710025. Levels Of Game

Medium

Levels Of Game

Summary

The "Levels Of Game" problem involves designing a game where players can progress through levels, and each level has its own set of rules and rewards. The goal is to determine the maximum level that a player can reach based on their current score and the scores required for each level.

This problem requires an understanding of dynamic programming concepts, specifically optimization problems.

Detailed Explanation

The game consists of multiple levels, each with a specific score requirement to progress. We are given the current score of the player and the scores required for each level. Our task is to find the maximum level that the player can reach based on their current score.

To solve this problem, we can use dynamic programming. We initialize an array dp where dp[i] represents the maximum level that a player with a score i can reach.

We iterate through each level and update the dp array accordingly. For each level i, if the current score is greater than or equal to the score required for level i, then we set dp[i] = i. If not, we set dp[i] = dp[i-1].

The maximum level that a player can reach is the last index of the dp array where dp[i] is non-zero.

Here's a step-by-step breakdown:

  • Initialize dp array with all elements set to 0.
  • Iterate through each level:
    • If current score >= score required for level i, then dp[i] = i.
    • Otherwise, dp[i] = dp[i-1].
  • Return the maximum level that a player can reach, which is the last index of the dp array where dp[i] is non-zero.

Time complexity: O(n), where n is the number of levels. Space complexity: O(n), for the dynamic programming array.

Optimized Solutions

Java

View on GeeksforGeeks
public class LevelsOfGame {
    public static int maxLevel(int[] scores, int score) {
        int[] dp = new int[scores.length];
        int maxLevel = 0;
        
        for (int i = 0; i < scores.length; i++) {
            if (score >= scores[i]) {
                dp[i] = i + 1;
                maxLevel = Math.max(maxLevel, dp[i]);
            } else {
                dp[i] = dp[i - 1];
            }
        }
        
        return maxLevel;
    }
}

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