LeetCode 375: Guess Number Higher or Lower II Solution
Master LeetCode problem 375 (Guess Number Higher or Lower II), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
375. Guess Number Higher or Lower II
Problem Explanation
Explanation:
To solve this problem, we can use dynamic programming. Let's define a function dp(i, j) which represents the minimum cost to guess a number between i and j. We need to find dp(1, n).
The key observation is that for any x chosen as the guess, the maximum cost is x + max(dp(i, x-1), dp(x+1, j)). Therefore, we can iterate x from i to j and take the minimum of all such costs.
The final answer will be the minimum cost over all choices of x.
Time Complexity:
The time complexity of this dynamic programming approach is O(n^3) where n is the given number.
Space Complexity:
The space complexity is O(n^2) for storing the DP table.
:
Solution Code
class Solution {
public int getMoneyAmount(int n) {
int[][] dp = new int[n+1][n+1];
for (int len = 2; len <= n; len++) {
for (int start = 1; start <= n - len + 1; start++) {
int minCost = Integer.MAX_VALUE;
for (int guess = start; guess < start + len - 1; guess++) {
int cost = guess + Math.max(dp[start][guess - 1], dp[guess + 1][start + len - 1]);
minCost = Math.min(minCost, cost);
}
dp[start][start + len - 1] = minCost;
}
}
return dp[1][n];
}
}Try It Yourself
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