0496-Next Greater Element I.md

Next Greater Element I

You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2.
Find all the next greater numbers for nums1's elements in the corresponding places of nums2.

The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2.
If it does not exist, output -1 for this number.

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
Output: [-1,3,-1]
Explanation:
    For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
    For number 1 in the first array, the next greater number for it in the second array is 3.
    For number 2 in the first array, there is no next greater number for it in the second array, so output -1.

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4].
Output: [3,-1]
Explanation:
    For number 2 in the first array, the next greater number for it in the second array is 3.
    For number 4 in the first array, there is no next greater number for it in the second array, so output -1.

Note:

1. All elements in nums1 and nums2 are unique.
2. The length of both nums1 and nums2 would not exceed 1000.

Solution

O(nums1*nums2) Time, O(1) Space - Brute-Force:

class Solution {
    func nextGreaterElement(_ nums1: [Int], _ nums2: [Int]) -> [Int] {
        var result = Array(repeating: -1, count: nums1.count)

        // For each element in nums1
        for i in 0..<nums1.count {

            // First find the index in nums1 with element corresponding 
            // to the same value in nums2 (target), then find the next value 
            // in nums2 that's greater than such value (nextGreater)
            var target, nextGreater : Int?
            for j in 0..<nums2.count {
                if nums2[j] == nums1[i] {
                    target = j
                    continue
                }
                if let target = target, nums2[j] > nums2[target] {
                    result[i] = nums2[j]
                    break
                }
            }
        }
        return result
    }
}

O(nums1*nums2/2) Time, O(nums2) Space - Improved Brute-Force:

class Solution {
    func nextGreaterElement(_ nums1: [Int], _ nums2: [Int]) -> [Int] {
        var result = Array(repeating: -1, count: nums1.count)

        // Store values of nums2 with respect to their indices
        var lookup : [Int: Int] = [:]
        for (index, value) in nums2.enumerated() {
            lookup[value] = index
        }

        // Lookup in nums2 the index of value corresponding to nums1
        // Search in nums2 starting from such index to find the next greater value
        for (index, value) in nums1.enumerated() {
            if let target = lookup[value] {
                for j in target..<nums2.count {
                    if nums2[j] > value {
                        result[index] = nums2[j]
                        break
                    }
                }
            }
        }
        return result
    }
}

O(nums1+nums2) Time, O(nums2) Space - Stack:

class Solution {
    func nextGreaterElement(_ nums1: [Int], _ nums2: [Int]) -> [Int] {
        var stack : [Int] = [], nextGreater : [Int: Int] = [:]

        for i in stride(from: nums2.count-1, through: 0, by: -1) {
            let num = nums2[i]

            // Remove all elements on the stack smaller than num
            while let last = stack.last, last < num {
                _ = stack.removeLast()
            }

            // If stack is not empty, the next element in nums2 greater 
            // than num is on the top of the stack, otherwise there are no
            // next elements greater than num (-1)
            // Insert the next greater element corresponding to num into the lookup
            if let last = stack.last {
                nextGreater[num] = last
            } else {
                nextGreater[num] = -1
            }

            // Push num onto the stack
            stack.append(num)
        }

        // For each element in nums1, lookup its next greater element in nums2
        return nums1.map { nextGreater[$0]! }
    }
}