You are given an integer n. An 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.
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 3.
Input: n = 2
Output: 1
Explanation: According to the given rules, the maximum between nums[0], nums[1], and nums[2] is 1.
Input: n = 3
Output: 2
Explanation: According to the given rules, the maximum between nums[0], nums[1], nums[2], and nums[3] is 2.
0 <= n <= 100
Solutions (Click to expand)
Similar to the fibonacci, this problem is dependent on calculating numbers before it. We can visualize it with this tree for calculating example number nums[7]
which is using rules for generating a number provided.
nums[7]
/ \
nums[4] + nums[3]
/ / \
nums[2] nums[1] + nums[2]
/ \
nums[1] nums[1]
To find the greatest we can generate each number of the array and find the greatest of all of them. To minimize the number of operations needed to calculate a number in an array, we can store the result for nums[i]
in an array for larger numbers to use. This will prevent us from having to recalculate numbers but at the expense for extra space.