Easy Solution in Java
Intuition\nTo find the maximum product of three numbers from the array, there are two possible candidates:\n\n1.The product of the largest three numbers in the 
# Intuition\nTo find the maximum product of three numbers from the array, there are two possible candidates:\n\n1.The product of the largest three numbers in the array.\n2.The product of the two smallest numbers (possibly negative) and the largest number (as multiplying two negatives results in a positive product).\nThus, the result will be the maximum of these two products.\n\n# Approach\n1.Sort the Array: Sort the array to make it easier to select the largest and smallest numbers.\n2.Consider Two Cases:\n    i)The product of the largest three numbers.\n    ii)The product of the two smallest numbers and the largest number.\n3.Return the Maximum: Return the maximum of these two products.\n\n# Complexity\n- Time complexity:\n1.Sorting the array takes O(nlogn), where n is the number of elements in the array.\n2.Accessing the elements to compute the two possible products takes constant time, i.e., O(1).\nTherefore, the overall time complexity is O(nlogn).\n\n- Space complexity:\nThe space complexity is O(1) as we are only using a few variables for storing intermediate results, aside from the input array.\n\n# Code\n```java []\nclass Solution {\n    public int maximumProduct(int[] nums) {\n        Arrays.sort(nums);\n        int n=nums.length;\n        int a=nums[0]*nums[1]*nums[n-1];\n        int b=nums[n-1]*nums[n-2]*nums[n-3];\n        return Math.max(a,b);\n    }\n}
['Array', 'Math', 'Sorting', 'Java']