You’re given an array A[] of measurement N, the duty is to divide the array into precisely three subarrays such that each aspect belongs to precisely one subarray such that the product of the sum of the subarrays is the utmost.
Examples:
Enter: N = 4, A[] = { 1, 2, 2, 3}
Output: 18
Clarification: The optimum partitions are {1, 2}, {2}, {3}Enter: N = 3, A[] = { 3, 5, 7}
Output: 105
Clarification: There is just one attainable partition {3}, {5}, {7}.
Method: This downside will be solved utilizing the idea of sliding window and prefix-suffix array.
First, calculate the utmost product of two subarrays contemplating the scale of the array from 0 to N ranging from proper. Now as soon as we now have the utmost product of two subarrays then the third subarray might be on the left facet.
For instance, if we now have calculated the utmost product of two subarrays for all of the arrays ranging from i to N the place 0 < i < N – 1 then the third subarray might be subarray from 0 to i. And now we will calculate the utmost product of those two subarrays to get most product of three subarrays.
Comply with the steps talked about under to implement the thought.
- Create a suffix array of measurement N.
- Create two variables x and y to retailer the sum of the primary two subarrays and initialize them as A[N-2] and A[N-1].
- Initialize suff[N-1] because the product of x and y.
- Now broaden the subarray with sum x by 1 and maintain sliding the subarray with sum y in the direction of the subarray with sum x till suff[i] is lower than x*y and replace suff[i] as x*y.
- Lastly, run a loop to calculate the utmost product between subarray 0 to i and the utmost product of two subarrays from i to N i.e. suff[i].
Beneath is the implementation of the above strategy:
C++
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Time Complexity: O(N)
Auxiliary Area: O(N)
