Description#
Given the array of integers nums, you will choose two different indices i and j of that array. Return the maximum value of (nums[i]-1)*(nums[j]-1).
Example 1:
Input: nums = [3,4,5,2]
Output: 12
Explanation: If you choose the indices i=1 and j=2 (indexed from 0), you will get the maximum value, that is, (nums[1]-1)*(nums[2]-1) = (4-1)*(5-1) = 3*4 = 12.
Example 2:
Input: nums = [1,5,4,5]
Output: 16
Explanation: Choosing the indices i=1 and j=3 (indexed from 0), you will get the maximum value of (5-1)*(5-1) = 16.
Example 3:
Input: nums = [3,7]
Output: 12
Constraints:
2 <= nums.length <= 5001 <= nums[i] <= 10^3
Solutions#
Solution 1#
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| class Solution:
def maxProduct(self, nums: List[int]) -> int:
ans = 0
for i, a in enumerate(nums):
for b in nums[i + 1 :]:
ans = max(ans, (a - 1) * (b - 1))
return ans
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| class Solution {
public int maxProduct(int[] nums) {
int ans = 0;
int n = nums.length;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
ans = Math.max(ans, (nums[i] - 1) * (nums[j] - 1));
}
}
return ans;
}
}
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| class Solution {
public:
int maxProduct(vector<int>& nums) {
int ans = 0;
int n = nums.size();
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
ans = max(ans, (nums[i] - 1) * (nums[j] - 1));
}
}
return ans;
}
};
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| func maxProduct(nums []int) int {
ans := 0
for i, a := range nums {
for _, b := range nums[i+1:] {
t := (a - 1) * (b - 1)
if ans < t {
ans = t
}
}
}
return ans
}
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| function maxProduct(nums: number[]): number {
const n = nums.length;
for (let i = 0; i < 2; i++) {
let maxIdx = i;
for (let j = i + 1; j < n; j++) {
if (nums[j] > nums[maxIdx]) {
maxIdx = j;
}
}
[nums[i], nums[maxIdx]] = [nums[maxIdx], nums[i]];
}
return (nums[0] - 1) * (nums[1] - 1);
}
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| impl Solution {
pub fn max_product(nums: Vec<i32>) -> i32 {
let mut max = 0;
let mut submax = 0;
for &num in nums.iter() {
if num > max {
submax = max;
max = num;
} else if num > submax {
submax = num;
}
}
(max - 1) * (submax - 1)
}
}
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| class Solution {
/**
* @param Integer[] $nums
* @return Integer
*/
function maxProduct($nums) {
$max = 0;
$submax = 0;
for ($i = 0; $i < count($nums); $i++) {
if ($nums[$i] > $max) {
$submax = $max;
$max = $nums[$i];
} elseif ($nums[$i] > $submax) {
$submax = $nums[$i];
}
}
return ($max - 1) * ($submax - 1);
}
}
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| int maxProduct(int* nums, int numsSize) {
int max = 0;
int submax = 0;
for (int i = 0; i < numsSize; i++) {
int num = nums[i];
if (num > max) {
submax = max;
max = num;
} else if (num > submax) {
submax = num;
}
}
return (max - 1) * (submax - 1);
}
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Solution 2#
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| class Solution:
def maxProduct(self, nums: List[int]) -> int:
nums.sort()
return (nums[-1] - 1) * (nums[-2] - 1)
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| class Solution {
public int maxProduct(int[] nums) {
Arrays.sort(nums);
int n = nums.length;
return (nums[n - 1] - 1) * (nums[n - 2] - 1);
}
}
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| class Solution {
public:
int maxProduct(vector<int>& nums) {
sort(nums.rbegin(), nums.rend());
return (nums[0] - 1) * (nums[1] - 1);
}
};
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| func maxProduct(nums []int) int {
sort.Ints(nums)
n := len(nums)
return (nums[n-1] - 1) * (nums[n-2] - 1)
}
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| function maxProduct(nums: number[]): number {
let max = 0;
let submax = 0;
for (const num of nums) {
if (num > max) {
submax = max;
max = num;
} else if (num > submax) {
submax = num;
}
}
return (max - 1) * (submax - 1);
}
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Solution 3#
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| class Solution:
def maxProduct(self, nums: List[int]) -> int:
a = b = 0
for v in nums:
if v > a:
a, b = v, a
elif v > b:
b = v
return (a - 1) * (b - 1)
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| class Solution {
public int maxProduct(int[] nums) {
int a = 0, b = 0;
for (int v : nums) {
if (v > a) {
b = a;
a = v;
} else if (v > b) {
b = v;
}
}
return (a - 1) * (b - 1);
}
}
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| class Solution {
public:
int maxProduct(vector<int>& nums) {
int a = 0, b = 0;
for (int v : nums) {
if (v > a) {
b = a;
a = v;
} else if (v > b) {
b = v;
}
}
return (a - 1) * (b - 1);
}
};
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| func maxProduct(nums []int) int {
a, b := 0, 0
for _, v := range nums {
if v > a {
b, a = a, v
} else if v > b {
b = v
}
}
return (a - 1) * (b - 1)
}
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