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This approach involves sorting the array, and then choosing the maximum product by examining either the product of the three largest numbers or the product of the two smallest numbers and the largest number.
Time Complexity: O(n log n) due to sorting.
Space Complexity: O(1) as we use in-place sorting.
1function maximumProduct(nums) {
2 nums.sort((a, b) => a - b);
3 let n = nums.length;
4 return Math.max(nums[n-1] * nums[n-2] * nums[n-3], nums[0] * nums[1] * nums[n-1]);
5}
6
7console.log(maximumProduct([1, 2, 3, 4]));We start by sorting the array, then check the product of the last three elements and the product of the first two and last element, and return the maximum one.
This approach involves finding the largest three and smallest two numbers in a single traversal of the array. This avoids sorting and gives a more optimal solution for time complexity.
Time Complexity: O(n) as there's only a single pass through the array.
Space Complexity: O(1).
1
We maintain variables for the three largest and two smallest numbers as we iterate. This allows us to compute the potential maximum products and choose the maximum one without needing to sort the array.