Leetcode 2708: Maximum Strength of a Group
You are given an integer array representing the scores of students in an exam. Your task is to form a non-empty group of students such that the group’s strength, defined as the product of their scores, is maximized. The goal is to return the maximum possible strength that can be achieved by choosing an optimal group of students.
Problem
Approach
Steps
Complexity
Input: The input consists of a list of integers representing the scores of the students.
Example: Input: nums = [1, -3, 2, -1, 4]
Constraints:
• 1 <= nums.length <= 13
• -9 <= nums[i] <= 9
Output: Return the maximum possible strength of a group of students formed from the input array.
Example: Output: 72
Constraints:
• The result must be the maximum product of selected scores.
Goal: Maximize the product of scores by selecting an optimal subset of the array elements.
Steps:
• Step 1: Calculate the product of positive numbers and the product of negative numbers in the array.
• Step 2: Handle special cases where the number of negative numbers is odd.
• Step 3: If there are zeros, consider excluding the negative numbers or zeros to maximize the product.
Goal: The input array will have a maximum length of 13, and each element will be an integer between -9 and 9.
Steps:
• The array length is between 1 and 13.
• Each score is between -9 and 9.
Assumptions:
• The array is non-empty.
• Input: Input: nums = [3, -2, 4, -1]
• Explanation: We can select the elements [3, -2, 4] to maximize the product. The strength would be 3 * -2 * 4 = -24.
• Input: Input: nums = [5, -6, 3, -2, -1]
• Explanation: The maximum strength can be achieved by selecting the elements [5, -6, 3], with a product of 5 * -6 * 3 = 90.
Approach: To solve this problem, we need to consider the potential of both positive and negative numbers in the array to maximize the product.
Observations:
• A product of negative numbers can be positive if the number of negative numbers is even.
• If there is a zero in the array, we might use it to avoid negative products.
• The optimal solution will involve careful consideration of negative and positive numbers, and possibly excluding negative products if they would decrease the overall strength.
Steps:
• Step 1: Calculate the total product of all the numbers.
• Step 2: If the number of negative numbers is odd, try excluding one negative number with the smallest absolute value to maximize the product.
• Step 3: If there are zero values, consider if removing negative values might be more beneficial.
Empty Inputs:
• The input array will never be empty, as per constraints.
Large Inputs:
• The solution should work within the constraints, considering arrays with up to 13 elements.
Special Values:
• If all numbers are zero, the result should be 0.
• If there is only one element, return that element.
Constraints:
• Ensure handling of negative numbers and zeros in the array.
long long maxStrength(vector<int>& nums) {
bool zero = false, allneg = true, allzero = true, pos = false;
int ncnt = 0, pcnt = 0;
long long ans = 1, neg = 1;
int mxn = -10;
int n = nums.size();
for(int i = 0; i < n; i++) {
if(nums[i] > 0) {
pcnt++;
pos = true;
allzero = false;
allneg = false;
ans *= nums[i];
} else if(nums[i] == 0) {
zero = true;
} else {
ncnt++;
allzero = false;
mxn= max(mxn, nums[i]);
neg *= nums[i];
}
}
if(allzero) return 0;
if(ncnt > 0 && (ncnt %2)) {
if(pcnt == 0 && ncnt == 1) return zero? 0: neg;
neg = neg / mxn;
}
if(pcnt == 0) {
return neg;
}
return ans * neg;
}
1 : Function Declaration
long long maxStrength(vector<int>& nums) {
Start of the `maxStrength` function that takes a vector of integers `nums` and returns a long long value representing the maximum strength.
2 : Variable Initialization
bool zero = false, allneg = true, allzero = true, pos = false;
Initialize variables: `zero` to track if there are any zeros, `allneg` to check if all numbers are negative, `allzero` to check if all numbers are zero, and `pos` to check if there are any positive numbers.
3 : Counter Variables
int ncnt = 0, pcnt = 0;
Initialize counters: `ncnt` for negative numbers and `pcnt` for positive numbers.
4 : Initialization of Answer Variables
long long ans = 1, neg = 1;
Initialize variables `ans` and `neg` to 1 to hold the product of positive numbers and negative numbers, respectively.
5 : Empty Line
int mxn = -10;
Initialize `mxn` to -10 to keep track of the largest negative number.
6 : Initialize Array Size
int n = nums.size();
Get the size of the input vector `nums` and store it in the variable `n`.
7 : Loop Through Array
for(int i = 0; i < n; i++) {
Start a loop to iterate over each element of the vector `nums`.
8 : Check Positive Value
if(nums[i] > 0) {
Check if the current number is positive.
9 : Increment Positive Counter
pcnt++;
Increment the `pcnt` counter for positive numbers.
10 : Set Positive Flag
pos = true;
Set the `pos` flag to true indicating that a positive number has been found.
11 : Update Zero Flag
allzero = false;
Set `allzero` to false as we have found a non-zero number.
12 : Update Negative Flag
allneg = false;
Set `allneg` to false as we have found a non-negative number.
13 : Multiply Positive Value
ans *= nums[i];
Multiply the current positive number with `ans` to accumulate the product.
14 : Check for Zero
} else if(nums[i] == 0) {
Check if the current number is zero.
15 : Set Zero Flag
zero = true;
Set the `zero` flag to true indicating that a zero has been found.
16 : Handle Negative Value
} else {
Handle the case where the current number is negative.
17 : Increment Negative Counter
ncnt++;
Increment the `ncnt` counter for negative numbers.
18 : Update Zero Flag Again
allzero = false;
Set `allzero` to false again as we found a non-zero number.
19 : Update Maximum Negative
mxn= max(mxn, nums[i]);
Update `mxn` to keep track of the largest negative number.
20 : Multiply Negative Value
neg *= nums[i];
Multiply the current negative number with `neg` to accumulate the product of negative numbers.
21 : Check if All Zeroes
if(allzero) return 0;
Check if all numbers in the array were zero and return 0 in that case.
22 : Handle Odd Number of Negatives
if(ncnt > 0 && (ncnt %2)) {
If there are negative numbers and their count is odd, handle by excluding the largest negative number.
23 : Handle Special Case for One Negative
if(pcnt == 0 && ncnt == 1) return zero? 0: neg;
If there are no positive numbers and exactly one negative number, return the product (either zero or the negative value).
24 : Update Negative Product
neg = neg / mxn;
Remove the largest negative number from the product of negative numbers.
25 : Check for No Positive Numbers
if(pcnt == 0) {
Check if there are no positive numbers in the array.
26 : Return Negative Product
return neg;
Return the product of negative numbers if there are no positive numbers.
27 : Return Maximum Strength
return ans * neg;
Return the final product of positive and negative values as the maximum strength.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) because we are iterating through the array only once to compute the necessary product and handle special cases.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1), as we are only using a constant amount of extra space.
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