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DP_ContiguousSubArrayMaxProduct.java
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DP_ContiguousSubArrayMaxProduct.java
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import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
// https://practice.geeksforgeeks.org/problems/maximum-product-subarray3604/1
// https://leetcode.com/problems/maximum-product-subarray/
// https://leetcode.com/problems/maximum-product-subarray/discuss/48230/Possibly-simplest-solution-with-O(n)-time-complexity (BEST Resource) (I followed this)
// https://leetcode.com/problems/maximum-product-subarray/discuss/48230/Possibly-simplest-solution-with-O(n)-time-complexity/48320 (BEST Explanation)
// https://www.youtube.com/watch?v=vtJvbRlHqTA&ab_channel=SahilThakur (A good video explanation) (His algo is bit different)
// https://www.youtube.com/watch?v=hJ_Uy2DzE08&ab_channel=KnowledgeCenter (A good video explanation)
// https://www.geeksforgeeks.org/maximum-product-subarray/
// https://practice.geeksforgeeks.org/problems/kadanes-algorithm/0 (Similar question)
// https://leetcode.com/problems/maximum-subarray/ (Similar question)
public class DP_ContiguousSubArrayMaxProduct {
/* this is very similar to the "DP Contiguous SubArray Max Sum Kadane Algo" problem.
here you keep 2 values: the max cumulative product UP TO current element starting from SOMEWHERE in the past,
and the minimum cumulative product UP TO current element .
it would be easier to see the DP structure if we store these 2 values for each index, like maxProduct[i],minProduct[i] .
At each new element,
1. u could either add the new element to the existing product, or
2. start fresh the product from current index (wipe out previous results), hence the 2 Math.max() lines.
if we see a negative number, the "candidate" for max should instead become the previous min product,
because a bigger number multiplied by negative becomes smaller, hence the swap() */
static long getContiguousSubArrMaxProduct(int[] arr, int size) {
long currMax = arr[0];
long currMin = arr[0];
long answer = arr[0]; /* store the result that is the max we have found so far */
for (int i = 1; i < size; i++) {
if (arr[i] < 0) {
/* multiplied by a negative makes big number smaller, small number bigger so we redefine the extremums by swapping them*/
long temp = currMax;
currMax = currMin;
currMin = temp;
}
/* max/min product for the current number is either the current number itself or the max/min by the previous number times the current one */
currMax = Math.max(currMax * arr[i], arr[i]);
currMin = Math.min(currMin * arr[i], arr[i]);
answer = Math.max(answer, currMax); /* the newly computed max value is a candidate for our global result */
}
return answer;
}
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine().trim());
while(t-- > 0) {
int size = Integer.parseInt(br.readLine().trim());
String[] ip = br.readLine().trim().split("\\s+");
int[] arr = new int[size];
for (int i = 0; i < size; i++) {
arr[i] = Integer.parseInt(ip[i]);
}
System.out.println(getContiguousSubArrMaxProduct(arr, size));
}
}
}