Maximum Strength of a Group solution and explanation
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The problem
You are given a 0-indexed integer array nums representing the score of students in an exam. The teacher would like to form one non-empty group of students with maximal strength, where the strength of a group of students of indices i0, i1, i2, ... , ik is defined as nums[i0] * nums[i1] * nums[i2] * ... * nums[ik].
Return the maximum strength of a group the teacher can create.
Example 1:
Input: nums = [3,-1,-5,2,5,-9] Output: 1350 Explanation: One way to form a group of maximal strength is to group the students at indices [0,2,3,4,5]. Their strength is 3 * (-5) * 2 * 5 * (-9) = 1350, which we can show is optimal.
Example 2:
Input: nums = [-4,-5,-4] Output: 20 Explanation: Group the students at indices [0, 1] . Then, we’ll have a resulting strength of 20. We cannot achieve greater strength.
Constraints:
1 <= nums.length <= 13-9 <= nums[i] <= 9
Approach overview
Problem Overview: You are given an integer array where each value represents the strength of a member. The task is to select a non-empty subset whose product is maximized. Because the array can contain positive numbers, negative numbers, and zeros, choosing which elements to multiply becomes a careful balancing act.
Approach 1: Backtracking (O(2^n) time, O(n) space)
This method enumerates every possible non-empty subset and computes the product for each. You recursively decide whether to include or exclude each element while maintaining the current product. The maximum product seen during enumeration becomes the result. This approach is conceptually simple and demonstrates the full search space, but it becomes impractical as the array grows because the number of subsets doubles with every new element.
Backtracking is useful when explaining the brute-force baseline during interviews. It also helps illustrate why smarter strategies are needed for larger inputs. The recursion depth is at most n, giving O(n) auxiliary space. However, the exponential O(2^n) runtime makes it unsuitable for production scenarios.
Approach 2: Sorting and Greedy Multiplication (O(n log n) time, O(1) space)
The optimal strategy relies on how multiplication behaves with positive and negative numbers. Positive numbers always increase the product, while negative numbers can increase it only when paired together. Start by sorting the array so negatives and positives are grouped. Multiply all positive numbers. For negatives, pair them from the smallest values (largest magnitude) because the product of two negatives becomes positive.
If there is an odd count of negatives, skip the one closest to zero because including it would flip the sign of the product. Zeros act as neutralizers: if every other choice leads to a negative product, selecting zero can yield a better result. This greedy reasoning ensures you keep only values that contribute positively to the final product.
This approach leverages properties of multiplication rather than exploring every subset. Sorting simplifies pairing logic and ensures the strongest negatives are combined first. The algorithm runs in O(n log n) time due to sorting and uses constant extra space. It commonly appears in problems involving greedy selection and sign management within array processing. Some solutions also connect this reasoning to state transitions similar to dynamic programming maximum-product problems.
Recommended for interviews: Interviewers expect the greedy reasoning. Starting with the backtracking solution shows you understand the brute-force search space, but transitioning to the sorting-based greedy method demonstrates algorithmic insight and the ability to exploit mathematical properties of multiplication.
Approach 1: Approach 1: Backtracking
This approach explores all possible subsets of the given array using backtracking to compute the maximum product (strength). The idea is to explore each index, either including it in the product calculation or not, while keeping track of the current product. If the subset is non-empty, update the maximum strength encountered.
The backtracking function will recursively select or skip each element, updating the product when the element is included, and backtracking to try the next configuration.
