Minimum Levels to Gain More Points (LeetCode)

Problem Description

Problem Statement	Solution Link	LeetCode Profile
Minimum Levels to Gain More Points	Minimum Levels to Gain More Points Solution on LeetCode	vaishu_1904

Problem Description

You are given a binary array possible of length n.

Alice and Bob are playing a game that consists of n levels. Some of the levels in the game are impossible to clear while others can always be cleared. In particular, if possible[i] == 0, then the ith level is impossible to clear for both players. A player gains 1 point on clearing a level and loses 1 point if the player fails to clear it.

At the start of the game, Alice will play some levels in the given order starting from the 0th level, after which Bob will play for the rest of the levels.

Alice wants to know the minimum number of levels she should play to gain more points than Bob, if both players play optimally to maximize their points.

Return the minimum number of levels Alice should play to gain more points. If this is not possible, return -1.

Note that each player must play at least 1 level.

Example 1

• Input: possible = [1,0,1,0]
• Output: 1
• Explanation:
  • If Alice plays only level 0 and Bob plays the rest of the levels, Alice has 1 point, while Bob has -1 + 1 - 1 = -1 point.
  • If Alice plays till level 1 and Bob plays the rest of the levels, Alice has 0 points, while Bob has 0 points.
  • If Alice plays till level 2 and Bob plays the rest of the levels, Alice has 1 point, while Bob has -1 point.
  • Alice must play a minimum of 1 level to gain more points.

Example 2

• Input: possible = [1,1,1,1,1]
• Output: 3
• Explanation:
  • If Alice plays only level 0 and Bob plays the rest of the levels, Alice has 1 point, while Bob has 4 points.
  • If Alice plays till level 1 and Bob plays the rest of the levels, Alice has 2 points, while Bob has 3 points.
  • If Alice plays till level 2 and Bob plays the rest of the levels, Alice has 3 points, while Bob has 2 points.
  • If Alice plays till level 3 and Bob plays the rest of the levels, Alice has 4 points, while Bob has 1 point.
  • Alice must play a minimum of 3 levels to gain more points.

Example 3

• Input: possible = [0,0]
• Output: -1
• Explanation: The only possible way is for both players to play 1 level each. Alice plays level 0 and loses 1 point. Bob plays level 1 and loses 1 point. As both players have equal points, Alice can't gain more points than Bob.

Constraints

• 2 <= n == possible.length <= 10^5
• possible[i] is either 0 or 1.

Approach

To solve this problem, we can use a greedy approach combined with a two-pointer technique:

1. Traverse through the possible array and determine the total number of levels that can be played (total_possible_levels).
2. Initialize two pointers: alice_played and bob_played to simulate both players playing optimally.
3. Start by considering alice_played from 1 level up to total_possible_levels - 1.
4. Calculate the points for Alice and Bob after Alice plays alice_played levels and Bob plays the rest.
5. Track the minimum number of levels Alice needs to play to have more points than Bob.
6. Return the result based on the tracked minimum number of levels or -1 if it's not possible for Alice to gain more points.

Solution Code

Python

class Solution:
    def minLevelsToGainMorePoints(self, possible: List[int]) -> int:
        n = len(possible)
        total_possible_levels = sum(possible)
        
        if total_possible_levels < 2:
            return -1
        
        alice_played = 0
        min_levels = float('inf')
        
        alice_points = 0
        bob_points = 0
        
        for i in range(n):
            if possible[i] == 1:
                bob_points += 1
        
        for i in range(n):
            if possible[i] == 1:
                alice_played += 1
                alice_points += 1
                bob_points -= 1
            
            if alice_played >= 1 and alice_points > bob_points:
                min_levels = min(min_levels, alice_played)
        
        return min_levels if min_levels != float('inf') else -1

C++

#include <vector>
using namespace std;

class Solution {
public:
    int minLevelsToGainMorePoints(vector<int>& possible) {
        int n = possible.size();
        int total_possible_levels = 0;
        
        for (int i = 0; i < n; ++i) {
            total_possible_levels += possible[i];
        }
        
        if (total_possible_levels < 2) {
            return -1;
        }
        
        int alice_played = 0;
        int min_levels = INT_MAX;
        
        int alice_points = 0;
        int bob_points = 0;
        
        for (int i = 0; i < n; ++i) {
            if (possible[i] == 1) {
                bob_points++;
            }
        }
        
        for (int i = 0; i < n; ++i) {
            if (possible[i] == 1) {
                alice_played++;
                alice_points++;
                bob_points--;
            }
            
            if (alice_played >= 1 && alice_points > bob_points) {
                min_levels = min(min_levels, alice_played);
            }
        }
        
        return min_levels != INT_MAX ? min_levels : -1;
    }
};

Java

class Solution {
    public int minLevelsToGainMorePoints(int[] possible) {
        int n = possible.length;
        int total_possible_levels = 0;
        
        for (int i = 0; i < n; ++i) {
            total_possible_levels += possible[i];
        }
        
        if (total_possible_levels < 2) {
            return -1;
        }
        
        int alice_played = 0;
        int min_levels = Integer.MAX_VALUE;
        
        int alice_points = 0;
        int bob_points = 0;
        
        for (int i = 0; i < n; ++i) {
            if (possible[i] == 1) {
                bob_points++;
            }
        }
        
        for (int i = 0; i < n; ++i) {
            if (possible[i] == 1) {
                alice_played++;
                alice_points++;
                bob_points--;
            }
            
            if (alice_played >= 1 && alice_points > bob_points) {
                min_levels = Math.min(min_levels, alice_played);
            }
        }
        
        return min_levels != Integer.MAX_VALUE ? min_levels : -1;
    }
}

Conclusion

The provided solutions utilize a greedy approach combined with a two-pointer technique to determine the minimum number of levels Alice should play to gain more points than Bob. By iterating through possible levels Alice can play and computing the resultant points for both players, we can efficiently determine the solution to this problem.