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Logic:
Let array index start from 0 to n-1 (where n is the number of elements).
Let i = n - 1
Step 1. Find the largest number from index = i to index = 2, let position of the number (n1) found = pos1
Step2. Find the largest number from index = pos1 - 1 to index = 1, let position of the number (n2) found = pos2
Step3. Find the largest number form index = pos2 - 1 to index = 0. Let the position of the number (n3) found = pos3
Step4. if n1 > n2 and n2 > n3 then find the product n1 * n2 * n3.
Step5. If product is more then the prev. product (i.e. max_prod), update max_prod
Step6. Do i-- and repeat step 1 to step 6 till i >= 2
Step 7. Required result is stored in max_prod.
- Sanjay Agarwal April 09, 2013