152. Maximum Product Subarray
Explanation:
To solve this problem, we can use a dynamic programming approach. We will keep track of both the maximum and minimum products ending at each position in the array. This is because a negative number multiplied by a negative number can become a positive number, so we need to keep track of both the maximum and minimum products.
At each position, we update the maximum and minimum products by considering three possibilities:
- Current element is positive: we multiply it with the maximum product ending at the previous position.
- Current element is negative: we multiply it with the minimum product ending at the previous position.
- Current element resets the subarray: we start a new subarray from the current element.
Finally, we return the maximum product encountered during these iterations.
class Solution {
public int maxProduct(int[] nums) {
int maxProd = nums[0];
int minProd = nums[0];
int result = nums[0];
for (int i = 1; i < nums.length; i++) {
int temp = maxProd;
maxProd = Math.max(nums[i], Math.max(maxProd * nums[i], minProd * nums[i]));
minProd = Math.min(nums[i], Math.min(temp * nums[i], minProd * nums[i]));
result = Math.max(result, maxProd);
}
return result;
}
}Code Editor (Testing phase)
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