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This approach involves iterating over each element of nums1 and finding the corresponding element in nums2. Once located, search for the next greater element to its right. This straightforward method checks each pair and ensures correctness but may not be optimal for larger arrays.
The time complexity of this brute force approach is O(n * m), where n is the number of elements in nums1, and m is the number of elements in nums2. The space complexity is O(1) apart from the output array.
1function nextGreaterElement(nums1, nums2) {
2 let result = [];
3 nums1.forEach(num1 => {
4 let index = nums2.indexOf(num1);
5 let nextGreater = -1;
6 for (let i = index + 1; i < nums2.length; i++) {
7 if (nums2[i] > num1) {
8 nextGreater = nums2[i];
9 break;
10 }
11 }
12 result.push(nextGreater);
13 });
14 return result;
15}
16
17let nums1 = [4, 1, 2];
18let nums2 = [1, 3, 4, 2];
19console.log(nextGreaterElement(nums1, nums2));This JavaScript code adopts a brute force strategy by first locating each element in nums1 within nums2 and subsequently finding its next larger counterpart. It incorporates array methods for improved readability but remains inefficient with larger inputs.
This approach utilizes a stack and a hashmap to efficiently solve the problem. As we traverse nums2, we use the stack to store elements for which the next greater element hasn't been found yet. Whenever a greater element is found, it's recorded in the hashmap against the elements in the stack. This technique is optimal and runs in linear time.
The time complexity is O(n + m), approaching linear time with respect to input size and space is approximately O(m) for hashmap tracking.
1
Using both stack and hashmap, this Python solution processes nums2 entirely, updating the hashmap with the next greater elements as soon as they are found. The transformation into nums1 results is rapid due to pre-built mappings.