def max_strength(nums):
def backtrack(index, current_product, selected):
if index == len(nums):
if selected > 0:
self.max_product = max(self.max_product, current_product)
return
backtrack(index + 1, current_product, selected)
backtrack(index + 1, current_product * nums[index], selected + 1)
self.max_product = float('-inf')
backtrack(0, 1, 0)
return self.max_product
nums = [3, -1, -5, 2, 5, -9]
print(f"Max Strength: {max_strength(nums)}")import java.util.*;
public class MaxStrength {
public static void backtrack(int[] nums, int index, long currentProduct, long[] maxProduct, int selected) {
if (index == nums.length) {
if (selected > 0) {
maxProduct[0] = Math.max(maxProduct[0], currentProduct);
}
return;
}
backtrack(nums, index + 1, currentProduct, maxProduct, selected);
backtrack(nums, index + 1, currentProduct * nums[index], maxProduct, selected + 1);
}
public static long maxStrength(int[] nums) {
long[] maxProduct = {Long.MIN_VALUE};
backtrack(nums, 0, 1, maxProduct, 0);
return maxProduct[0];
}
public static void main(String[] args) {
int[] nums = {3, -1, -5, 2, 5, -9};
System.out.println("Max Strength: " + maxStrength(nums));
}
}#include <iostream>
#include <vector>
#include <climits>
using namespace std;
void backtrack(const vector<int>& nums, int index, int currentProduct, int& maxProduct, int selected) {
if (index == nums.size()) {
if (selected > 0) {
maxProduct = max(maxProduct, currentProduct);
}
return;
}
backtrack(nums, index + 1, currentProduct, maxProduct, selected);
backtrack(nums, index + 1, currentProduct * nums[index], maxProduct, selected + 1);
}
int maxStrength(const vector<int>& nums) {
int maxProduct = INT_MIN;
backtrack(nums, 0, 1, maxProduct, 0);
return maxProduct;
}
int main() {
vector<int> nums = {3, -1, -5, 2, 5, -9};
cout << "Max Strength: " << maxStrength(nums) << endl;
return 0;
}function maxStrength(nums) {
let maxProduct = Number.NEGATIVE_INFINITY;
function backtrack(index, currentProduct, selected) {
if (index === nums.length) {
if (selected > 0) {
maxProduct = Math.max(maxProduct, currentProduct);
}
return;
}
backtrack(index + 1, currentProduct, selected);
backtrack(index + 1, currentProduct * nums[index], selected + 1);
}
backtrack(0, 1, 0);
return maxProduct;
}
const nums = [3, -1, -5, 2, 5, -9];
console.log(`Max Strength: ${maxStrength(nums)}`);#include <stdio.h>
#include <limits.h>
void backtrack(int *nums, int numsSize, int index, int currentProduct, int *maxProduct, int selected) {
if (index == numsSize) {
if (selected > 0) {
if (currentProduct > *maxProduct) {
*maxProduct = currentProduct;
}
}
return;
}
backtrack(nums, numsSize, index + 1, currentProduct, maxProduct, selected);
backtrack(nums, numsSize, index + 1, currentProduct * nums[index], maxProduct, selected + 1);
}
int maxStrength(int* nums, int numsSize) {
int maxProduct = INT_MIN;
backtrack(nums, numsSize, 0, 1, &maxProduct, 0);
return maxProduct;
}
int main() {
int nums[] = {3, -1, -5, 2, 5, -9};
int numsSize = sizeof(nums) / sizeof(nums[0]);
printf("Max Strength: %d\n", maxStrength(nums, numsSize));
return 0;
}using System;
class MaxStrength
{
public static void Backtrack(int[] nums, int index, long currentProduct, ref long maxProduct, int selected)
{
if (index == nums.Length)
{
if (selected > 0)
{
maxProduct = Math.Max(maxProduct, currentProduct);
}
return;
}
Backtrack(nums, index + 1, currentProduct, ref maxProduct, selected);
Backtrack(nums, index + 1, currentProduct * nums[index], ref maxProduct, selected + 1);
}
public static long MaxStrengthFunc(int[] nums)
{
long maxProduct = long.MinValue;
Backtrack(nums, 0, 1, ref maxProduct, 0);
return maxProduct;
}
public static void Main()
{
int[] nums = {3, -1, -5, 2, 5, -9};
Console.WriteLine("Max Strength: " + MaxStrengthFunc(nums));
}
}The Python solution uses a nested backtracking function to find the maximum product from any non-empty subset of the given numbers. The recursive calls ensure that all possible combinations of elements are considered for the product computation.
Time Complexity: O(2^n), due to the exploration of all subsets.
Space Complexity: O(n), accounting for the recursive call stack.
Approach 2: Approach 2: Sorting and Greedy Multiplication
This approach involves sorting the array first to easily group the largest elements to maximize the product, taking care of negative numbers' interaction. The strategy involves first taking care of zeroes, then pairing negatives for positive products, and finally selecting the largest positive numbers.
After sorting, we use a greedy technique to pair negatives from the start (most negative) to make products positive, add largest positives, and consider if one last negative or zero as standalone can improve the product depending on the overall product value.
