Find the number of players cannot win the game?

We are given n players, each player has 3 values assigned A, B and C.

A player i cannot win if there exists another player j with all 3 values A[j] > A[i], B[j] > B[i] and C[j] > C[i]. We are asked to find number of players cannot win.

I tried this problem using brute force, which is a linear search over players array. But it's showing TLE.

For each player i, I am traversing the complete array to find if there exists any other player j for which the above condition holds true.

Code :

int count_players_cannot_win(vector<vector<int>> values) {
    int c = 0;
    int n = values.size();
    for(int i = 0; i < n; i++) {
        for(int j = 0; j!= i && j < n; j++) {
            if(values[i][0] < values[j][0] && values[i][1] < values[j][1] && values[i][2] < values[j][2]) { 
                c += 1;
                break;
           }
        }    
    }
    return c;
}

And this approach is O(n^2), as for every player we are traversing the complete array. Thus it is giving the TLE.

Sample testcase :

Sample Input

3(number of players)
A B C
1 4 2
4 3 2
2 5 3

Sample Output :

1 

Explanation :
Only player1 cannot win as there exists player3 whose all 3 values(A, B and C) are greater than that of player1.

Contraints :

n(number of players) <= 10^5

What would be optimal way to solve this problem?

Solution:

int n;
const int N = 4e5 + 1;
int tree[N];

int get_max(int i, int l, int r, int L) {  // range query of max in range v[B+1: n]
    if(r < L || n <= l)
        return numeric_limits<int>::min();
    else if(L <= l)
        return tree[i];
    int m = (l + r)/2;
    return max(get_max(2*i+1, l, m, L), get_max(2*i+2, m+1, r, L));
}
void update(int i, int l, int r, int on, int v) { // point update in tree[on]
    if(r < on || on < l) 
        return;
    else if(l == r) {
        tree[i] = max(tree[i], v);
        return;
    }
    int m = (l + r)/2;
    update(2*i+1, l, m, on, v);
    update(2*i+2, m + 1, r, on, v);
    tree[i] = max(tree[2*i+1], tree[2*i+2]);
}

bool comp(vector<int> a, vector<int> b) {
    return a[0] != b[0] ? a[0] > b[0] : a[1] < b[1];  
}

int solve(vector<vector<int>> &v) {
    n = v.size();
    vector<int> b(n, 0);           // reduce the scale of range from [0,10^9] to [0,10^5]
    for(int i = 0; i < n; i++) {
        b[i] = v[i][1];
    }
    for(int i = 0; i < n; i++) {
        cin >> v[i][2];
    }

//     sort on 0th col in reverse order
    sort(v.begin(), v.end(), comp);
    sort(b.begin(), b.end());
    
    int ans = 0;
    for(int i = 0; i < n;) {
        int j = i;
        while(j < n &&  v[j][0] == v[i][0]) {    
            int B = v[j][1];
            int pos = lower_bound(b.begin(), b.end(), B) - b.begin(); // position of B in b[]
            int mx = get_max(0, 0, n - 1, pos + 1);
            if(mx > v[j][2])
                ans += 1;
            j++;
        }
        while(i < j) {
            int B = v[i][1];
            int C = v[i][2];
            int pos = lower_bound(b.begin(), b.end(), B) - b.begin(); // position of B in b[]
            update(0, 0, n - 1, pos, C);
            i++;
        }
    }
    return ans;
}

This solution uses segment tree, and thus solves the problem in time O(n*log(n)) and space O(n).

Approach is explained in the accepted answer by @Primusa.