1913. Maximum product difference between two number pairs - no sorting required

1913. The maximum product difference between two number pairs - No sorting

Two pairs of numbers (a, b) and (c, d) Between The product difference Defined as (a * b) - (c * d) .

 for example ,(5, 6)  and  (2, 7)  The product difference between is  (5 * 6) - (2 * 7) = 16 .

Give you an array of integers nums , Choose four Different Subscript w、x、y and z , Make number pair (nums[w], nums[x]) and (nums[y], nums[z]) Between The product difference Fetch Maximum .

Returns... Of the product differences obtained in this way Maximum .

Example 1：

Input ：nums = [5,6,2,7,4]
Output ：34
explain ： You can choose the subscript as 1 and 3 The elements of form the first number pair (6, 7) And subscripts 2 and 4 Make up the second number pair (2, 4)
The product difference is (6 * 7) - (2 * 4) = 34

Example 2：

Input ：nums = [4,2,5,9,7,4,8]
Output ：64
explain ： You can choose the subscript as 3 and 6 The elements of form the first number pair (9, 8) And subscripts 1 and 5 Make up the second number pair (2, 4)
The product difference is (9 * 8) - (2 * 4) = 64

See this topic , The first reaction of many people may be sorting , But in fact , If you use sorting, it will be slow , We don't need to sort , You can also find the two largest numbers and the two smallest numbers , The solution code is as follows ：

int maxProductDifference(int* nums, int numsSize){
    
    int min1=0,max1=0;
    int min2=1,max2=1;
    for(int i=2;i<numsSize;i++){
    
        if(nums[i]>nums[max1]&&nums[i]>nums[max2]){
    
            if(nums[max1]>nums[max2]){
    
              max2=i;
            }
            else{
    
                max1=i;
            }
        }

        else if(nums[i]>nums[max1]&&nums[i]<=nums[max2]){
    
            max1=i;
        }
         else if(nums[i]<=nums[max1]&&nums[i]>nums[max2]){
    
            max2=i;
        }
         if(nums[i]<nums[min1]&&nums[i]<nums[min2]){
    
            if(nums[min1]>nums[min2]){
    
              min1=i;
            }
            else{
    
                min2=i;
            }
        }
        else if(nums[i]<nums[min1]&&nums[i]>=nums[min2]){
    
            min1=i;
        }
        else  if(nums[i]>=nums[min1]&&nums[i]<nums[min2]){
    
            min2=i;
        }
    }
    printf ("mx %d %d %d %d",min1,min2,max1,max2);

    return nums[max1]*nums[max2]-nums[min1]*nums[min2];

}