def max_strength(nums):
nums.sort()
product = 1
negatives, zeros, positives = 0, 0, 0
for num in nums:
if num < 0:
negatives += 1
elif num == 0:
zeros += 1
else:
positives += 1
if positives == 0 and (negatives % 2 == 0 or negatives == 0):
return 0
if negatives % 2 != 0:
negatives -= 1
for i in range(negatives):
product *= nums[i]
for i in range(len(nums) - 1, negatives + zeros - 1, -1):
product *= nums[i]
return product
nums = [3, -1, -5, 2, 5, -9]
print(f"Max Strength: {max_strength(nums)}")import java.util.*;
public class MaxStrength {
public static int maxStrength(int[] nums) {
Arrays.sort(nums);
int product = 1;
int negatives = 0, zeros = 0, positives = 0;
for (int num : nums) {
if (num < 0) negatives++;
if (num == 0) zeros++;
if (num > 0) positives++;
}
if (positives == 0 && (negatives % 2 == 0 || negatives == 0)) return 0;
if (negatives % 2 != 0) negatives--;
for (int i = 0; i < negatives; i++) product *= nums[i];
for (int i = nums.length - 1; i >= negatives + zeros; i--) product *= nums[i];
return product;
}
public static void main(String[] args) {
int[] nums = {3, -1, -5, 2, 5, -9};
System.out.println("Max Strength: " + maxStrength(nums));
}
}#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int maxStrength(vector<int>& nums) {
sort(nums.begin(), nums.end());
int product = 1;
int negatives = 0, zeros = 0, positives = 0;
for (int num : nums) {
if (num < 0) negatives++;
if (num == 0) zeros++;
if (num > 0) positives++;
}
if (positives == 0 && (negatives % 2 == 0 || negatives == 0)) return 0;
if (negatives % 2 != 0) negatives--;
for (int i = 0; i < negatives; i++) product *= nums[i];
for (int i = nums.size() - 1; i >= negatives + zeros; i--) product *= nums[i];
return product;
}
int main() {
vector<int> nums = {3, -1, -5, 2, 5, -9};
cout << "Max Strength: " << maxStrength(nums) << endl;
return 0;
}function maxStrength(nums) {
nums.sort((a, b) => a - b);
let product = 1;
let negatives = 0, zeros = 0, positives = 0;
for (const num of nums) {
if (num < 0) negatives++;
if (num === 0) zeros++;
if (num > 0) positives++;
}
if (positives === 0 && (negatives % 2 === 0 || negatives === 0)) return 0;
if (negatives % 2 !== 0) negatives--;
for (let i = 0; i < negatives; i++) product *= nums[i];
for (let i = nums.length - 1; i >= negatives + zeros; i--) product *= nums[i];
return product;
}
const nums = [3, -1, -5, 2, 5, -9];
console.log(`Max Strength: ${maxStrength(nums)}`);#include <stdio.h>
#include <stdlib.h>
// comparator function used for sorting
int compare(const void *a, const void *b) {
return (*(int*)a - *(int*)b);
}
int maxStrength(int* nums, int numsSize) {
qsort(nums, numsSize, sizeof(int), compare);
int product = 1;
int negatives = 0, zeros = 0, positives = 0;
for (int i = 0; i < numsSize; i++) {
if (nums[i] < 0) negatives++;
if (nums[i] == 0) zeros++;
if (nums[i] > 0) positives++;
}
if (positives == 0 && (negatives % 2 == 0 || negatives == 0)) return 0;
if (negatives % 2 != 0) negatives--; // ignore least neg value if odd
for (int i = 0; i < negatives; i++) product *= nums[i];
for (int i = numsSize - 1; i >= negatives + zeros; i--) product *= nums[i];
return product;
}
int main() {
int nums[] = {3, -1, -5, 2, 5, -9};
int numsSize = sizeof(nums) / sizeof(nums[0]);
printf("Max Strength: %d\n", maxStrength(nums, numsSize));
return 0;
}using System;
class MaxStrength {
public static int MaxStrengthFunc(int[] nums) {
Array.Sort(nums);
int product = 1;
int negatives = 0, zeros = 0, positives = 0;
foreach (int num in nums) {
if (num < 0) negatives++;
if (num == 0) zeros++;
if (num > 0) positives++;
}
if (positives == 0 && (negatives % 2 == 0 || negatives == 0)) return 0;
if (negatives % 2 != 0) negatives--;
for (int i = 0; i < negatives; i++) product *= nums[i];
for (int i = nums.Length - 1; i >= negatives + zeros; i--) product *= nums[i];
return product;
}
public static void Main() {
int[] nums = {3, -1, -5, 2, 5, -9};
Console.WriteLine("Max Strength: " + MaxStrengthFunc(nums));
}
}The Python solution leverages sorting to iterate through numbers, making calculated decisions for pairing negative numbers to achieve the maximum product subset. This method captures the essence of handling edge cases (like zero).
Time Complexity: O(n log n) for sorting, leveraging time-efficient sub-selection for the product.
Space Complexity: O(1).
Approach 3: Binary Enumeration
The problem is actually to find the maximum product of all subsets. Since the length of the array does not exceed 13, we can consider using the method of binary enumeration.
We enumerate all subsets in the range of [1, 2^n), and for each subset, we calculate its product, and finally return the maximum value.
