This problem gives an array and asks for the maximum product of a subarray, considering positive numbers, negative numbers, and zeros. The hint suggests computing a max and min for each value, then using them to generate the maximum and minimum including the current element. I used a simulation approach instead: multiply all numbers together — if the product is positive, it’s the maximum; if negative, compare the results of dropping the leftmost or rightmost negative number. To handle zeros, first split the array by zeros.
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| class Solution {
public:
int maxProductNoZero(vector<int>& nums) {
int max = 1;
int left = nums.size();
int right = -1;
for (int i = 0; i < nums.size(); i++) {
max *= nums[i];
if (nums[i] < 0 && i < left) {
left = i;
}
if (nums[i] < 0 && i > right) {
right = i;
}
}
if (max < 0 && nums.size() > 1) {
int max_left = 1;
for (int i = left + 1; i < nums.size(); i++) {
max_left *= nums[i];
}
int max_right = 1;
for (int i = 0; i < right; i++) {
max_right *= nums[i];
}
max = max_left > max_right ? max_left : max_right;
}
return max;
}
int maxProduct(vector<int>& nums) {
int max = nums[0];
vector<int> zero;
zero.clear();
zero.push_back(-1);
for (int i = 0; i < nums.size(); i++) {
if (max < nums[i]) {
max = nums[i];
}
if (0 == nums[i]) {
zero.push_back(i);
}
}
zero.push_back(nums.size());
for (int i = 0; i < zero.size() - 1; i++) {
if (zero[i + 1] - 1 >= zero[i] + 1) {
vector<int> sub;
sub.clear();
for (int j = zero[i] + 1; j <= zero[i + 1] - 1; j++) {
sub.push_back(nums[j]);
}
int tmp = maxProductNoZero(sub);
if (tmp > max) max = tmp;
}
}
return max;
}
};
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