Given an array of integers For example, if the array is This problem can be efficiently approached using a stack. By traversing the entire array only once, we can use the stack to solve the issues occurring on the right side of each element. This reduces the time complexity.<\/p>\n The code is structured as follows:<\/p>\narr<\/code>, each element in the array is a non-negative integer. For each element, the task is to find the index of the first element to its right that is greater than itself, and create a new array with those indices. If there is no such element, store -1.<\/p>\n[2, 1, 5, 3, 6, 4]<\/code>, the result will be [2, 1, 1, 1, -1, -1]<\/code>.<\/p>\nApproach to the Problem<\/h2>\n
Algorithm Explanation<\/h3>\n
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result<\/code> to store the results.<\/li>\n\n
-1<\/code> in the result array.<\/li>\nCode Implementation<\/h2>\n
func nextGreaterElement(arr: [Int]) -> [Int] {\n var result = Array(repeating: -1, count: arr.count)\n var stack: [Int] = []\n\n for i in (0..Code Explanation<\/h2>\n
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result<\/code>: The result array is initialized to -1.<\/li>\nstack<\/code>: The stack is initialized to store indices.<\/li>\nreversed<\/code>: The array is traversed in reverse. This is done to compare each element with the elements to its right.<\/li>\nTime Complexity<\/h2>\n