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:
1. I: Given an index I, update the value of AI to zero.
2. 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 j1 assdgsdgasghsf
#define tm sdfjahlfasfh
#define lr asgasgash
#define norm asdfasdgasdgsd

#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 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;
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;
}```