Maximum Product Subarray
Problem Statement :
Given an integer array nums, find a subarray that has the largest product, and return the product. The test cases are generated so that the answer will fit in a 32-bit integer. Example 1: Input: nums = [2,3,-2,4] Output: 6 Explanation: [2,3] has the largest product 6. Example 2: Input: nums = [-2,0,-1] Output: 0 Explanation: The result cannot be 2, because [-2,-1] is not a subarray. Constraints: 1 <= nums.length <= 2 * 104 -10 <= nums[i] <= 10 The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.
Solution :
Solution in C :
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
int maxProduct(int* nums, int N)
{
int *pos = calloc(N, sizeof(int));
int *neg = calloc(N, sizeof(int));
int i, best = INT_MIN;
if (N == 1)
return nums[0];
for (i = 0; i < N; i++) {
int n = nums[i];
if (i == 0) {
if (n < 0) {
neg[0] = n;
pos[0] = 0;
} else {
neg[0] = 0;
pos[0] = n;
}
} else {
if (n < 0) {
neg[i] = MIN(pos[i - 1] * n, n);
pos[i] = neg[i - 1] * n;
} else {
neg[i] = neg[i - 1] * n;
pos[i] = MAX(pos[i - 1] * n, n);
}
}
best = MAX(best, pos[i]);
}
return best;
}
Solution in C++ :
class Solution {
public:
int maxProduct(vector<int>& nums) {
int n = nums.size();
int maxProduct = INT_MIN;
//applying Kadane's algo from Left to Right
int currProd_Left_to_Right = 1;
for(int i=0; i<n; i++){
currProd_Left_to_Right*=nums[i];
maxProduct = max(maxProduct, currProd_Left_to_Right);
if(currProd_Left_to_Right == 0){
currProd_Left_to_Right = 1;
}
}
//applying Kadane's algo from Right to Left
int currProd_Right_to_Left = 1;
for(int i=n-1; i>=0; i--){
currProd_Right_to_Left*=nums[i];
maxProduct = max(maxProduct, currProd_Right_to_Left);
if(currProd_Right_to_Left == 0){
currProd_Right_to_Left = 1;
}
}
return maxProduct;
}
};
Solution in Java :
class Solution {
public int maxProduct(int[] nums) {
int n = nums.length;
int maxProduct = Integer.MIN_VALUE;
//applying Kadane's algo from Left to Right
int currProd_Left_to_Right = 1;
for(int i=0; i<n; i++){
currProd_Left_to_Right*=nums[i];
maxProduct = Math.max(maxProduct, currProd_Left_to_Right);
if(currProd_Left_to_Right == 0){
currProd_Left_to_Right = 1;
}
}
//applying Kadane's algo from Right to Left
int currProd_Right_to_Left = 1;
for(int i=n-1; i>=0; i--){
currProd_Right_to_Left*=nums[i];
maxProduct = Math.max(maxProduct, currProd_Right_to_Left);
if(currProd_Right_to_Left == 0){
currProd_Right_to_Left = 1;
}
}
return maxProduct;
}
}
Solution in Python :
class Solution:
def maxProduct(self, nums: List[int]) -> int:
n = len(nums)
maxProduct = float('-inf')
#applying Kadane's algo from Left to Right
currProd_Left_to_Right = 1
for i in range(0,n):
currProd_Left_to_Right*=nums[i]
maxProduct = max(maxProduct, currProd_Left_to_Right)
if currProd_Left_to_Right == 0:
currProd_Left_to_Right = 1
#applying Kadane's algo from Right to Left
currProd_Right_to_Left = 1
for i in range(n-1,-1,-1):
currProd_Right_to_Left*=nums[i]
maxProduct = max(maxProduct, currProd_Right_to_Left)
if currProd_Right_to_Left == 0:
currProd_Right_to_Left = 1
return maxProduct
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