HackerEarth Zeros and Ones problem solution

In this HackerEarth zero and One’s problem solution, You are given an array A which contains initially only ones. You can perform two operations:

I: Given an index I, update the value of AI to zero.
K: Given the value K, print the index of Kth one in the array A. If there is no such index then print 1.

HackerEarth Zeros and One’s problem solution.

#include <iostream>
using namespace std;
int tree[4000000];

void build(int node, int st, int en) {
    if (st == en) {
        tree[node] = 1;
        return;
    }
    int mid = (st+en)/2;
    build(2*node, st, mid);
    build(2*node+1, mid+1, en);
    tree[node] = tree[2*node] + tree[2*node+1];
}

void update(int node, int st, int en, int index) {
    if (st == index && en == index) {
        tree[node] = 0;
        return;
    }

    int mid = (st+en)/2;
    if(index >= st && index <= mid) {
        update(2*node, st, mid, index);
    } else {
        update(2*node+1, mid+1, en, index);
    }
    tree[node] = tree[2*node] + tree[2*node+1];
}

int query(int node, int st, int en, int k, int n) {
    if (st < 1 || en > n || tree[node] < k) {
        return -1;
    }
    if (st == en && k == 1) {
        return st;
    }
    int val = tree[node];
    int mid = (st+en)/2;
    int leftNode = tree[2*node];
    int rightNode = tree[2*node+1];
    if (k >leftNode) {
        return query(2*node+1, mid+1, en, k-leftNode, n);
    } else {
        return query(2*node, st, mid, k, n);
    }
}

int main() {
    int N, K, Q, type, I;
    cin >> N;
    build(1, 1, N);
    cin >> Q;
    while(Q--) {
        cin >> type;
        if (type) {
            cin >> K;
            cout << query(1, 1, N, K, N) << endl;
        } else {
            cin >> I;
            update(1, 1, N, I);
        }
    }
    return 0;
}

Second solution

#pragma GCC optimize("O3")
#include <bits/stdc++.h>

#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define be begin()
#define en end()
#define all(x) (x).begin(),(x).end()
#define alli(a, n, k) (a+k),(a+n+k)
#define REP(i, a, b, k) for(__typeof(a) i = a;i < b;i += k)
#define REPI(i, a, b, k) for(__typeof(a) i = a;i > b;i -= k)
#define REPITER(it, a) for(__typeof(a.begin()) it = a.begin();it != a.end(); ++it)

#define y0 sdkfaslhagaklsldk
#define y1 aasdfasdfasdf
#define yn askfhwqriuperikldjk
#define j1 assdgsdgasghsf
#define tm sdfjahlfasfh
#define lr asgasgash
#define norm asdfasdgasdgsd
#define have adsgagshdshfhds

#define eps 1e-6
#define pi 3.141592653589793

using namespace std;

template<class T> inline T gcd(T a, T b) { while(b) b ^= a ^= b ^= a %= b; return a; }
template<class T> inline T mod(T x) { if(x < 0) return -x; else return x; }

typedef vector<int> VII;
typedef vector<ll> VLL;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef pair<int, PII > PPII;
typedef vector< PII > VPII;
typedef vector< PPII > VPPI;

const int MOD = 1e9 + 7;
const int INF = 1e9;

// Template End

const int MAX = 1e6 + 5;
int tree[MAX], n;
bool A[MAX];

int read(int idx) {
    int sum = 0;
    while (idx > 0) {
        sum += tree[idx];
        idx -= (idx & -idx);
    }
    return sum;
}


void update(int idx, int val) {
    while (idx <= n) {
        tree[idx] += val;
        idx += (idx & -idx);
    }
}

int binary(int k) {
    int l, r, mid, y;
    l = 1, r = n;
    while (l < r) {
        mid = (l + r) >> 1;
        y = read(mid);
        if (y == k) r = mid;
        else if (y < k) l = mid + 1;
        else r = mid-1;
    }
    if (read(l) == k) return l;
    else return -1;
}

int main(int argc, char* argv[]) {
    if(argc == 2 or argc == 3) freopen(argv[1], "r", stdin);
    if(argc == 3) freopen(argv[2], "w", stdout);
    ios::sync_with_stdio(false);
    int q, x, y;
    cin >> n >> q;
    assert(1 <= n and n <= 1000000);
    assert(1 <= q and q <= 1000000);

    REP(i, 1, n+1, 1) update(i, 1);

    REP(i, 0, q, 1) {
        cin >> x >> y;
        assert(0 <= x and x <= 1);
        assert(1 <= y and y <= n);
        if (x == 0) {
            if (A[y] == false) update(y, -1);
            A[y] = true;
        }
        else {
            cout << binary(y) << endl;
        }
    }
    return 0;
}

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