Maximum product of any pair
Hard
Java
Arrays
21 views
Problem Statement
Task: values can be negative. Return maximum product of any two elements.
Input Format
An integer n (size of array)
Then n integers (array elements, can be negative)
Then n integers (array elements, can be negative)
Output Format
Maximum product of any two elements.
Example
5 -10 -3 5 6 -2
30
Constraints
• n ≥ 2
• Elements can be negative, zero, or positive
• Elements can be negative, zero, or positive
Concept Explanation
The maximum product can come from:
• the two largest positive numbers, or
• the two smallest (most negative) numbers
(because negative × negative = positive).
• the two largest positive numbers, or
• the two smallest (most negative) numbers
(because negative × negative = positive).
Step-by-Step Explanation
1.Read array size n and elements.
2.Initialize:
• max1 = Integer.MIN_VALUE
• max2 = Integer.MIN_VALUE
• min1 = Integer.MAX_VALUE
• min2 = Integer.MAX_VALUE
3.Loop through the array once:
• Update the two largest values (max1, max2).
• Update the two smallest values (min1, min2).
4.Compute:
• product1 = max1 * max2
• product2 = min1 * min2
5.The answer is max(product1, product2).
6.Print the result.
2.Initialize:
• max1 = Integer.MIN_VALUE
• max2 = Integer.MIN_VALUE
• min1 = Integer.MAX_VALUE
• min2 = Integer.MAX_VALUE
3.Loop through the array once:
• Update the two largest values (max1, max2).
• Update the two smallest values (min1, min2).
4.Compute:
• product1 = max1 * max2
• product2 = min1 * min2
5.The answer is max(product1, product2).
6.Print the result.
Concept Explanation
The maximum product can come from:
• the two largest positive numbers, or
• the two smallest (most negative) numbers
(because negative × negative = positive).
• the two largest positive numbers, or
• the two smallest (most negative) numbers
(because negative × negative = positive).
Step-by-Step Explanation
1.Read array size n and elements.
2.Initialize:
• max1 = Integer.MIN_VALUE
• max2 = Integer.MIN_VALUE
• min1 = Integer.MAX_VALUE
• min2 = Integer.MAX_VALUE
3.Loop through the array once:
• Update the two largest values (max1, max2).
• Update the two smallest values (min1, min2).
4.Compute:
• product1 = max1 * max2
• product2 = min1 * min2
5.The answer is max(product1, product2).
6.Print the result.
2.Initialize:
• max1 = Integer.MIN_VALUE
• max2 = Integer.MIN_VALUE
• min1 = Integer.MAX_VALUE
• min2 = Integer.MAX_VALUE
3.Loop through the array once:
• Update the two largest values (max1, max2).
• Update the two smallest values (min1, min2).
4.Compute:
• product1 = max1 * max2
• product2 = min1 * min2
5.The answer is max(product1, product2).
6.Print the result.
Input / Output Format
Input Format
An integer n (size of array)
Then n integers (array elements, can be negative)
Then n integers (array elements, can be negative)
Output Format
Maximum product of any two elements.
Constraints
• n ≥ 2
• Elements can be negative, zero, or positive
• Elements can be negative, zero, or positive
Examples
Input:
5
-10 -3 5 6 -2
Output:
30
Example Solution (Public)
static long maxPairProduct(int[] a){int max1=Integer.MIN_VALUE,max2=Integer.MIN_VALUE,min1=Integer.MAX_VALUE,min2=Integer.MAX_VALUE;for(int x:a){if(x>max1){max2=max1;max1=x;}else if(x>max2){max2=x;}if(x<min1){min2=min1;min1=x;}else if(x<min2){min2=x;}}long p1=1L*max1*max2;long p2=1L*min1*min2;return p1>p2?p1:p2;}
Official Solution Code
static long maxPairProduct(int[] a){int max1=Integer.MIN_VALUE,max2=Integer.MIN_VALUE,min1=Integer.MAX_VALUE,min2=Integer.MAX_VALUE;for(int x:a){if(x>max1){max2=max1;max1=x;}else if(x>max2){max2=x;}if(x<min1){min2=min1;min1=x;}else if(x<min2){min2=x;}}long p1=1L*max1*max2;long p2=1L*min1*min2;return p1>p2?p1:p2;}
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