Write a Python program to find the first duplicate element in a given array of integers. Return -1 if there are no such elements.Sample Solution:Python Code :def find_first_duplicate(nums): num_set = set() no_duplicate = -1 for i in range(len(nums)): if nums[i] in num_set: re...
Learn how to find the lost element from a duplicated array in JavaScript with our comprehensive guide and example code.
{ Element = 5, Count = 2 }, { Element = -1, Count = 2 } Alternatively, if you need a dictionary with the duplicate element as a key and the duplicate element’s count as its value, do as follows: 1 2 3 4 5 6 7 8
array = [1 2 3 4 5 6] % find() will get the index of element % store it in the index index = find(array==3) 输出: 注意:如果数组包含重复项,则 find(X) 函数将返回该整数的所有索引。 示例2: MATLAB % MATLAB code for if the array contains % duplicate elements array = [1 2 3 ...
How to find duplicate values in a JavaScript array - In this tutorial, we will discuss how we can find duplicate or repeating values in a JavaScript array using different methods or approaches to reach the solution to this problem. Below is the list of t
Find the minimum element. Example Given[4, 5, 6, 7, 0, 1, 2]return0 Note You may assume no duplicate exists in the array. 题解 如前节所述,对于旋转数组的分析可使用画图的方法,如下图所示,升序数组经旋转后可能为如下两种形式。 最小值可能在上图中的两种位置出现,如果仍然使用数组首部元素作为...
the algorithm follows fairly simply. First, we sort the array, and then we compare each element to the previous element. Because there is exactly one duplicated element in the array, we know that the array is of at least length 2, and we can return the duplicate element as soon as we ...
In the callback function, we again use the indexOf() method to compare the current element index with other elements in the array. If both the indexes are the same, it means that the current item is not duplicate:const numbers = [1, 2, 3, 2, 4, 5, 5, 6]; const isDuplicate =...
element in this array\n"; else cout << "The peak element(maximum number) is " << peakNumber << "\n"; return; } //recursive binary search int findThePeakEfficientRecur(vector<int>& arr, int start, int end) { //base cases //only one element in array if (start ==...
There is only one duplicate number in the array, but it could be repeated more than once. ##思路 本题给定一个大小为n + 1 n+1n+1的数组,数组里面的值的值域为[ 1 , n ] [1,n][1,n];那么请找出数组里面的重复项。这道题目如果仔细分析的话,本质上就是个快慢指针的问题。首先什么是快慢指针...