtin_popcount(n)
#define bil(n) __builtin_popcountll(n)


typedef long long ll;
typedef pair<int,int> pii;
const int N = 100010;
ll mod = 1e9+7;
ll fmod(ll b,ll exp){
    ll res =1;
    while(exp){if(exp&1ll)res=(res*b)%mod;
        b =(b*b)%mod;exp/=2ll;
    }
    return res;
}
int A[N],B[N];
ll C[20*N];

struct node{
    int mxbest,mxlft,mxrt;
    int mnbest,mnlft,mnrt;
    int sm;
    node(){};
    node(int val){
        sm = val;mnlft = val;mxlft = val;
        mxrt = val; mnrt = val;mxbest = val;
        mnbest = val;
    }
    void flip(){
        sm = -sm;
        swap(mxbest,mnbest);
        swap(mnlft,mxlft);swap(mxrt,mnrt);
        mxbest = -mxbest;mxlft = -mxlft;
        mxrt  = -mxrt;mnbest = -mnbest;
        mnlft = -mnlft; mnrt = -mnrt;
    }

};
node tr[4*N];
int lazy[4*N];
node merge(node &a,node &b)
{
    node p;
    p.mxbest = max(a.mxbest,max(b.mxbest,a.mxrt+b.mxlft));
    p.mxlft = max(a.mxlft,a.sm+b.mxlft);
    p.mxrt = max(b.mxrt, b.sm+a.mxrt);
    p.mnlft = min(a.mnlft,a.sm+b.mnlft);
    p.mnrt = min(b.mnrt,b.sm+a.mnrt);
    p.mnbest = min(a.mnbest,min(b.mnbest,a.mnrt+b.mnlft));
    p.sm = a.sm + b.sm ;
    return p;
}


void build(int i,int st,int en){
    lazy[i]=0;
    if(st == en){
        tr[i] = node(B[st]);
        return;
    }
    int mid = (st+en)>>1;
    build(i+i,st,mid);build(i+i+1,mid+1,en);
    tr[i]= merge(tr[i+i],tr[i+i+1]);
}

inline void prop(int i,int st,int en){
    if(lazy[i] && st!=en){
        tr[i+i].flip();
        lazy[i+i]^=1;
        tr[i+i+1].flip();
        lazy[i+i+1]^=1;
    }
    lazy[i]=0;
}
void upd(int i,int st,int en,int l,int r){
    if(st >r || l>en || st>en)return;
    if(st>=l && en<=r){
        tr[i].flip();
        lazy[i]^=1;
        return;
    }
    if(lazy[i])prop(i,st,en);
    int mid = (st+en)>>1;
    if(l<= mid )upd(i+i,st,mid,l,r);
    if(r>mid)upd(i+i+1,mid+1,en,l,r);
    tr[i]= merge(tr[i+i],tr[i+i+1]);
}
int getLft(int i,int st,int en,int val){
    if(tr[i].sm == val)return st;
    int mid = (st+en)>>1;
    if(lazy[i])prop(i,st,en);
    if(tr[i+i+1].mxrt == val)return getLft(i+i+1,mid+1,en,val);
    return getLft(i+i,st,mid,val-tr[i+i+1].sm);
}
int getRt(int i,int st,int en,int val){
    if(tr[i].sm == val)return en;
    int mid = (st+en)>>1;
    if(lazy[i])prop(i,st,en);
    if(tr[i+i].mxlft == val)return getRt(i+i,st,mid,val);
    return getRt(i+i+1,mid+1,en,val-tr[i+i].sm);
}
pair<int,int> getAnswer(int i,int st,int en,int val){

    if(tr[i].sm == val)return mp(st,en);
    if(lazy[i])prop(i,st,en);
    int mid = (st+en)>>1;
    if(tr[i+i].mxbest == val)return getAnswer(i+i,st,mid,val);
    if(tr[i+i+1].mxbest == val)return getAnswer(i+i+1,mid+1,en,val);
    int l1 = getLft(i+i,st,mid,val-tr[i+i+1].mxlft);
    int r1 = getRt(i+i+1,mid+1,en,val-tr[i+i].mxrt);
    return mp(l1,r1);
}





int main()
{
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    //ios_base::sync_with_stdio(false);cin.tie(NULL);
    int n,m,i,j,k,x,q;
    int t=1;
    cin>>n>>k;
    ll ans = 0;
    int len = 0;
    rep(i,1,n){
        cin>>A[i];
        ans += A[i];
    }
    for(int bit=19;bit>=0;bit--){
        rep(i,1,n)if((1<<bit)&A[i])B[i]=1;
        else B[i]=-1;
        build(1,1,n);
        while(tr[1].mxbest>0)
        {
            int cur = tr[1].mxbest;
            pair<int,int> ind = getAnswer(1,1,n,cur);
            upd(1,1,n,ind.f,ind.s);
            C[++len] = (1ll<<bit)*1ll*cur;
        }
    }
    sort(C+1,C+len+1);
    reverse(C+1,C+len+1);
    rep(i,1,min(k,len)){
       // cout<<C[i]<<"n";
        ans-=C[i];
    }
    cout<<ans<<"n";
    return 0;
}