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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

• Java
• C++
• Python
• Go
• TypeScript

• 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);
      }
  }
  

• 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);
      }
  };
  

• 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])
  
  

• 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)
  }
  

• 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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