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628. Maximum Product of Three Numbers

Description

Given an integer array nums, find three numbers whose product is maximum and return the maximum product.

 

Example 1:

Input: nums = [1,2,3]
Output: 6

Example 2:

Input: nums = [1,2,3,4]
Output: 24

Example 3:

Input: nums = [-1,-2,-3]
Output: -6

 

Constraints:

  • 3 <= nums.length <= 104
  • -1000 <= nums[i] <= 1000

Solutions

Solution 1

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class Solution:
    def maximumProduct(self, nums: List[int]) -> int:
        nums.sort()
        a = nums[-1] * nums[-2] * nums[-3]
        b = nums[-1] * nums[0] * nums[1]
        return max(a, b)

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class Solution {
    public int maximumProduct(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        int a = nums[n - 1] * nums[n - 2] * nums[n - 3];
        int b = nums[n - 1] * nums[0] * nums[1];
        return Math.max(a, b);
    }
}

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class Solution {
public:
    int maximumProduct(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        int a = nums[n - 1] * nums[n - 2] * nums[n - 3];
        int b = nums[n - 1] * nums[0] * nums[1];
        return max(a, b);
    }
};

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func maximumProduct(nums []int) int {
    sort.Ints(nums)
    n := len(nums)
    a := nums[n-1] * nums[n-2] * nums[n-3]
    b := nums[n-1] * nums[0] * nums[1]
    if a > b {
        return a
    }
    return b
}

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function maximumProduct(nums: number[]): number {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    const a = nums[n - 1] * nums[n - 2] * nums[n - 3];
    const b = nums[n - 1] * nums[0] * nums[1];
    return Math.max(a, b);
}

Solution 2

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class Solution:
    def maximumProduct(self, nums: List[int]) -> int:
        top3 = nlargest(3, nums)
        bottom2 = nlargest(2, nums, key=lambda x: -x)
        return max(top3[0] * top3[1] * top3[2], top3[0] * bottom2[0] * bottom2[1])

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class Solution {
    public int maximumProduct(int[] nums) {
        final int inf = 1 << 30;
        int mi1 = inf, mi2 = inf;
        int mx1 = -inf, mx2 = -inf, mx3 = -inf;
        for (int x : nums) {
            if (x < mi1) {
                mi2 = mi1;
                mi1 = x;
            } else if (x < mi2) {
                mi2 = x;
            }
            if (x > mx1) {
                mx3 = mx2;
                mx2 = mx1;
                mx1 = x;
            } else if (x > mx2) {
                mx3 = mx2;
                mx2 = x;
            } else if (x > mx3) {
                mx3 = x;
            }
        }
        return Math.max(mi1 * mi2 * mx1, mx1 * mx2 * mx3);
    }
}

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class Solution {
public:
    int maximumProduct(vector<int>& nums) {
        const int inf = 1 << 30;
        int mi1 = inf, mi2 = inf;
        int mx1 = -inf, mx2 = -inf, mx3 = -inf;
        for (int x : nums) {
            if (x < mi1) {
                mi2 = mi1;
                mi1 = x;
            } else if (x < mi2) {
                mi2 = x;
            }
            if (x > mx1) {
                mx3 = mx2;
                mx2 = mx1;
                mx1 = x;
            } else if (x > mx2) {
                mx3 = mx2;
                mx2 = x;
            } else if (x > mx3) {
                mx3 = x;
            }
        }
        return max(mi1 * mi2 * mx1, mx1 * mx2 * mx3);
    }
};

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func maximumProduct(nums []int) int {
    const inf = 1 << 30
    mi1, mi2 := inf, inf
    mx1, mx2, mx3 := -inf, -inf, -inf
    for _, x := range nums {
        if x < mi1 {
            mi1, mi2 = x, mi1
        } else if x < mi2 {
            mi2 = x
        }
        if x > mx1 {
            mx1, mx2, mx3 = x, mx1, mx2
        } else if x > mx2 {
            mx2, mx3 = x, mx2
        } else if x > mx3 {
            mx3 = x
        }
    }
    return max(mi1*mi2*mx1, mx1*mx2*mx3)
}

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function maximumProduct(nums: number[]): number {
    const inf = 1 << 30;
    let mi1 = inf,
        mi2 = inf;
    let mx1 = -inf,
        mx2 = -inf,
        mx3 = -inf;
    for (const x of nums) {
        if (x < mi1) {
            mi2 = mi1;
            mi1 = x;
        } else if (x < mi2) {
            mi2 = x;
        }
        if (x > mx1) {
            mx3 = mx2;
            mx2 = mx1;
            mx1 = x;
        } else if (x > mx2) {
            mx3 = mx2;
            mx2 = x;
        } else if (x > mx3) {
            mx3 = x;
        }
    }
    return Math.max(mi1 * mi2 * mx1, mx1 * mx2 * mx3);
}

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