The time complexity is O(2^n times n), where n is the length of the array. The space complexity is O(1).
function maxStrength(nums: number[]): number {
let ans = -Infinity;
const n = nums.length;
for (let i = 1; i < 1 << n; ++i) {
let t = 1;
for (let j = 0; j < n; ++j) {
if ((i >> j) & 1) {
t *= nums[j];
}
}
ans = Math.max(ans, t);
}
return ans;
}func maxStrength(nums []int) int64 {
ans := int64(-1e14)
for i := 1; i < 1<<len(nums); i++ {
var t int64 = 1
for j, x := range nums {
if i>>j&1 == 1 {
t *= int64(x)
}
}
ans = max(ans, t)
}
return ans
}Approach 4: Sorting + Greedy
First, we can sort the array. Based on the characteristics of the array, we can draw the following conclusions:
- If there is only one element in the array, then the maximum strength value is this element.
- If there are two or more elements in the array, and
nums[1] = nums[n - 1] = 0, then the maximum strength value is0. - Otherwise, we traverse the array from small to large. If the current element is less than
0and the next element is also less than0, then we multiply these two elements and accumulate the product into the answer. Otherwise, if the current element is less than or equal to0, we skip it directly. If the current element is greater than0, we multiply this element into the answer. Finally, we return the answer.
The time complexity is O(n times log n), and the space complexity is O(log n). Where n is the length of the array.
function maxStrength(nums: number[]): number {
nums.sort((a, b) => a - b);
const n = nums.length;
if (n === 1) {
return nums[0];
}
if (nums[1] === 0 && nums[n - 1] === 0) {
return 0;
}
let ans = 1;
for (let i = 0; i < n; ++i) {
if (nums[i] < 0 && i + 1 < n && nums[i + 1] < 0) {
ans *= nums[i] * nums[i + 1];
++i;
} else if (nums[i] > 0) {
ans *= nums[i];
}
}
return ans;
}func maxStrength(nums []int) int64 {
sort.Ints(nums)
n := len(nums)
if n == 1 {
return int64(nums[0])
}
if nums[1] == 0 && nums[n-1] == 0 {
return 0
}
ans := int64(1)
for i := 0; i < n; i++ {
if nums[i] < 0 && i+1 < n && nums[i+1] < 0 {
ans *= int64(nums[i] * nums[i+1])
i++
} else if nums[i] > 0 {
ans *= int64(nums[i])
}
}
return ans
}Complexity comparison
| Approach | Complexity |
|---|---|
| Approach 1: Backtracking | Time Complexity: O(2^n), due to the exploration of all subsets. |
| Approach 2: Sorting and Greedy Multiplication | Time Complexity: O(n log n) for sorting, leveraging time-efficient sub-selection for the product. |
| Binary Enumeration | – |
| Sorting + Greedy | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Backtracking Enumeration | O(2^n) | O(n) | Understanding the brute-force search space or when n is very small |
| Sorting + Greedy Multiplication | O(n log n) | O(1) | General case; efficiently handles positives, negatives, and zeros |
Frequently asked questions
Is Maximum Strength of a Group easy or hard?
Maximum Strength of a Group is rated Medium difficulty. The implementation is short, but identifying the correct greedy rule for handling negative numbers and zeros requires careful reasoning about how multiplication affects the final product.
How to solve Maximum Strength of a Group in O(n)?
You can achieve O(n) time by scanning the array once to count negatives, positives, and zeros while tracking the largest negative (closest to zero). Multiply all positives and pair negatives except one if their count is odd. This avoids sorting but requires careful handling of edge cases like all negatives or zeros.
What is the best approach for Maximum Strength of a Group?
The most practical approach is a greedy strategy combined with sorting. Multiply all positive numbers, pair negative numbers so their product becomes positive, and skip a leftover negative if the count is odd. This method runs in O(n log n) time due to sorting and uses constant extra space.
Is Maximum Strength of a Group asked at Google/Amazon/Meta?
Problems involving maximum product subsets and sign handling frequently appear in interviews at large tech companies such as Amazon, Google, and Meta. While the exact problem number may vary, the core concept of pairing negatives and maximizing multiplicative gain is a common interview pattern.
What data structure is used in Maximum Strength of a Group?
The problem primarily uses arrays for input storage. The optimal solution relies on sorting the array and applying greedy multiplication logic. No advanced data structures are required beyond simple iteration and arithmetic operations.
What is the time complexity of Maximum Strength of a Group?
The optimal greedy solution runs in O(n log n) time because the array is sorted before pairing negatives and multiplying positives. Space complexity is O(1) aside from the input array. A brute-force backtracking solution requires O(2^n) time since it evaluates every possible subset.
Maximum Strength of a Group Python or Java solution approach?
In Python or Java, the typical solution sorts the array, multiplies all positive numbers, and pairs negative numbers from the smallest values. If the negative count is odd, the least harmful negative is skipped. The implementation is straightforward using loops and basic arithmetic operations.