Last updated on October 6th, 2024 at 02:04 pm

Here, We see Guess Number Higher or Lower II LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Guess Number Higher or Lower II LeetCode Solution

Problem Statement

We are playing the Guessing Game. The game will work as follows:

1. I pick a number between 1 and n.
2. You guess a number.
3. If you guess the right number, you win the game.
4. If you guess the wrong number, then I will tell you whether the number I picked is higher or lower, and you will continue guessing.
5. Every time you guess a wrong number x, you will pay x dollars. If you run out of money, you lose the game.

Given a particular n, return the minimum amount of money you need to guarantee a win regardless of what number I pick.

Example 1:

Input: n = 10
Output: 16

Explanation: The winning strategy is as follows:
– The range is [1,10]. Guess 7.  
– If this is my number, your total is $0. Otherwise, you pay $7.  
– If my number is higher, the range is [8,10]. Guess 9.  
– If this is my number, your total is $7. Otherwise, you pay $9.  
– If my number is higher, it must be 10. Guess 10. Your total is $7 + $9 = $16.  

– If my number is lower, it must be 8. Guess 8. Your total is $7 + $9 = $16.  
– If my number is lower, the range is [1,6]. Guess 3.  
– If this is my number, your total is $7. Otherwise, you pay $3.  
– If my number is higher, the range is [4,6]. Guess 5.  
– If this is my number, your total is $7 + $3 = $10. Otherwise, you pay $5.  

– If my number is higher, it must be 6. Guess 6. Your total is $7 + $3 + $5 = $15.  
– If my number is lower, it must be 4. Guess 4. Your total is $7 + $3 + $5 = $15.  
– If my number is lower, the range is [1,2]. Guess 1.  
– If this is my number, your total is $7 + $3 = $10. Otherwise, you pay $1.  
– If my number is higher, it must be 2. Guess 2. Your total is $7 + $3 + $1 = $11. The worst case in all these scenarios is that you pay $16. Hence, you only need $16 to guarantee a win.

Example 2:
Input: n = 1
Output: 0
Explanation: There is only one possible number, so you can guess 1 and not have to pay anything.

Example 3:
Input: n = 2
Output: 1
Explanation: There are two possible numbers, 1 and 2.
– Guess 1.  
– If this is my number, your total is $0. Otherwise, you pay $1.  
– If my number is higher, it must be 2. Guess 2. Your total is $1. The worst case is that you pay $1.

1. Guess Number Higher or Lower II LeetCode Solution C++

class Solution {
public:
    int getMoneyAmount(int n) {
        vector<vector<int>> dp(n+1, vector<int>(n+1, 0));
        for(int len=2; len<=n; ++len){
            for(int begin=0; begin<=n-len; ++begin){
                int end = begin + len;
                for(int i=begin; i<end; ++i){
                    int numPicked = i+1;
                    if(i == begin){
                        dp[begin][end] = numPicked + dp[begin+1][end];
                    } else {
                        dp[begin][end] = min(dp[begin][end], max(dp[begin][i], dp[i+1][end]) + numPicked);
                    }
                }
            }
        }
        return dp[0][n];
    }
};

2. Guess Number Higher or Lower II LeetCode Solution Java

class Solution {
    public int getMoneyAmount(int n) {
	    int[][] f = new int[n + 1][n + 1];
	    Deque<Integer[]> q;
	    int a, b, k0, v, f1, f2;
	    for (b = 2; b <= n; b++) {
		    k0 = b - 1;
		    q = new LinkedList<Integer[]>();
		    for (a = b - 1; a > 0; a--) {
                while (f[a][k0 - 1] > f[k0 + 1][b])
                    k0--;
                while (!q.isEmpty() && q.peekFirst()[0] > k0)
                    q.pollFirst();
                v = f[a + 1][b] + a;
                while (!q.isEmpty() && v < q.peekLast()[1])
                    q.pollLast();
                q.offerLast(new Integer[] { a, v });
                f1 = q.peekFirst()[1];
                f2 = f[a][k0] + k0 + 1;
                f[a][b] = Math.min(f1, f2);
            }
        }
        return f[1][n];
    }
}

3. Guess Number Higher or Lower II LeetCode Solution JavaScript

var getMoneyAmount = function (n) {
  const dp = Array(n + 1).fill(null).map(() => Array(n + 1).fill(Infinity))
  const minimax = (l, r) => {
    if (l >= r) return 0
    if (dp[l][r] !== Infinity) return dp[l][r]
    for (let i = l; i <= r; i++) {
      dp[l][r] = Math.min(dp[l][r], i + Math.max(minimax(i + 1, r), minimax(l, i - 1)))
    }
    return dp[l][r]
  }
  return minimax(1, n)
};

4. Guess Number Higher or Lower II LeetCode Solution Python

class Solution(object):
    def getMoneyAmount(self, n):
        need = [[0] * (n+1) for _ in range(n+1)]
        for lo in range(n, 0, -1):
            for hi in range(lo+1, n+1):
                need[lo][hi] = min(x + max(need[lo][x-1], need[x+1][hi])
                                for x in range(lo, hi))
        return need[1][n]