array<\/a> of size n and q queries. Your task is to find the next greater element for each query. Each query consists of an integer representing an index. If there is no greater element for any index, then print -1.<\/p>\nInput<\/strong>: Two integer arrays of size n and q.
\nOutput<\/strong>: Integer array of size q.<\/p>\nTest cases<\/strong>:
\nInput<\/strong>:
\nArr = [9, 4, 2, 3, 8, 6, 10, 1]
\nQueries = [1, 2, 5, 6]<\/p>\nOutput<\/strong>:
\n[8, 3, 10, -1]<\/p>\nExplanation:<\/h4>\n\n- Output[0] = 8 because the next element greater than arr[1] = 4 is 8.<\/li>\n
- Output[1] = 3 because the next element greater than arr[2] = 2 is 3.<\/li>\n
- Output[2] = 10 because the next element greater than arr[5] = 6 is 10.<\/li>\n
- Output[3] = -1 because there is no element greater than arr[6] on its left side.<\/li>\n<\/ul>\n
Naive Approach – Brute Force<\/h3>\n
For each query, we will iterate the right side of the input array, starting from the index represented by this query. The answer of any query will be the first element greater than itself. If we don’t find any greater element than the answer of that element will be -1.<\/p>\n
Algorithm<\/h4>\n1. Initialize an empty array\n2. For each i in 0 to length(queries):\n a. index = queries[i]\n b. For each j in index + 1 to length(arr)\n i. If arr[j] > arr[index], then ans[i] = arr[j]\n c. If we does not found the greater element, then ans[i] = -1\n3. Return ans array.<\/code><\/pre>\nCode Implementation:<\/h3>\n\t\t\t\t\t\t
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