Intuition

Since all values are positive, (a-1)*(b-1) is maximized by choosing the two largest numbers in the array. Track those two values in one pass, then return (first - 1) * (second - 1).

Approach: Track Top Two Values

1. Initialize firstMax and secondMax to 0.
2. For each num in nums:
3. If num > firstMax, shift firstMax into secondMax and update firstMax.
4. Else if num > secondMax, update secondMax.
5. Return (firstMax - 1) * (secondMax - 1).

Complexity

• Time complexity: $$O(n)$$, where n is nums.length.
• Space complexity: $$O(1)$$.

Code

Go

func maxProduct(nums []int) int {
    firstMax, secondMax := 0, 0
    for _, num := range nums {
        if num > firstMax {
            secondMax, firstMax = firstMax, num
        } else if num > secondMax {
            secondMax = num
        }
    }
    return (firstMax - 1) * (secondMax - 1)
}

Rust

impl Solution {
    pub fn max_product(nums: Vec<i32>) -> i32 {
        let (mut first, mut second) = (0, 0);
        for num in nums {
            if num > first {
                second = first;
                first = num;
            } else if num > second {
                second = num;
            }
        }
        (first - 1) * (second - 1)
    }
}