Get Maximum in Generated Array

Problem Description

You are given an integer n. A 0-indexed integer array nums of length n + 1 is generated in the following way:

• nums[0] = 0
• nums[1] = 1
• nums[2 * i] = nums[i] when 2 <= 2 * i <= n
• nums[2 * i + 1] = nums[i] + nums[i + 1] when 2 <= 2 * i + 1 <= n

Return the maximum integer in the array nums.

Examples

Example 1:

Input: n = 7
Output: 3
Explanation: According to the given rules:
  nums[0] = 0
  nums[1] = 1
  nums[(1 * 2) = 2] = nums[1] = 1
  nums[(1 * 2) + 1 = 3] = nums[1] + nums[2] = 1 + 1 = 2
  nums[(2 * 2) = 4] = nums[2] = 1
  nums[(2 * 2) + 1 = 5] = nums[2] + nums[3] = 1 + 2 = 3
  nums[(3 * 2) = 6] = nums[3] = 2
  nums[(3 * 2) + 1 = 7] = nums[3] + nums[4] = 2 + 1 = 3
Hence, nums = [0,1,1,2,1,3,2,3], and the maximum is max(0,1,1,2,1,3,2,3) = 3.

Example 2:

Input: n = 2
Output: 1
Explanation: According to the given rules, nums = [0,1,1]. The maximum is max(0,1,1) = 1.

Example 3:

Input: n = 3
Output: 2
Explanation: According to the given rules, nums = [0,1,1,2]. The maximum is max(0,1,1,2) = 2.

Constraints

• 0 <= n <= 100

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Solution for Get Maximum in Generated Array Problem

Approach: Dynamic Programming

The problem can be solved efficiently using a dynamic programming approach. We generate the array according to the given rules and keep track of the maximum value.

Code in Different Languages

C++

Written by @ImmidiSivani

class Solution {
public:
    int getMaximumGenerated(int n) {
        if (n == 0) return 0;
        if (n == 1) return 1;
        
        vector<int> nums(n + 1);
        nums[0] = 0;
        nums[1] = 1;
        int maxVal = 1;
        
        for (int i = 2; i <= n; ++i) {
            if (i % 2 == 0) {
                nums[i] = nums[i / 2];
            } else {
                nums[i] = nums[i / 2] + nums[i / 2 + 1];
            }
            maxVal = max(maxVal, nums[i]);
        }
        
        return maxVal;
    }
};

Java

Written by @ImmidiSivani

class Solution {
    public int getMaximumGenerated(int n) {
        if (n == 0) return 0;
        if (n == 1) return 1;

        int[] nums = new int[n + 1];
        nums[0] = 0;
        nums[1] = 1;
        int maxVal = 1;

        for (int i = 2; i <= n; ++i) {
            if (i % 2 == 0) {
                nums[i] = nums[i / 2];
            } else {
                nums[i] = nums[i / 2] + nums[i / 2 + 1];
            }
            maxVal = Math.max(maxVal, nums[i]);
        }

        return maxVal;
    }
}

Python

Written by @ImmidiSivani

class Solution:
    def getMaximumGenerated(self, n: int) -> int:
        if n == 0:
            return 0
        if n == 1:
            return 1
        
        nums = [0] * (n + 1)
        nums[0] = 0
        nums[1] = 1
        max_val = 1
        
        for i in range(2, n + 1):
            if i % 2 == 0:
                nums[i] = nums[i // 2]
            else:
                nums[i] = nums[i // 2] + nums[i // 2 + 1]
            max_val = max(max_val, nums[i])
        
        return max_val

Complexity Analysis

• Time Complexity: $O(n)$
  • We iterate through the array once to fill in the values.
• Space Complexity: $O(n)$
  • We use additional space to store the array nums of length n + 1.

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Authors:

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